5E Steps to Solving General Linear Equations

A linear equation is an equation where the variable is to the first power. If a variable has an exponent of 2 (square) or 3 (cube) or higher, then it will require more advanced methods to solve the problem. In this chapter we are focusing on solving linear equations, but first we need to talk about the different types of equations.

3 Types of Equations

There are 3 types of equations: conditional, contradiction and identity.

Conditional equations are only true sometimes. Look at the equation $x+4=11$. This type of equation is a conditional equation because it is only true if x = 7 and is not true if x is any other number. When an equation has a finite, nonzero number of solutions, it is conditional. Most equations are conditional. Our job is to find the number we can plug in to make the equation true.

Contradiction equations are never true. Look at the equation w+3 = w+5. This equation is never true. No matter what number we replace w with, we always get a false statement. When an equation has no solution it is called a contradiction equation. When asked to solve a contradiction, something weird will happen. When we subtract w from both sides we are left with 3 = 5. This is not true. So you know the equation has “no solution”. Look at another example $3b+7=3b-2$. A technique in solving equations is to bring the variable terms to one side. But if we subtract 3b from both sides we get +7 = -2. That is never true!! This tells us that the equation is never true no matter what. Hence this is a contradiction equation and the answer is “No Solution”.

The third type of equation is an identity equation. This is one that is always true. Look at y+4 = y+4. We can plug in any number we want for y and it will be true. For example we could plug in 70 and see that 70+4=70+4 (true). We could plug in -549 and see that -549+4 = -549+4. Every time we plug in any number we get a true statement. The solution to an identity equation is “all real numbers”, since it has infinitely many solutions. When solving an identity equation like y+4 = y+4, we will subtract y from both sides, but then all the variables are gone and we are left with the true statement 4=4. When that happens you know the answer is “all real numbers”. Look at another example $5a+1=5a+1$. Did you notice the two sides are exactly the same? If not, we can try to bring the variable terms to one side. But if we subtract 5a from both sides we get +1 = +1. That is always true!! This tells us that the equation is true no matter what we plug in for a. Hence this is an identity equation and the answer is “All Real Numbers”.

General Equation Solving

We have seen that we can solve equations by guessing the answer. If we cannot guess we can use the multiplication and addition properties to help us figure out the answer. Let’s now look at some more complicated equations and the steps to solving them.

Steps to Solving a Linear Equation (It is Vital to Memorize These!!)

1. Eliminate parenthesis by using the distributive property.
2. Eliminate fractions by multiplying both sides of the equation by the LCD.
3. Eliminate decimals by multiplying both sides of the equation by a power of 10 (10, 100, 1000…)
4. Use the addition property to eliminate variable terms so that there are only variables on one side of the equation.
5. Use the addition property to eliminate constants so that there are only constants on one side of the equation. The constants should be on the opposite side of the variables.
6. Use the multiplication property to multiply or divide both sides of the equation in order to isolate the variable by creating a coefficient of 1 for the variable.
7. Check your answer by plugging it into the original equation and see if the two sides are equal.

Let’s look at another example equation $5x+7=4x-3$. When dealing with an equation like this, our goal is to bring letters to one side and the constant numbers to the other side. If we want to eliminate the 4x on the right side, we can subtract 4x from both sides.

$$\begin{array}{rcl}5x+7& =& 4x-3\\ -4x& & -4x\\ x+7& =& 0-3\\ x+7& =& -3\end{array}$$

Notice that there are only x variables on the left side. Can you guess the answer now? If not we can get rid of the 7 by subtracting 7 (adding -7) to both sides.

$$\begin{array}{rcl}x+7& =& -3\\ -7& & -7\\ x+0& =& -10\\ x& =& -10\end{array}$$

Notice the number we can plug in for x that makes the equation true is -10. Check if that is the correct answer. Plugging in -10 into the original equation we get the following:

$$\begin{array}{l}5(-10)+7=4(-10)-3\\ -50+7=-40-3\\ -43=-43\end{array}$$

So when we plug in -10, we do get a true statement. Hence -10 is the solution. Note that the two sides were equal and both equal to -43, but -43 is not the solution. The solution is the number we replaced the letter with that made the two sides equal. Also notice this was a conditional equation. It was only true when x = -10 and false otherwise.

Let’s look at an example $$-6(2w-8)+4w=3w-(7w+9)$$

Step 1: Our first step is to distribute and eliminate parenthesis, so we will multiply the -6 times both the 2w and the -8. We will also distribute the negative to the 7w and the 9 and eliminate that parenthesis as well. We should only distribute to the terms in the parenthesis. For example we do not distribute the -6 to the 4w since the 4w is not in the parenthesis.

$$\begin{array}{l}-6(2w-8)+4w=3w-(7w+9)\\ -12w+48+4w=3w-7w-9\end{array}$$

Always look to simplify after each step. For example in this problem the -12w and 4w are like terms on the same side. Also the 3w and -7w are also like terms on the same side. If you struggle with negatives, you can convert the -7w to adding the opposite. Be careful. Do not add or subtract terms on opposite sides of the equation.

$$\begin{array}{l}-6(2w-8)+4w=3w-(7w+9)\\ -12w+48+4w=3w-7w-9\\ -8w+48=-4w-9\end{array}$$

Step 2 and 3: There are no fractions or decimals so we proceed directly to step 4.

Step 4: We need to bring the variables to one side. You can bring variables to either side, but many students like to bring the variables to the left side only. So we will need to eliminate the -4w on the right side. Hence we will add the opposite +4w to both sides.

$$\begin{array}{rcl}-8w+48& =& -4w-9\\ +4w& & +4w\\ -4w+48& =& 0-9\\ -4w+48& =& -9\end{array}$$

Step 5: We need to bring the constants to the opposite side as the variables. So we need to get rid of the +48. Hence we will subtract 48 (add -48) to both sides. We are then left with -4w= -57

$$\begin{array}{rcl}-4w+48& =& -9\\ -48& & -48\\ -4w+0& =& -57\\ -4w& =& -57\end{array}$$

Step 6: We now need to get the w by itself. Since the w is being multiplied by -4, we will divide both sides by -4 to get our answer of 57/4. Notice the answer can be written three ways and all are equally correct $(\frac{57}{4}\text{or}14⁤\frac{1}{4}\text{or}14.25)$

$$\begin{array}{l}-4w=-57\\ \frac{{}^1-4}{-4}w=\frac{-57}{-4}\\ 1w=14.25\\ w=14.25\end{array}$$

Step 7: Let’s check our answer by plugging into the original equation. Notice that all of the w’s have to be replaced with 14.25 and don’t forget to use the order of operations when simplifying each side. As you can see, checking your answer can be just as much work as solving the equation.

$$\begin{array}{l}-6(2w-8)+4w=3w-(7w+9)\\ -6(2\times 14.25-8)+4\times 14.25=3\times 14.25-(7\times 14.25+9)\\ -6(28.5-8)+4\times 14.25=3\times 14.25-(99.75+9)\\ -6\left(20.5\right)+4\times 14.25=3\times 14.25-(108.75)\\ -123+57=42.75-(108.75)\\ -66=-66\end{array}$$

Let’s try another example. Look at $\frac{1}{3}c-\frac{3}{5}=\frac{1}{2}c+4$

Step 1: There are no parenthesis so we proceed to step 2.

Step 2: To eliminate fractions we find the LCD. Since the denominators are 3,5 and 2 the LCD is 30. Hence we will multiply everything on both sides by 30. This will eliminate the fractions. Remember all the terms must be multiplied by 30. When multiplying a whole number (30) by fractions it is good to write the whole number as a fraction (30/1).

$$\begin{array}{l}\frac{30}{1}(\frac{1}{3}c-\frac{3}{5})=\frac{30}{1}(\frac{1}{2}c+4)\\ \frac{30}{1}\times \frac{1}{3}c-\frac{30}{1}\times \frac{3}{5}=\frac{30}{1}\times \frac{1}{2}c+\frac{30}{1}\times 4\\ \frac{30}{3}c-\frac{90}{5}=\frac{30}{2}c+\frac{120}{1}\\ 10c-18=15c+120\end{array}$$

Notice we are now left with an equation without fractions.

Step 3: There are no decimals, so we proceed to step 4.

Step 4: We bring all the variables to the left side by subtracting 15c (adding -15c) to both sides.

$$\begin{array}{rcl}10c-18& =& 15c+120\\ -15c& & -15c\\ -5c-18& =& 0+120\\ -5c-18& =& 120\end{array}$$

Step 5: We bring all the constants to the opposite side. We can eliminate the -18 by adding +18 to both sides.

$$\begin{array}{rcl}-5c-18& =& 120\\ +18& & +18\\ -5c+0& =& 138\\ -5c& =& 138\end{array}$$

Step 6: Isolate the variable. Since we are multiplying c by -5, we divide both sides by -5.

$$\begin{array}{l}-5c=138\\ \frac{{}^1-5}{-5}c=\frac{138}{-5}\\ 1c=-\frac{138}{5}\\ c=-27⁤\frac{3}{5}\text{ or }-27.6\end{array}$$

Step 7: Check your answer. By plugging $-27.6$ into the original equation, the two sides are equal.

Let’s try a third example: $$0.24(3m-1)=0.82m-0.24-0.1m$$

Step 1: We will distribute the 0.24 to the 3m and -1 to eliminate the parenthesis. We will make sure to combine any like terms that are on the same side. Notice 0.82m and -0.1m are like terms on the same side, so we can add them.

$$\begin{array}{l}0.24(3m-1)=0.82m-0.24-0.1m\\ 0.24\times 3m+0.24\times -1=0.82m-0.24-0.1m\\ 0.72m-0.24=0.72m-0.24\end{array}$$

Step 2: There are no fractions, so we proceed to step 3.

Step 3: Since the most decimal places to the right is two (hundredths place), we will multiply everything on both sides by 100. If one of the decimals had ended in the thousandths place, we would multiply by 1000 and so on. Notice all the decimals are gone.

$$\begin{array}{l}0.72m-0.24=0.72m-0.24\\ 100(0.72m-0.24)=100(0.72m-0.24)\\ 100\times 0.72m+100\times -0.24=100\times 0.72m+100\times -0.24\\ 72m-24=72m-24\end{array}$$

Step 4: Bring the variables to one side by subtracting 72m from both sides. Notice all variables cancel and we are left with -24 = -24 which is a true statement.

$$\begin{array}{rcl}72m-24& =& 72m-24\\ -72m& & -72m\\ 0-24& =& 0-24\\ -24& =& -24\end{array}$$

Since we are left with a true statement and all the variables are gone, we need go no further. This is an always true equation (identity). So the answer is “All Real Numbers”.

Try to solve the following equations with your instructor.

Practice Problems Section 5E

Solve the following equations. If your answer is a fraction, be sure to simplify it completely. For #39, #40 and #47, remember to eliminate parenthesis before eliminating fractions or decimals.

1. $$7y-15=20$$

   Check answer

   $$y=5$$

2. $$-18=4p+2$$

3. $$12=-1a-17$$

   Check answer

   $$a=-29$$

4. $$9+-7n=8$$

5. $$-12x+7=-9x-11$$

   Check answer

   $$x=6$$

6. $$5y-14=-3y+18$$

7. $$-1a+27=-6a-13$$

   Check answer

   $$a=-8$$

8. $$-2b+13=9b+7$$

9. $$6-13v=-5+-11v$$

   Check answer

   $$v=5⁤\frac{1}{2}\text{ or }5.5$$

10. $$-23+5m=-19-18m$$

11. $$13-y=-1y+17-4$$

    Check answer

    All Real Numbers

12. $$14g-3=-3g+8$$

13. $$6a=5a-7$$

    Check answer

    $$a=-7$$

14. $$8b-13=8b+2$$

15. $$17v=16v-9$$

    Check answer

    $$v=-9$$

16. $$-8b=-9b+4$$

17. $$-8x++8x=13-13$$

    Check answer

    All Real Numbers

18. $$-5w=-6w+-11$$

19. $$-9a-13=-10a+4$$

    Check answer

    $$a=17$$

20. $$1.3x-2.7=0.3x-3.4$$

21. $$16w+9-15w=w+9$$

    Check answer

    All Real Numbers

22. $$-5.2x+7.3=-6.2x-3.6$$

23. $$6a-\frac{1}{4}=5a+\frac{3}{4}$$

    Check answer

    $$a=1$$

24. $$0.8b-2.57=-0.2b+2.9$$

25. $$a-0.004=-0.5a+0.053+0.5a$$

    Check answer

    $$a=0.057$$

26. $$\frac{8}{5}x-\frac{1}{3}=\frac{3}{5}x+\frac{1}{3}$$

27. $$26x+19-21x=5x-17$$

    Check answer

    No Solution

28. $$-5.2d+7.3=-6.2d-3.6$$

29. $$-31m+17+28m=-3m-10+27$$

    Check answer

    All Real Numbers

30. $$25+5n-24=-3n+18+8n$$

31. $$-2(4x+3)=-6x$$

    Check answer

    $$x=-3$$

32. $$7(y-3)+2y=6$$

33. $$-4(5\mathrm{a}-1)+18\mathrm{a}=-2(-3\mathrm{a}+2)$$

    Check answer

    $$a=1$$

34. $$11b-(7b+9)=4(b-3)$$

35. $$\frac{1}{2}m+1=-\frac{1}{3}m-2$$

    Check answer

    $$m=-3⁤\frac{3}{5}\text{ or }-3.6$$

36. $$\frac{2}{3}n-\frac{1}{5}=\frac{2}{5}n$$

37. $$\frac{3}{4}c+2=1c-\frac{1}{2}$$

    Check answer

    $$c=10$$

38. $$\frac{1}{6}d-\frac{3}{2}=\frac{1}{2}d+\frac{5}{6}$$

39. $$\frac{1}{5}(2p+\frac{1}{3})=\frac{1}{2}(\frac{1}{3}p-1)$$

    Check answer

    $$p=-2⁤\frac{3}{7}$$

40. $$\frac{2}{3}(w+6)=\frac{1}{4}(w-8)$$

41. $$0.5x-0.3=0.8x+0.9$$

    Check answer

    $$x=-4$$

42. $$0.09p-0.04=0.09p+0.06$$

43. $$0.23x-0.4=0.38x+0.2$$

    Check answer

    $$x=-4$$

44. $$0.007y-0.03=0.009y+0.06$$

45. $$0.2m+0.41=0.18m+0.67$$

    Check answer

    $$m=13$$

46. $$0.6n-2=n+0.3$$

47. $$-0.4(2a+1)=-0.9a+0.5+0.1a-0.9$$

    Check answer

    All Real Numbers

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