6D Classic Algebraic Problem Solving

Algebra can be a useful tool for solving even when we do not know the formula for the situation. Here are some general steps to using algebra to solve word problems.

Steps to Solving Problems with Algebra

1. Read and reread the problem and write down all of the unknowns in the problem on a piece of paper.
2. Let your variable (x) stand for the unknown that you have the least amount of information about.
3. Use the information given in the problem to write algebraic expressions for all of your other unknowns.
4. Once you have algebraic expressions for all your unknowns, then you can use them to write an equation.
5. Solve the equation for the variable (x).
6. Go back and plug in the value of the variable (x) into all the algebraic expressions and find all the unknowns.
7. Check and see if your answers make sense in the context of the problem.

It is also good to make a list of words and their meaning .

Addition words: Sum, more than, increased by, total

Subtraction words: Subtract from, less than, less, decreased by, difference

Multiplication words: Multiply, product, times, percent of (multiply by percent), fraction of (multiply by fraction)

Division words: Quotient, ratio, divided by

Words that mean “Equals”: Is, Is equal to, Is equivalent to, Is the same as

Let’s look at an example. When seven is subtracted from the product of a number and four, the result is 69. Find the number.

Step 1-3 - Since there is only one unknown, we will let x be the number.

Step 4 - Now we will translate the words into an equation. Subtracted from means subtract in the opposite order. The product of a number and four means multiply 4 and x. So we get the following.

$$4x-7=69$$

Step 5 - Now solve the equation. Since there is only one number we are looking for, x is the answer.

$$\begin{array}{rcl}4x-7& =& 69\\ +7& & +7\\ 4x+0& =& 76\\ 4x& =& 76\\ \frac{{}^{1}4x}{4}& =& \frac{76}{4}\\ x& =& 19\end{array}$$

Steps 6&7 – The only unknown is the number 19 and that is our answer. The answer does make sense since 4 times the number is 76 and then 7 less than it is 69.

Let’s look at another example. Suppose a political science club is divided into three groups: liberal, moderate and conservative. There are twice as many liberals as moderates and there are seven more conservatives than moderates. If the club has a total of 75 members, how many are in each group?

Step 1 – Write down all unknowns. You should write down the following on a piece of paper.

# of liberals =

# of moderates =

# of conservatives =

Step 2 – Since we know something about liberals and conservatives but nothing about moderates we will let x be the number of moderates.

Step 3 – Twice as many liberals as moderates means we can represent liberals as 2x. Seven more conservatives means we can represent conservatives with x+7.

# of liberals = 2x

# of moderates = x

# of conservatives = x+7

Step 4 – Never try to write an equation until you have algebraic expressions for all your unknowns. Since we know the total is 75 we will add our algebraic expressions and set it equal to 75.

2x + x + x+7 = 75

Step 5 – Now we solve the equation for x using the techniques we learned in the last chapter. Notice that 2x, x and x are like terms on the same side so we can simply add them.

$$\begin{array}{rcl}2x+x+x+7& =& 75\\ 4x+7& =& 75\\ -7& & -7\\ 4x+0& =& 68\\ 4x& =& 68\\ \frac{{}^{1}4x}{4}& =& \frac{68}{4}\\ x& =& 17\end{array}$$

Step 6 – Now that we know x we can find all the solutions. Go back to your unknowns and algebraic expressions. If you plug in 17 for x in all of them you will have your answers.

# of liberals = 2x = 2(17) = 34

# of moderates = x = 17

# of conservatives = x+7 = 17+7 = 24

Step 7 – So our answer is 34 liberals, 17 moderates and 24 conservatives. Does this answer make sense? Yes. There are twice as many liberals and seven more conservatives than moderates and the total is 75.

Let’s look at another example. Jimmy has a bunch of coins in the cushions of his couch. As he is out of money and needs gas in his car, he is searching for as many coins as he can find. The coins he found had a total value of $7.50. He only found quarters, dimes and nickels. There were three times as many dimes as nickels and eighteen more quarters than nickels. How many of each type of coin did he find?

Step 1 – Write down all unknowns. You should write down the following on a piece of paper. Again the key thing to remember is that for coin problems, the number of coins is not money. If we have 3 quarters, that does not mean I have $3. You need to multiply the number of the coins times the value of the coin. So 3 quarters means 3 x 0.25 = $0.75. Not only do we need to know how many coins but also we need algebraic expressions that represent the amount of money for each coin.

# of nickels =

# of dimes =

# of quarters =

$ in nickels =

$ in dimes =

$ in quarters =

Step 2 – Since we know something about quarters and dimes but nothing about nickels we will let x be the number of nickels.

Step 3 – Three times as many dimes as nickels means we can represent the number of dimes as 3x. Eighteen more quarters than nickels means we can represent quarters with x + 18. Remember to get the amount of money, we will multiply the number of nickels times its value ($0.05), the number of dimes times its value ($0.10) and the number of quarters times its value ($0.25). We can simplify each expression. For the quarters we will use the distributive property.

# of nickels = x

# of dimes = 3x

# of quarters = x + 18

$ in nickels = 0.05x

$ in dimes = 0.10 (3x) = 0.30x

$ in quarters = 0.25 (x + 18) = 0.25x + 4.5

Step 4 – Never try to write an equation until you have algebraic expressions for all your unknowns. Since we know the total amount of money is $7.50 we will add our algebraic expressions for money and set it equal to 7.5

$$0.05x+0.30x+0.25x+4.5=7.5$$

Step 5 – Now we solve the equation for x using the techniques we learned in the last chapter. Notice that the 0.05x, 0.30x and 0.25x are like terms on the same side so we can simply add them.

$$\begin{array}{rcl}0.05x+0.30x+0.25x+4.5& =& 7.5\\ 0.6x+4.5& =& 7.5\\ 10(0.6x+4.5)& =& 10(7.5)\\ 6x+45& =& 75\\ -45& & -45\\ 6x+0& =& 30\\ 6x& =& 30\\ \frac{{}^{1}6x}{6}& =& \frac{30}{6}\\ x& =& 5\end{array}$$

Step 6 – Now that we know x we can find all the solutions. Go back to your unknowns and algebraic expressions. If you plug in 5 for x in all of them you will have your answers.

# of nickels = x = 5 nickels

# of dimes = 3x = 3(5) = 15 dimes

# of quarters = x + 18 = 5 + 18 = 23 quarters

$ in nickels = 0.05x = 0.05 (5) = $0.25

$ in dimes = 0.10 (3x) = 0.30x = 0.3 (5) = $1.50

$ in quarters = 0.25 (x + 18) = 0.25x + 4.5 = 0.25 (5) + 4.5 = 1.25+4.5 = $5.75

Step 7 – So our answer is 5 nickels, 15 dimes and 23 quarters. Does this answer make sense? Yes. There are three times as many dimes as nickels and 18 more quarters than nickels. The amount of money in nickels ($0.25) + amount of money in dimes ($1.50) + amount of money in quarters ($5.75) = $7.50.

Try the following examples with your instructor:

Practice Problems Section 6D

1. The sum of twice a number and nine is the same as the number subtracted from twenty-four. Find the number.

   Check answer

   5

2. The product of a number and six is equivalent to the sum of the number and forty. Find the number.

   Check answer

   8

3. Two less than the quotient of a number and five is the same as eleven. Find the number.

   Check answer

   65

4. The smaller of two numbers is six less than the larger. The sum of the two numbers is 118. Find both numbers.

   Check answer

   Smaller number: 56, Larger number: 62

5. The sum of half a number and seven is nineteen. Find the number.

   Check answer

   24

6. Twelve less than a number is eight. Find the number.

7. The product of a number and 4 is equal to the sum of twice the number and fourteen. Find the number.

   Check answer

   7

8. Four less than twice a number is 20. Find the number.

9. A number divided by six added to four is equivalent to the number decreased by 71.

   Check answer

   90

10. Ten times a number is sixteen more than twice the number. Find the number.

11. Jim has a three-number code to unlock the padlock on his garage door. The second number is seven less than the third number. The first number is two more than the third number. Find all three numbers if the sum of the numbers adds up to 34.

    Check answer

    First number: 15, Second number: 6, Third number 13

12. Carrie loves to plant flowers in her garden. Her favorites are daisies, roses and sunflowers. She has three times as many daisies as sunflowers and twice as many roses as sunflowers. She has a total of 78 flowers in her garden. How many of each type of flower does she have?

    Check answer

    39 daisies, 26 roses, 13 sunflowers

13. The sum of the angles of any triangle add up to 180 degrees. The largest angle of a triangle is four times as large as the smallest angle. The middle angle is 42 degrees greater than the smallest angle. Find all three angles.

    Check answer

    Largest angle: 92 degrees, Middle angle: 65 degrees, Smallest angle: 23 degrees

14. Mrs. Smith has a total of 43 children in her class. The number of girls is one more than twice the number of boys. How many boys and girls does she have in her class?

    Check answer

    14 boys, 29 girls

15. The students in a political science class are divided up into three categories: liberal, conservative, or moderate. There are five more conservatives than moderates, and twice as many liberals as conservatives. How many of each type are there if there is a total of 87 students in the class?

    Check answer

    Liberals: 46 students, Conservatives: 23 students, Moderates: 18 students

16. Jimmy collects baseball and football trading cards. The number of baseball cards is one less than three times as many as the football cards. He has a total of 267 cards in his collection. How many baseball cards does he have? How many football cards does he have?

    Check answer

    200 baseball cards, 67 football cards

17. Gary owns a car dealership that sells cars, SUVs and minivans. Gary has three times as many SUVs as Minivans. He has eight more cars than SUVs. If he has a total of 127 vehicles on his lot, how many of each type does he have?

    Check answer

    59 cars, 17 minivans, 51 SUVs.

18. Harry has a total of $237 in his wallet in twenty dollar bills, five dollar bills, and 1 dollar bills. He has twice as many five dollar bills as one dollar bills. He has one more twenty dollar bill than one dollar bills. How many of each type does he have?

    Check answer

    8 twenty dollar bills, 14 five dollar bills, 7 one dollar bills

19. Lucia has a total of $2.60 in quarters, dimes and nickels in her purse. She has three more dimes than quarters and six more nickels than quarters. How many of each type does she have?

    Check answer

    5 quarters, 8 dimes, 11 nickels

20. Marcos collects pennies, dimes and nickels. He has twice as many pennies as dimes and three more nickels than dimes. If the total amount of money is $6.95, how many of each type of coin does he have?

    Check answer

    80 pennies, 43 nickels, 40 dimes

21. A golden rectangle has length approximately 1.618 times its width. For this exercise, suppose a rectangular wall has these proportions. If its perimeter is 261.8 feet, what is the width and length of the wall?

    Check answer

    Length: 80.9 ft, Width: 50 ft

Adapted from Preparing for Algebra and Statistics, Third Edition, by Matt Teachout. License: CC BY 4.0. Copyright and sources.