#set document(title: "5.5 Solve Mixture Applications with Systems of Equations", author: "OpenStax / XYZ Homework")
#set page(width: 8.5in, height: auto, margin: 1in)
#import "@preview/cetz:0.5.2"
#set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en")
#show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math"))
#set par(justify: true, leading: 0.62em, spacing: 0.9em)
#set enum(spacing: 1.1em)   // room between list items so tall inline fractions don't collide
#set list(spacing: 1.1em)
#set table(stroke: 0.5pt + rgb("#c7ccd3"))

#let BLUE = rgb("#183B6F")     // brand navy — section bars + example/solution labels (white on navy 11.09:1)
#let ORANGE = rgb("#A94509")   // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text)
#let RED = rgb("#DC2626")      // brand error-600
#let GREEN = rgb("#059669")    // brand success-600 (decoration only; small green text uses green-text #007942)

#show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt,
  fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt,
  text(fill: white, weight: "bold", size: 19pt, it.body))
#show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt,
  fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt,
  text(fill: white, weight: "bold", size: 12pt, it.body))
#show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body)
#show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body)

#let examplebox(label, title, body) = block(width: 100%, breakable: true,
  fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt,
  above: 12pt, below: 12pt)[
  #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt,
    text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]]
  #body]

// rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade
#let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint,
    stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt),
    above: 11pt, below: 11pt)[
    #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)]
    #linebreak() #body]

#let solutionbox(body) = block(above: 4pt, below: 8pt)[
  #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body]

#let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"),
  stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[
  #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))]

// Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a
// consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the
// print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a
// degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero.
#let _STD_W = 3.5
#let _WIDE_W = 5.6
#let _MAX_H = 3.4
#let _ASPECT_WIDE = 2.2
#let _UPSCALE_MAX = 1.15
#let stdfig(body) = context {
  let m = measure(body)
  let w = calc.max(m.width / 1in, 0.01)
  let h = calc.max(m.height / 1in, 0.01)
  let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W }
  let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX)
  align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body)))
}

#show figure: set block(breakable: false)
#set figure(gap: 8pt)
#show figure.caption: set text(size: 8.5pt, fill: rgb("#555"))


== 5.5#h(0.6em)Solve Mixture Applications with Systems of Equations

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Before you get started, take this readiness quiz.

Multiply #math.equation(block: false, alt: "4.025 (1,562).")[$4.025 "(1,562)."$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "6, 287.05")[$6, 287.05$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write 8.2% as a decimal. #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "0.082")[$0.082$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Earl’s dinner bill came to \$32.50 and he wanted to leave an 18% tip. How much should the tip be? #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "$5.85")[$\$5.85$]
]
]


=== Solve Mixture Applications

When we solved mixture applications with coins and tickets earlier, we started by creating a table so we could organize the information. For a coin example with nickels and dimes, the table looked like this:

#figure(figph[This is a table with three rows and four columns. The first row of the table is a header row, and each cell names the column or columns below it. The first cell from the left is named “Type.” The second cell contains the equation “Number” times “Value” equals “Total Value,” with one column corresponding to “Number,” one column corresponding to “Value,” and one column corresponding to total value. Hence the content of the “Number” column times the content of the “Value” column equals the content of the “Total Value” column. In the second row of the table, the “Type” column contains “nickels,” the “Number” column is blank, the “Value” column contains 0.05, and the “Total Value” column is blank. In the third row of the table, the “Type” column contains “dimes,” the “Number” column is blank, the “Value column contains 0.10, and the “Total Value” column is blank.], alt: "This is a table with three rows and four columns. The first row of the table is a header row, and each cell names the column or columns below it. The first cell from the left is named “Type.” The second cell contains the equation “Number” times “Value” equals “Total Value,” with one column corresponding to “Number,” one column corresponding to “Value,” and one column corresponding to total value. Hence the content of the “Number” column times the content of the “Value” column equals the content of the “Total Value” column. In the second row of the table, the “Type” column contains “nickels,” the “Number” column is blank, the “Value” column contains 0.05, and the “Total Value” column is blank. In the third row of the table, the “Type” column contains “dimes,” the “Number” column is blank, the “Value column contains 0.10, and the “Total Value” column is blank.", caption: none)

Using one variable meant that we had to relate the number of nickels and the number of dimes. We had to decide if we were going to let #emph[n] be the number of nickels and then write the number of dimes in terms of #emph[n], or if we would let #emph[d] be the number of dimes and write the number of nickels in terms of #emph[d].

Now that we know how to solve systems of equations with two variables, we’ll just let #emph[n] be the number of nickels and #emph[d] be the number of dimes. We’ll write one equation based on the total value column, like we did before, and the other equation will come from the number column.

For the first example, we’ll do a ticket problem where the ticket prices are in whole dollars, so we won’t need to use decimals just yet.

#examplebox("Example 1")[][
Translate to a system of equations and solve:

The box office at a movie theater sold 147 tickets for the evening show, and receipts totaled \$1,302. How many \$11 adult and how many \$8 child tickets were sold?

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#strong[Step 1. Read] the problem.], [We will create a table to organize the information.]),
  [#strong[Step 2. Identify] what we are looking for.], [We are looking for the number of adult tickets #linebreak() and the number of child tickets sold.],
  [#strong[Step 3. Name] what we are looking for.], [Let #math.equation(block: false, alt: "a equals")[$a =$] the number of adult tickets. #linebreak() #math.equation(block: false, alt: "c equals")[$#h(1.5em) c =$] the number of child tickets],
  [A table will help us organize the data. #linebreak() We have two types of tickets: adult and child.], [Write #emph[a] and #emph[c] for the number of tickets.],
  [Write the total number of tickets sold at the #linebreak() bottom of the Number column.], [Altogether 147 were sold.],
  [Write the value of each type of ticket in the #linebreak() Value column.], [The value of each adult ticket is \$11. #linebreak() The value of each child tickets is \$8.],
  [The number times the value gives the total #linebreak() value, so the total value of adult tickets is #linebreak() #math.equation(block: false, alt: "a times 11 equals 11 a")[$a · 11 = 11 a$], and the total value of child #linebreak() tickets is #math.equation(block: false, alt: "c times 8 equals 8 c .")[$c · 8 = 8 c .$]], [#figure(figph[A math table for a word problem: 'a' adults at \$11 each, 'c' children at \$8 each. Total 147 people, total value \$1302. Used to find 'a' and 'c' for system of equations.], alt: "A math table for a word problem: 'a' adults at $11 each, 'c' children at $8 each. Total 147 people, total value $1302. Used to find 'a' and 'c' for system of equations.", caption: none)],
  [Altogether the total value of the tickets was #linebreak() \$1,302.], [Fill in the Total Value column.],
  [#strong[Step 4. Translate] into a system of equations.], [],
  [The Number column and the Total Value #linebreak() column give us the system of equations. #linebreak() We will use the elimination method to solve #linebreak() this system.], [#figure(figph[A system of two linear equations is displayed, featuring the variables 'a' and 'c'. The equations are: a + c = 147 and 11a + 8c = 1302.], alt: "A system of two linear equations is displayed, featuring the variables 'a' and 'c'. The equations are: a + c = 147 and 11a + 8c = 1302.", caption: none)],
  [Multiply the first equation by −8.], [#figure(figph[A system of two linear equations is presented, with the first equation being -8(a + c) = -8(147) and the second equation being 11a + 8c = 1302.], alt: "A system of two linear equations is presented, with the first equation being -8(a + c) = -8(147) and the second equation being 11a + 8c = 1302.", caption: none)],
  [Simplify and add, then solve for #emph[a].], [#figure(figph[A system of two linear equations is presented: -8a + 8c = -1176 and 11a + 8c = 1302. Below these equations, a line shows a resulting partial equation: 3a = 126.], alt: "A system of two linear equations is presented: -8a + 8c = -1176 and 11a + 8c = 1302. Below these equations, a line shows a resulting partial equation: 3a = 126.", caption: none)],
  [], [#figure(figph[Two algebraic equations are displayed: 'a = 42' and 'a + c = 147'. A red circle highlights '42', and a red arrow points from '42' to the 'a' in the second equation, indicating substitution.], alt: "Two algebraic equations are displayed: 'a = 42' and 'a + c = 147'. A red circle highlights '42', and a red arrow points from '42' to the 'a' in the second equation, indicating substitution.", caption: none)],
  [Substitute #emph[a] = 42 into the first equation, #linebreak() then solve for #emph[c].], [#figure(figph[A clear, concise image displays the mathematical equation '42 + c = 147', with the number 42 highlighted in red, indicating a problem or question where a variable 'c' needs to be solved.], alt: "A clear, concise image displays the mathematical equation '42 + c = 147', with the number 42 highlighted in red, indicating a problem or question where a variable 'c' needs to be solved.", caption: none)],
  [], [#figure(figph[The equation 'c = 105' is clearly displayed in the center of a plain white background, presenting a simple numerical assignment.], alt: "The equation 'c = 105' is clearly displayed in the center of a plain white background, presenting a simple numerical assignment.", caption: none)],
  [#strong[Step 5. Check] the answer in the problem. #linebreak()  #linebreak()   42 adult tickets at \$11 per ticket makes \$462 #linebreak()   105 child tickets at \$8 per ticket makes \$840. #linebreak()   The total receipts are \$1,302.✓], [],
  [#strong[Step 6. Answer] the question.], [The movie theater sold 42 adult tickets and 105 child tickets.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Translate to a system of equations and solve:

The ticket office at the zoo sold 553 tickets one day. The receipts totaled \$3,936. How many \$9 adult tickets and how many \$6 child tickets were sold?

#solutionbox[
There were 206 adult tickets sold and 347 children tickets sold.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Translate to a system of equations and solve:

A science center sold 1,363 tickets on a busy weekend. The receipts totaled \$12,146. How many \$12 adult tickets and how many \$7 child tickets were sold?

#solutionbox[
There were 521 adult tickets sold and 842 children tickets sold.
]
]

In  we’ll solve a coin problem. Now that we know how to work with systems of two variables, naming the variables in the ‘number’ column will be easy.

#examplebox("Example 2")[][
Translate to a system of equations and solve:

Priam has a collection of nickels and quarters, with a total value of \$7.30. The number of nickels is six less than three times the number of quarters. How many nickels and how many quarters does he have?

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#strong[Step 1. Read] the problem.], [We will create a table to organize the information.]),
  [#strong[Step 2. Identify] what we are looking for.], [We are looking for the number of nickels #linebreak() and the number of quarters.],
  [#strong[Step 3. Name] what we are looking for.], [Let #math.equation(block: false, alt: "n equals")[$n =$] the number of nickels. #linebreak() #math.equation(block: false, alt: "q equals")[$#h(1.5em) q =$] the number of quarters],
  [A table will help us organize the data. #linebreak() We have two types of coins, nickels #linebreak() and quarters.], [Write #emph[n] and #emph[q] for the number of each type of coin.],
  [Fill in the Value column with the value of each #linebreak() type of coin.], [The value of each nickel is \$0.05. #linebreak() The value of each quarter is \$0.25.],
  [The number times the value gives the total #linebreak() value, so, the total value of the nickels is #linebreak() #emph[n] (0.05) = 0.05#emph[n] and the total value of #linebreak() quarters is #emph[q] (0.25) = 0.25#emph[q]. #linebreak() Altogether the total value of the coins #linebreak() is \$7.30.], [#figure(figph[A table showing the type, number, value, and total value for nickels and quarters. Nickels are valued at \$0.05, quarters at \$0.25, with a total value of \$7.30.], alt: "A table showing the type, number, value, and total value for nickels and quarters. Nickels are valued at $0.05, quarters at $0.25, with a total value of $7.30.", caption: none)],
  [#strong[Step 4. Translate] into a system of equations.], [],
  [The Total value column gives one equation.], [#figure(figph[A mathematical equation is displayed: 0.05n + 0.25q = 7.30.], alt: "A mathematical equation is displayed: 0.05n + 0.25q = 7.30.", caption: none)],
  [We also know the number of nickels is six less #linebreak() than three times the number of quarters. #linebreak() Translate to get the second equation.], [#figure(figph[The image shows the algebraic equation n = 3q - 6, presented on a plain white background.], alt: "The image shows the algebraic equation n = 3q - 6, presented on a plain white background.", caption: none)],
  [Now we have the system to solve.], [#figure(figph[A system of two linear equations is presented, with the first equation being 0.05n + 0.25q = 7.30, and the second equation being n = 3q - 6.], alt: "A system of two linear equations is presented, with the first equation being 0.05n + 0.25q = 7.30, and the second equation being n = 3q - 6.", caption: none)],
  [#strong[Step 5. Solve] the system of equations #linebreak() We will use the substitution method. #linebreak() Substitute #emph[n] = 3#emph[q] − 6 into the first equation. #linebreak() Simplify and solve for #emph[q].], [#figure(figph[A mathematical equation is displayed: 0.05n + 0.25q = 7.30, likely representing the total value of nickels (n) and quarters (q) summing to \$7.30.], alt: "A mathematical equation is displayed: 0.05n + 0.25q = 7.30, likely representing the total value of nickels (n) and quarters (q) summing to $7.30.", caption: none)],
  [], [#figure(figph[A mathematical equation is displayed on a white background, which reads '0.05(3q - 6) + 0.25q = 7.3'.], alt: "A mathematical equation is displayed on a white background, which reads '0.05(3q - 6) + 0.25q = 7.3'.", caption: none)],
  [], [#figure(figph[A mathematical equation is displayed, showing '0.15q - 0.3 + 0.25q = 7.3' in white text on a blank background.], alt: "A mathematical equation is displayed, showing '0.15q - 0.3 + 0.25q = 7.3' in white text on a blank background.", caption: none)],
  [], [#figure(figph[A mathematical equation is displayed against a white background, reading '0.4q - 0.3 = 7.3'. The equation involves a variable 'q' with decimal coefficients and constants.], alt: "A mathematical equation is displayed against a white background, reading '0.4q - 0.3 = 7.3'. The equation involves a variable 'q' with decimal coefficients and constants.", caption: none)],
  [], [#figure(figph[A mathematical equation is displayed against a white background, reading '0.4q = 7.6'.], alt: "A mathematical equation is displayed against a white background, reading '0.4q = 7.6'.", caption: none)],
  [], [#figure(figph[The mathematical equation 'q = 19' is displayed in a simple, clear font against a white background.], alt: "The mathematical equation 'q = 19' is displayed in a simple, clear font against a white background.", caption: none)],
  [To find the number of nickels, substitute #linebreak() #emph[q] = 19 into the second equation.], [#figure(figph[The image shows the algebraic equation n = 3q - 6, presented in a clear, standard mathematical notation on a white background.], alt: "The image shows the algebraic equation n = 3q - 6, presented in a clear, standard mathematical notation on a white background.", caption: none)],
  [], [#figure(figph[A mathematical equation shows 'n = 3  ⋅ 19 - 6' with the number 19 highlighted in red.], alt: "A mathematical equation shows 'n = 3 ⋅ 19 - 6' with the number 19 highlighted in red.", caption: none)],
  [], [#figure(figph[The image displays the equation 'n = 51' in the center, written in black text against a plain white background.], alt: "The image displays the equation 'n = 51' in the center, written in black text against a plain white background.", caption: none)],
  [#strong[Step 6. Check] the answer in the problem. #linebreak()  #linebreak() #math.equation(block: false, alt: "19 quarters at $ 0.25, equals, $ 4.75; 51 nickels at $ 0.05, equals, $ 2.55; Total, equals, $ 7.30 ✓; 3 times 19 minus 16, equals, 51 ✓")[$19 #h(0.2em) "quarters at" #h(0.2em) \$ 0.25 & = & \$ 4.75 \ 51 " nickels at" #h(0.2em) \$ 0.05 & = & \$ 2.55 \ "Total" & = & \$ 7.30 #h(0.2em) ✓ \ 3 ⋅ 19 − 16 & = & 51 #h(0.2em) ✓$]], [],
  [#strong[Step 7. Answer] the question.], [Priam has 19 quarters and 51 nickels.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Translate to a system of equations and solve:

Matilda has a handful of quarters and dimes, with a total value of \$8.55. The number of quarters is 3 more than twice the number of dimes. How many dimes and how many quarters does she have?

#solutionbox[
Matilda has 13 dimes and 29 quarters.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Translate to a system of equations and solve:

Juan has a pocketful of nickels and dimes. The total value of the coins is \$8.10. The number of dimes is 9 less than twice the number of nickels. How many nickels and how many dimes does Juan have?

#solutionbox[
Juan has 36 nickels and 63 dimes.
]
]

Some mixture applications involve combining foods or drinks. Example situations might include combining raisins and nuts to make a trail mix or using two types of coffee beans to make a blend.

#examplebox("Example 3")[][
Translate to a system of equations and solve:

Carson wants to make 20 pounds of trail mix using nuts and chocolate chips. His budget requires that the trail mix costs him \$7.60 per pound. Nuts cost \$9.00 per pound and chocolate chips cost \$2.00 per pound. How many pounds of nuts and how many pounds of chocolate chips should he use?

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#strong[Step 1. Read] the problem.], [We will create a table to organize the information.]),
  [#strong[Step 2. Identify] what we are looking for.], [We are looking for the number of pounds of nuts #linebreak() and the number of pounds of chocolate chips.],
  [#strong[Step 3. Name] what we are looking for.], [Let #math.equation(block: false, alt: "n equals")[$n =$] the number of pound of nuts. #linebreak() #math.equation(block: false, alt: "c equals")[$#h(1.5em) c =$] the number of pounds of chips],
  [Carson will mix nuts and chocolate chips #linebreak() to get trail mix. #linebreak() Write in #emph[n] and #emph[c] for the number of pounds #linebreak() of nuts and chocolate chips. #linebreak()  #linebreak() There will be 20 pounds of trail mix. #linebreak() Put the price per pound of each item in #linebreak() the Value column. #linebreak() Fill in the last column using], [#figure(figph[A table displays data for nuts, chocolate chips, and trail mix. It includes columns for Type, Number of pounds, Value (\$) per pound, and Total Value (\$). Variables 'n' and 'c' are used for pounds of nuts and chocolate chips, respectively.], alt: "A table displays data for nuts, chocolate chips, and trail mix. It includes columns for Type, Number of pounds, Value ($) per pound, and Total Value ($). Variables 'n' and 'c' are used for pounds of nuts and chocolate chips, respectively.", caption: none)],
  [Number · Value = Total Value], [],
  [#strong[Step 4. Translate] into a system of equations. #linebreak() We get the equations from the Number #linebreak() and Total Value columns.], [#figure(figph[A system of two linear equations is presented: n + c = 20 and 9n + 2c = 152. These equations are typically solved for the values of 'n' and 'c' using substitution or elimination methods.], alt: "A system of two linear equations is presented: n + c = 20 and 9n + 2c = 152. These equations are typically solved for the values of 'n' and 'c' using substitution or elimination methods.", caption: none)],
  [#strong[Step 5. Solve] the system of equations #linebreak() We will use elimination to solve the system.], [],
  [Multiply the first equation by −2 to eliminate #emph[c].], [#figure(figph[A system of two linear equations is shown: -2(n + c) = -2(20) and 9n + 2c = 152. The equations are enclosed by a curly brace on the left.], alt: "A system of two linear equations is shown: -2(n + c) = -2(20) and 9n + 2c = 152. The equations are enclosed by a curly brace on the left.", caption: none)],
  [Simplify and add. Solve for #emph[n].], [#figure(figph[A system of two linear equations, -2n - 2c = -40 and 9n + 2c = 152, is shown being solved by the elimination method, resulting in the simplified equation 7n = 112.], alt: "A system of two linear equations, -2n - 2c = -40 and 9n + 2c = 152, is shown being solved by the elimination method, resulting in the simplified equation 7n = 112.", caption: none)],
  [], [#figure(figph[The image displays the equation 'n = 16' in gray font centered on a white background.], alt: "The image displays the equation 'n = 16' in gray font centered on a white background.", caption: none)],
  [To find the number of pounds of #linebreak() chocolate chips, substitute #emph[n] = 16 into #linebreak() the first equation, then solve for #emph[c].], [#figure(figph[A mathematical equation is displayed, showing 'n + c = 20' in dark gray text against a plain white background.], alt: "A mathematical equation is displayed, showing 'n + c = 20' in dark gray text against a plain white background.", caption: none)

 #linebreak() #figure(figph[A simple algebraic equation is displayed, reading '16 + c = 20'. The numbers and the variable 'c' are written in a clear, dark font against a white background.], alt: "A simple algebraic equation is displayed, reading '16 + c = 20'. The numbers and the variable 'c' are written in a clear, dark font against a white background.", caption: none)],
  [], [#math.equation(block: false, alt: "c equals 4")[$c = 4$]],
  [#strong[Step 6. Check] the answer in the problem. #linebreak()  #linebreak() #math.equation(block: false, alt: "16 plus 4, equals, 20 ✓; 9 times 16 plus 2 times 4, equals, 152 ✓")[$16 + 4 & = & 20 #h(0.2em) ✓ \ 9 · 16 + 2 · 4 & = & 152 #h(0.2em) ✓$]], [],
  [#strong[Step 7. Answer] the question.], [Carson should mix 16 pounds of nuts with #linebreak() 4 pounds of chocolate chips to create the trail mix.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Translate to a system of equations and solve:

Greta wants to make 5 pounds of a nut mix using peanuts and cashews. Her budget requires the mixture to cost her \$6 per pound. Peanuts are \$4 per pound and cashews are \$9 per pound. How many pounds of peanuts and how many pounds of cashews should she use?

#solutionbox[
Greta should use 3 pounds of peanuts and 2 pounds of cashews.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Translate to a system of equations and solve:

Sammy has most of the ingredients he needs to make a large batch of chili. The only items he lacks are beans and ground beef. He needs a total of 20 pounds combined of beans and ground beef and has a budget of \$3 per pound. The price of beans is \$1 per pound and the price of ground beef is \$5 per pound. How many pounds of beans and how many pounds of ground beef should he purchase?

#solutionbox[
Sammy should purchase 10 pounds of beans and 10 pounds of ground beef.
]
]

Another application of mixture problems relates to concentrated cleaning supplies, other chemicals, and mixed drinks. The concentration is given as a percent. For example, a 20% concentrated household cleanser means that 20% of the total amount is cleanser, and the rest is water. To make 35 ounces of a 20% concentration, you mix 7 ounces (20% of 35) of the cleanser with 28 ounces of water.

For these kinds of mixture problems, we’ll use percent instead of value for one of the columns in our table.

#examplebox("Example 4")[][
Translate to a system of equations and solve:

Sasheena is a lab assistant at her community college. She needs to make 200 milliliters of a 40% solution of sulfuric acid for a lab experiment. The lab has only 25% and 50% solutions in the storeroom. How much should she mix of the 25% and the 50% solutions to make the 40% solution?

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#strong[Step 1. Read] the problem.], [A figure may help us visualize the situation, then we #linebreak() will create a table to organize the information.]),
  [Sasheena must mix some of the 25% #linebreak() solution and some of the 50% solution #linebreak() together to get 200 ml of the 40% solution.], [#figure(figph[Diagram illustrating two solutions, x (25%) and y (50%), being combined to create a final 200 ml solution with a 40% concentration, typically used in mixing problems.], alt: "Diagram illustrating two solutions, x (25%) and y (50%), being combined to create a final 200 ml solution with a 40% concentration, typically used in mixing problems.", caption: none)],
  [#strong[Step 2. Identify] what we are looking for.], [We are looking for how much of each solution #linebreak() she needs.],
  [#strong[Step 3. Name] what we are looking for.], [Let #math.equation(block: false, alt: "x equals")[$x =$] number of ml of 25% solution. #linebreak() #math.equation(block: false, alt: "y equals")[$#h(1.5em) y =$] number of ml of 50% solution],
  [A table will help us organize the data. #linebreak()  #linebreak() She will mix #emph[x] ml of 25% with #emph[y] ml of #linebreak() 50% to get 200 ml of 40% solution. #linebreak()  #linebreak() We write the percents as decimals in #linebreak() the chart. #linebreak()  #linebreak() We multiply the number of units times #linebreak() the concentration to get the total #linebreak() amount of sulfuric acid in each solution.], [#figure(figph[A mathematical table illustrating the calculation of 'Amount' by multiplying 'Number of units' by 'Concentration %' for 25% (x units), 50% (y units), and 40% (200 units).], alt: "A mathematical table illustrating the calculation of 'Amount' by multiplying 'Number of units' by 'Concentration %' for 25% (x units), 50% (y units), and 40% (200 units).", caption: none)],
  [#strong[Step 4. Translate] into a system of #linebreak() equations. We get the equations from #linebreak() the Number column and the Amount #linebreak() column.], [],
  [Now we have the system.], [#figure(figph[A system of two linear equations is presented, where the first equation is x + y = 200, and the second is 0.25x + 0.50y = 0.40(200), likely representing a mixture problem or similar application.], alt: "A system of two linear equations is presented, where the first equation is x + y = 200, and the second is 0.25x + 0.50y = 0.40(200), likely representing a mixture problem or similar application.", caption: none)],
  [#strong[Step 5. Solve] the system of equations. #linebreak() We will solve the system by elimination. #linebreak() Multiply the first equation by −0.5 to #linebreak() eliminate #emph[y].], [#figure(figph[A system of two linear equations is presented: -0.5(x + y) = -0.5(200) and 0.25x + 0.50y = 80. The constant -0.5 is highlighted in red in the first equation.], alt: "A system of two linear equations is presented: -0.5(x + y) = -0.5(200) and 0.25x + 0.50y = 80. The constant -0.5 is highlighted in red in the first equation.", caption: none)],
  [Simplify and add to solve for #emph[x].], [#figure(figph[A system of two linear equations is solved using the elimination method. By adding the two equations, the y-terms cancel out, leading to -0.25x = -20, which simplifies to x = 80.], alt: "A system of two linear equations is solved using the elimination method. By adding the two equations, the y-terms cancel out, leading to -0.25x = -20, which simplifies to x = 80.", caption: none)],
  [To solve for #emph[y], substitute #emph[x] = 80 into the #linebreak() first equation.], [#figure(figph[A simple algebraic equation, x + y = 200, is displayed in black text on a plain white background.], alt: "A simple algebraic equation, x + y = 200, is displayed in black text on a plain white background.", caption: none)],
  [], [#figure(figph[A mathematical equation is displayed on a white background, reading '80 + y = 200'. The number '80' is highlighted in red, while the rest of the equation is in black.], alt: "A mathematical equation is displayed on a white background, reading '80 + y = 200'. The number '80' is highlighted in red, while the rest of the equation is in black.", caption: none)],
  [], [#figure(figph[The image displays the mathematical equation 'y = 120' in black text against a plain white background.], alt: "The image displays the mathematical equation 'y = 120' in black text against a plain white background.", caption: none)],
  [#strong[Step 6. Check] the answer in the problem. #linebreak()  #linebreak() #math.equation(block: false, alt: "80 plus 120, equals, 120 ✓; 0.25 open parenthesis 80 close parenthesis plus 0.50 open parenthesis 120 close parenthesis, equals, 80 ✓; Yes!")[$80 + 120 & = & 120 #h(0.2em) ✓ \ 0.25 ( 80 ) + 0.50 ( 120 ) & = & 80 #h(0.2em) ✓ \  &  & "Yes!"$]], [],
  [#strong[Step 7. Answer] the question.], [Sasheena should mix 80 ml of the 25% solution #linebreak() with 120 ml of the 50% solution to get the 200 ml #linebreak() of the 40% solution.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Translate to a system of equations and solve:

LeBron needs 150 milliliters of a 30% solution of sulfuric acid for a lab experiment but only has access to a 25% and a 50% solution. How much of the 25% and how much of the 50% solution should he mix to make the 30% solution?

#solutionbox[
LeBron needs 120 ml of the 25% solution and 30 ml of the 50% solution.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Translate to a system of equations and solve:

Anatole needs to make 250 milliliters of a 25% solution of hydrochloric acid for a lab experiment. The lab only has a 10% solution and a 40% solution in the storeroom. How much of the 10% and how much of the 40% solutions should he mix to make the 25% solution?

#solutionbox[
Anatole should mix 125 ml of the 10% solution and 125 ml of the 40% solution.
]
]


=== Solve Interest Applications

The formula to model interest applications is #emph[I] = #emph[Prt]. Interest, #emph[I], is the product of the principal, #emph[P], the rate, #emph[r], and the time, #emph[t]. In our work here, we will calculate the interest earned in one year, so #emph[t] will be 1.

We modify the column titles in the mixture table to show the formula for interest, as you’ll see in .

#examplebox("Example 5")[][
Translate to a system of equations and solve:

Adnan has \$40,000 to invest and hopes to earn 7.1% interest per year. He will put some of the money into a stock fund that earns 8% per year and the rest into bonds that earns 3% per year. How much money should he put into each fund?

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#strong[Step 1. Read] the problem.], [A chart will help us organize the information.]),
  [#strong[Step 2. Identify] what we are looking for.], [We are looking for the amount to invest in each fund.],
  [#strong[Step 3. Name] what we are looking for.], [Let #math.equation(block: false, alt: "s equals")[$s =$] the amount invested in stocks. #linebreak() #math.equation(block: false, alt: "b equals")[$#h(1.5em) b =$] the amount invested in bonds.],
  [Write the interest rate as a decimal for #linebreak() each fund. #linebreak() Multiply: #linebreak() Principal · Rate · Time #linebreak() to get the Interest.], [#figure(figph[A table displays simple interest calculations for a stock fund and bonds. It shows principal (s, b, total 40,000), rates (0.08, 0.03, total 0.071), time (1), and interest (0.08s, 0.03b, total 0.071(40k)).], alt: "A table displays simple interest calculations for a stock fund and bonds. It shows principal (s, b, total 40,000), rates (0.08, 0.03, total 0.071), time (1), and interest (0.08s, 0.03b, total 0.071(40k)).", caption: none)],
  [#strong[Step 4. Translate] into a system of #linebreak() equations. #linebreak() We get our system of equations from #linebreak() the Principal column and the #linebreak() Interest column.], [#figure(figph[A system of two linear equations is presented. The first equation is s + b = 40,000, and the second is 0.08s + 0.03b = 0.071(40,000).], alt: "A system of two linear equations is presented. The first equation is s + b = 40,000, and the second is 0.08s + 0.03b = 0.071(40,000).", caption: none)],
  [#strong[Step 5. Solve] the system of equations #linebreak() Solve by elimination. #linebreak() Multiply the top equation by −0.03.], [#figure(figph[A system of two linear equations is displayed. The first equation is -0.03(s + b) = -0.03(40,000), and the second is 0.08s + 0.03b = 2,840. The equations involve variables s and b.], alt: "A system of two linear equations is displayed. The first equation is -0.03(s + b) = -0.03(40,000), and the second is 0.08s + 0.03b = 2,840. The equations involve variables s and b.", caption: none)],
  [Simplify and add to solve for #emph[s].], [#figure(figph[A system of two linear equations, -0.03s - 0.03b = -1,200 and 0.08s + 0.03b = 2,840, is shown with the elimination method applied, resulting in 0.05s = 1,640.], alt: "A system of two linear equations, -0.03s - 0.03b = -1,200 and 0.08s + 0.03b = 2,840, is shown with the elimination method applied, resulting in 0.05s = 1,640.", caption: none)],
  [], [#figure(figph[The image displays the equation S = 32,800 in a simple, clear font on a plain white background. The text is centrally aligned and easy to read, presenting a numerical value assigned to the variable 'S'.], alt: "The image displays the equation S = 32,800 in a simple, clear font on a plain white background. The text is centrally aligned and easy to read, presenting a numerical value assigned to the variable 'S'.", caption: none)],
  [To find #emph[b], substitute #emph[s] = 32,800 into the first equation.], [#figure(figph[A mathematical equation, 's + b = 40,000', is displayed against a white background.], alt: "A mathematical equation, 's + b = 40,000', is displayed against a white background.", caption: none)

 #linebreak() #figure(figph[A mathematical equation is displayed on a white background: 32,800 + b = 40,000. The number 32,800 is highlighted in red, while the rest of the equation is in black.], alt: "A mathematical equation is displayed on a white background: 32,800 + b = 40,000. The number 32,800 is highlighted in red, while the rest of the equation is in black.", caption: none)],
  [], [#figure(figph[A simple mathematical equation 'b = 7,200' is displayed in the center of a white background.], alt: "A simple mathematical equation 'b = 7,200' is displayed in the center of a white background.", caption: none)],
  [#strong[Step 6. Check] the answer in the problem.], [We leave the check to you.],
  [#strong[Step 7. Answer] the question.], [Adnan should invest \$32,800 in stock and #linebreak() \$7,200 in bonds.],
))

Did you notice that the Principal column represents the total amount of money invested while the Interest column represents only the interest earned? Likewise, the first equation in our system, #emph[s] + #emph[b] = 40,000, represents the total amount of money invested and the second equation, 0.08#emph[s] + 0.03#emph[b] = 0.071(40,000), represents the interest earned.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Translate to a system of equations and solve:

Leon had \$50,000 to invest and hopes to earn 6.2 % interest per year. He will put some of the money into a stock fund that earns 7% per year and the rest in to a savings account that earns 2% per year. How much money should he put into each fund?

#solutionbox[
Leon should put \$42,000 in the stock fund and \$8000 in the savings account.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Translate to a system of equations and solve:

Julius invested \$7,000 into two stock investments. One stock paid 11% interest and the other stock paid 13% interest. He earned 12.5% interest on the total investment. How much money did he put in each stock?

#solutionbox[
Julius invested \$1,750 at 11% and \$5,250 at 13%.
]
]

#examplebox("Example 6")[][
Translate to a system of equations and solve:

Rosie owes \$21,540 on her two student loans. The interest rate on her bank loan is 10.5% and the interest rate on the federal loan is 5.9%. The total amount of interest she paid last year was \$1,669.68. What was the principal for each loan?

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#strong[Step 1. Read] the problem.], [A chart will help us organize the information.]),
  [#strong[Step 2. Identify] what we are looking for.], [We are looking for the principal of each loan.],
  [#strong[Step 3. Name] what we are looking for.], [Let #math.equation(block: false, alt: "b equals")[$b =$] the principal for the bank loan. #linebreak() #math.equation(block: false, alt: "f equals")[$#h(1.5em) f =$] the principal on the federal loan],
  [The total loans are \$21,540.], [],
  [Record the interest rates as decimals #linebreak() in the chart.], [#figure(figph[A table showing simple interest calculations for two accounts, 'bank' and 'federal,' with respective principals 'b' and 'f', and interest rates of 0.105 and 0.059 over one year, totaling a principal of \$21,540 and interest of \$1669.68.], alt: "A table showing simple interest calculations for two accounts, 'bank' and 'federal,' with respective principals 'b' and 'f', and interest rates of 0.105 and 0.059 over one year, totaling a principal of $21,540 and interest of $1669.68.", caption: none)],
  [Multiply using the formula #emph[l] = Pr #emph[t] to #linebreak() get the Interest.], [],
  [#strong[Step 4. Translate] into a system of #linebreak() equations. #linebreak()  The system of equations comes from #linebreak() the Principal column and the Interest #linebreak() column.], [#figure(figph[A system of two linear equations with variables 'b' and 'f'. The first equation is b + f = 21,540, and the second is 0.105b + 0.059f = 1669.68.], alt: "A system of two linear equations with variables 'b' and 'f'. The first equation is b + f = 21,540, and the second is 0.105b + 0.059f = 1669.68.", caption: none)],
  [#strong[Step 5. Solve] the system of equations #linebreak() We will use substitution to solve. #linebreak() Solve the first equation for #emph[b].], [#figure(figph[Two lines of algebraic equations are shown. The first line is b + f = 21,540, and the second line is b = -f + 21,540, which is a rearrangement of the first equation.], alt: "Two lines of algebraic equations are shown. The first line is b + f = 21,540, and the second line is b = -f + 21,540, which is a rearrangement of the first equation.", caption: none)],
  [Substitute #emph[b] = −#emph[f] + 21,540 into the #linebreak() second equation.], [#figure(figph[Two linear equations are shown, with the second equation representing a substitution of 'b' in terms of 'f' into the first equation: 0.105b + 0.059f = 1669.68 and 0.105(-f + 21,540) + 0.059f = 1669.68.], alt: "Two linear equations are shown, with the second equation representing a substitution of 'b' in terms of 'f' into the first equation: 0.105b + 0.059f = 1669.68 and 0.105(-f + 21,540) + 0.059f = 1669.68.", caption: none)],
  [Simplify and solve for #emph[f].], [#figure(figph[A mathematical equation is displayed with the variables and constants: -0.105f + 2261.70 + 0.059f = 1669.68.], alt: "A mathematical equation is displayed with the variables and constants: -0.105f + 2261.70 + 0.059f = 1669.68.", caption: none)],
  [], [#figure(figph[A mathematical equation is displayed, showing '-0.046f + 2261.70 = 1669.68' in a dark gray font against a plain white background.], alt: "A mathematical equation is displayed, showing '-0.046f + 2261.70 = 1669.68' in a dark gray font against a plain white background.", caption: none)],
  [], [#figure(figph[A mathematical equation shows '-0.046f = -592.02' on a white background.], alt: "A mathematical equation shows '-0.046f = -592.02' on a white background.", caption: none)],
  [], [#figure(figph[The image displays the equation f = 12,870, presented in a clean, legible font against a plain white background, suggesting a numerical value assignment to the variable 'f'.], alt: "The image displays the equation f = 12,870, presented in a clean, legible font against a plain white background, suggesting a numerical value assignment to the variable 'f'.", caption: none)],
  [To find #emph[b], substitute #emph[f] = 12,870 into #linebreak() the first equation.], [#figure(figph[A white background displays the mathematical equation 'b + f = 21,540' in black font, with 'b' and 'f' as variables, the plus symbol indicating addition, and the equals sign preceding the number 21,540.], alt: "A white background displays the mathematical equation 'b + f = 21,540' in black font, with 'b' and 'f' as variables, the plus symbol indicating addition, and the equals sign preceding the number 21,540.", caption: none)

 #linebreak() #figure(figph[A mathematical equation is displayed on a white background, which reads '12,870 + b = 21,540'.], alt: "A mathematical equation is displayed on a white background, which reads '12,870 + b = 21,540'.", caption: none)],
  [], [#figure(figph[The image shows the mathematical equation b = 8,670 in black text on a plain white background, appearing as if part of a larger numerical or algebraic problem.], alt: "The image shows the mathematical equation b = 8,670 in black text on a plain white background, appearing as if part of a larger numerical or algebraic problem.", caption: none)],
  [#strong[Step 6. Check] the answer in the #linebreak() problem.], [We leave the check to you.],
  [#strong[Step 7. Answer] the question.], [The principal of the bank loan is \$12,870 and #linebreak() the principal for the federal loan is \$8,670.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Translate to a system of equations and solve:

Laura owes \$18,000 on her student loans. The interest rate on the bank loan is 2.5% and the interest rate on the federal loan is 6.9 %. The total amount of interest she paid last year was \$1,066. What was the principal for each loan?

#solutionbox[
The principal amount for the bank loan was \$4,000. The principal amount for the federal loan was \$14,000.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Translate to a system of equations and solve:

Jill’s Sandwich Shoppe owes \$65,200 on two business loans, one at 4.5% interest and the other at 7.2% interest. The total amount of interest owed last year was \$3,582. What was the principal for each loan?

#solutionbox[
The principal amount for was \$41,200 at 4.5%. The principal amount was, \$24,000 at 7.2%.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Access these online resources for additional instruction and practice with solving application problems with systems of linear equations.

- #link("https://www.openstax.org/l/25LinEqu1")[Cost and Mixture Word Problems]
- #link("https://www.openstax.org/l/25EqMixture")[Mixture Problems]
]


=== Key Concepts

- #strong[Table for coin and mixture applications] #linebreak() 
#figure(figph[This table is mostly blank. It has four columns and four rows. The last row is labeled “Total.” The first row labels each column as “Type,” and “Number times Value = Total Value.”], alt: "This table is mostly blank. It has four columns and four rows. The last row is labeled “Total.” The first row labels each column as “Type,” and “Number times Value = Total Value.”", caption: none)
- #strong[Table for concentration applications] #linebreak() 
#figure(figph[This table is mostly blank. It has four columns and four rows. The last row is labeled “Total.” The first row labels each column as “Type,” and “Number of units times Concentration = Amount.”], alt: "This table is mostly blank. It has four columns and four rows. The last row is labeled “Total.” The first row labels each column as “Type,” and “Number of units times Concentration = Amount.”", caption: none)
- #strong[Table for interest applications] #linebreak() 
#figure(figph[This table is mostly blank. It has five columns and four rows. The last row is labeled “Total.” The first row labels each column as “Type,” and “Principal times Rate times Time = Interest”], alt: "This table is mostly blank. It has five columns and four rows. The last row is labeled “Total.” The first row labels each column as “Type,” and “Principal times Rate times Time = Interest”", caption: none)


==== Practice Makes Perfect

#strong[Solve Mixture Applications]

In the following exercises, translate to a system of equations and solve.

Tickets to a Broadway show cost \$35 for adults and \$15 for children. The total receipts for 1650 tickets at one performance were \$47,150. How many adult and how many child tickets were sold?

#solutionbox[
There 1120 adult tickets and 530 child tickets sold.
]

Tickets for a show are \$70 for adults and \$50 for children. One evening performance had a total of 300 tickets sold and the receipts totaled \$17,200. How many adult and how many child tickets were sold?

Tickets for a train cost \$10 for children and \$22 for adults. Josie paid \$1,200 for a total of 72 tickets. How many children’s tickets and how many adult tickets did Josie buy?

#solutionbox[
Josie bought 40 adult tickets and 32 children tickets.
]

Tickets for a baseball game are \$69 for Main Level seats and \$39 for Terrace Level seats. A group of sixteen friends went to the game and spent a total of \$804 for the tickets. How many of Main Level and how many Terrace Level tickets did they buy?

Tickets for a dance recital cost \$15 for adults and \$7 for children. The dance company sold 253 tickets and the total receipts were \$2,771. How many adult tickets and how many child tickets were sold?

#solutionbox[
There were 125 adult tickets and 128 children tickets sold.
]

Tickets for the community fair cost \$12 for adults and \$5 dollars for children. On the first day of the fair, 312 tickets were sold for a total of \$2,204. How many adult tickets and how many child tickets were sold?

Brandon has a cup of quarters and dimes with a total value of \$3.80. The number of quarters is four less than twice the number of dimes. How many quarters and how many dimes does Brandon have?

#solutionbox[
Brandon has 12 quarters and 8 dimes.
]

Sherri saves nickels and dimes in a coin purse for her daughter. The total value of the coins in the purse is \$0.95. The number of nickels is two less than five times the number of dimes. How many nickels and how many dimes are in the coin purse?

Peter has been saving his loose change for several days. When he counted his quarters and dimes, he found they had a total value \$13.10. The number of quarters was fifteen more than three times the number of dimes. How many quarters and how many dimes did Peter have?

#solutionbox[
Peter had 11 dimes and 48 quarters.
]

Lucinda had a pocketful of dimes and quarters with a value of \$ \$6.20. The number of dimes is eighteen more than three times the number of quarters. How many dimes and how many quarters does Lucinda have?

A cashier has 30 bills, all of which are \$10 or \$20 bills. The total value of the money is \$460. How many of each type of bill does the cashier have?

#solutionbox[
The cashier has fourteen \$10 bills and sixteen \$20 bills.
]

A cashier has 54 bills, all of which are \$10 or \$20 bills. The total value of the money is \$910. How many of each type of bill does the cashier have?

Marissa wants to blend candy selling for \$1.80 per pound with candy costing \$1.20 per pound to get a mixture that costs her \$1.40 per pound to make. She wants to make 90 pounds of the candy blend. How many pounds of each type of candy should she use?

#solutionbox[
Marissa should use 60 pounds of the \$1.20/lb candy and 30 pounds of the \$1.80/lb candy.
]

How many pounds of nuts selling for \$6 per pound and raisins selling for \$3 per pound should Kurt combine to obtain 120 pounds of trail mix that cost him \$5 per pound?

Hannah has to make twenty-five gallons of punch for a potluck. The punch is made of soda and fruit drink. The cost of the soda is \$1.79 per gallon and the cost of the fruit drink is \$2.49 per gallon. Hannah’s budget requires that the punch cost \$2.21 per gallon. How many gallons of soda and how many gallons of fruit drink does she need?

#solutionbox[
Hannah needs 10 gallons of soda and 15 gallons of fruit drink.
]

Joseph would like to make 12 pounds of a coffee blend at a cost of \$6.25 per pound. He blends Ground Chicory at \$4.40 a pound with Jamaican Blue Mountain at \$8.84 per pound. How much of each type of coffee should he use?

Julia and her husband own a coffee shop. They experimented with mixing a City Roast Columbian coffee that cost \$7.80 per pound with French Roast Columbian coffee that cost \$8.10 per pound to make a 20 pound blend. Their blend should cost them \$7.92 per pound. How much of each type of coffee should they buy?

#solutionbox[
Julia and her husband should buy 12 pounds of City Roast Columbian coffee and 8 pounds of French Roast Columbian coffee.
]

Melody wants to sell bags of mixed candy at her lemonade stand. She will mix chocolate pieces that cost \$4.89 per bag with peanut butter pieces that cost \$3.79 per bag to get a total of twenty-five bags of mixed candy. Melody wants the bags of mixed candy to cost her \$4.23 a bag to make. How many bags of chocolate pieces and how many bags of peanut butter pieces should she use?

Jotham needs 70 liters of a 50% alcohol solution. He has a 30% and an 80% solution available. How many liters of the 30% and how many liters of the 80% solutions should he mix to make the 50% solution?

#solutionbox[
Jotham should mix 42 liters of the 30% solution and 28 liters of the 80% solution.
]

Joy is preparing 15 liters of a 25% saline solution. She only has 40% and 10% solution in her lab. How many liters of the 40% and how many liters of the 10% should she mix to make the 25% solution?

A scientist needs 65 liters of a 15% alcohol solution. She has available a 25% and a 12% solution. How many liters of the 25% and how many liters of the 12% solutions should she mix to make the 15% solution?

#solutionbox[
The scientist should mix 15 liters of the 25% solution and 50 liters of the 12% solution.
]

A scientist needs 120 liters of a 20% acid solution for an experiment. The lab has available a 25% and a 10% solution. How many liters of the 25% and how many liters of the 10% solutions should the scientist mix to make the 20% solution?

A 40% antifreeze solution is to be mixed with a 70% antifreeze solution to get 240 liters of a 50% solution. How many liters of the 40% and how many liters of the 70% solutions will be used?

#solutionbox[
160 liters of the 40% solution and 80 liters of the 70% solution will be used.
]

A 90% antifreeze solution is to be mixed with a 75% antifreeze solution to get 360 liters of a 85% solution. How many liters of the 90% and how many liters of the 75% solutions will be used?

#strong[Solve Interest Applications]

In the following exercises, translate to a system of equations and solve.

Hattie had \$3,000 to invest and wants to earn 10.6% interest per year. She will put some of the money into an account that earns 12% per year and the rest into an account that earns 10% per year. How much money should she put into each account?

#solutionbox[
Hattie should invest \$900 at 12% and \$2,100 at 10%.
]

Carol invested \$2,560 into two accounts. One account paid 8% interest and the other paid 6% interest. She earned 7.25% interest on the total investment. How much money did she put in each account?

Sam invested \$48,000, some at 6% interest and the rest at 10%. How much did he invest at each rate if he received \$4,000 in interest in one year?

#solutionbox[
Sam invested \$28,000 at 10% and \$20,000 at 6%.
]

Arnold invested \$64,000, some at 5.5% interest and the rest at 9%. How much did he invest at each rate if he received \$4,500 in interest in one year?

After four years in college, Josie owes \$65,800 in student loans. The interest rate on the federal loans is 4.5% and the rate on the private bank loans is 2%. The total interest she owed for one year was \$2,878.50. What is the amount of each loan?

#solutionbox[
The federal loan is \$62,500 and the bank loan is \$3,300.
]

Mark wants to invest \$10,000 to pay for his daughter’s wedding next year. He will invest some of the money in a short term CD that pays 12% interest and the rest in a money market savings account that pays 5% interest. How much should he invest at each rate if he wants to earn \$1,095 in interest in one year?

A trust fund worth \$25,000 is invested in two different portfolios. This year, one portfolio is expected to earn 5.25% interest and the other is expected to earn 4%. Plans are for the total interest on the fund to be \$1150 in one year. How much money should be invested at each rate?

#solutionbox[
\$12,000 should be invested at 5.25% and \$13,000 should be invested at 4%.
]

A business has two loans totaling \$85,000. One loan has a rate of 6% and the other has a rate of 4.5%. This year, the business expects to pay \$4650 in interest on the two loans. How much is each loan?


==== Everyday Math

In the following exercises, translate to a system of equations and solve.

Laurie was completing the treasurer’s report for her son’s Boy Scout troop at the end of the school year. She didn’t remember how many boys had paid the \$15 full-year registration fee and how many had paid the \$10 partial-year fee. She knew that the number of boys who paid for a full-year was ten more than the number who paid for a partial-year. If \$250 was collected for all the registrations, how many boys had paid the full-year fee and how many had paid the partial-year fee?

#solutionbox[
14 boys paid the full-year fee. 4 boys paid the partial-year fee,
]

As the treasurer of her daughter’s Girl Scout troop, Laney collected money for some girls and adults to go to a three-day camp. Each girl paid \$75 and each adult paid \$30. The total amount of money collected for camp was \$765. If the number of girls is three times the number of adults, how many girls and how many adults paid for camp?


==== Writing Exercises

Take a handful of two types of coins, and write a problem similar to  relating the total number of coins and their total value. Set up a system of equations to describe your situation and then solve it.

#solutionbox[
Answers will vary.
]

In  we solved the system of equations #math.equation(block: false, alt: "{ b plus f equals 21,540; 0.105 b plus 0.059 f equals 1669.68")[$\{ b + f = 21,540 \ 0.105 b + 0.059 f = 1669.68$] by substitution. Would you have used substitution or elimination to solve this system? Why?


==== Self Check

After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

#figure(figph[This figure shows a table with four rows and four columns. The columns are labeled, “I can…,” “Confidently.” “With some help.” and “No - I don’t get it.” The only column with filled in cells below it is labeled “I can…” It reads, “solve mixture applications.” “solve interest applications.”], alt: "This figure shows a table with four rows and four columns. The columns are labeled, “I can…,” “Confidently.” “With some help.” and “No - I don’t get it.” The only column with filled in cells below it is labeled “I can…” It reads, “solve mixture applications.” “solve interest applications.”", caption: none)

After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?