5.2: Solving Equations Part II
- Page ID
- 137922
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)We return to solving equations involving integers, only this time the equations will be a bit more advanced, requiring the use of the distributive property and skill at combining like terms. Let’s begin.
Example 1
Solve for x:
7x − 11x = 12
Solution
Combine like terms.
\[ \begin{aligned} 7x-11x=12 ~ & \textcolor{red}{ \text{ Original equation.}} \\ =4x=12 ~ & \textcolor{red}{ \text{ Combine like terms: } 7x-11x=-4x.} \end{aligned}\nonumber \]
To undo the effect of multiplying by −4, divide both sides of the last equation by −4.
\[ \begin{aligned} \frac{-4x}{-4} = \frac{12}{-4} ~ & \textcolor{red}{ \text{ Divide both sides by } -4.} \\ x = -3 ~ & \textcolor{red}{ \text{ Simplify: } 12/(-4) = -3.} \end{aligned}\nonumber \]
Check
Substitute −3 for x in the original equation.
\[ \begin{aligned} 7x-11x = 12 ~ & \textcolor{red}{ \text{ Original equation.}} \\ 7(-3)-11(-3)=12 ~ & \textcolor{red}{ \text{ Substitute } -3 \text{ for } x.} \\ -21 + 33 = 12 ~ & \textcolor{red}{ \text{ On the left, multiply first.}} \\ 12 = 12 ~ & \textcolor{red}{ \text{ On the left, add.}} \end{aligned}\nonumber \]
Because the last line of the check is a true statement, −3 is a solution of the original equation.
Exercise
Solve for x:
−6x − 5x = 22
- Answer
-
x = −2
Example 2
Solve for x:
12 = 5x − (4 + x)
Solution
To take the negative of a sum, negate each term in the sum (change each term to its opposite). Thus, −(4 + x) = −4 − x.
\[ \begin{aligned} 12 = 5x - (4+x) ~ & \textcolor{red}{ \text{ Original equation.}} \\ 12 = 5x - 4 - x ~ & \textcolor{red}{-(4+x)=-4-x.} \\ 12 = 4x-4 ~ & \textcolor{red}{ \text{ Combine like terms: } 5x - x = 4x.} \end{aligned}\nonumber \]
To undo the effect of subtracting 4, add 4 to both sides of the last equation.
\[ \begin{aligned} 12 + 4 = 4x-4+4 ~ & \textcolor{red}{ \text{ Add 4 to both sides.}} \\ 16 = 4x ~ & \textcolor{red}{ \text{ Simplify both sides.}} \end{aligned}\nonumber \]
To undo the effect of multiplying by 4, divide both sides of the last equation by 4.
\[ \begin{aligned} \frac{16}{4} = \frac{4x}{4} ~ & \textcolor{red}{ \text{ Divide both sides by 4.}} \\ 4 = x ~ & \textcolor{red}{ \text{ Simplify: 16/4=4.} \end{aligned}]
Check
Substitute 4 for x in the original equation.
\[ \begin{aligned} 12 = 5x - (4+x) ~ & \textcolor{red}{ \text{ Original equation.}} \\ 12 = 5(4) - (4+4) ~ & \textcolor{red}{ \text{ Substitute 4 for } x.} \\ 12 = 20 - 8 ~ & \textcolor{red}{ \text{ On the right, 5(4) = 20 and evaluate}} \\ ~ & \textcolor{red}{ \text{ parentheses: } 4+4=8.} \\ 12 = 12 ~ & \textcolor{red}{ \text{ Simplify.}} \end{aligned}\nonumber \]
Because the last line of the check is a true statement, 4 is a solution of the original equation.
Exercise
Solve for x:
11 = 3x − (1 − x)
- Answer
-
x = 3
Variables on Both Sides
Variables can occur on both sides of the equation.
Goal
Isolate the terms containing the variable you are solving for on one side of the equation.
Example 3
Solve for x:
5x = 3x − 18
Solution
To isolate the variables on one side of the equation, subtract 3x from both sides of the equation and simplify.
\[ \begin{aligned} 5x = 3x-18 ~ & \textcolor{red}{ \text{ Original equation.}} \\ 5x-3x=3x-18-3x ~ & \textcolor{red}{ \text{ Subtract } 3x \text{ from both sides.}} \\ 2x = -18 ~ & \textcolor{red}{ \text{ Combine like terms: } 5x - 3x = 2x } \\ ~ & \textcolor{red}{ \text{ and } 3x - 3x = 0.} \end{aligned}\nonumber \]
Note that the variable is now isolated on the left-hand side of the equation. To undo the effect of multiplying by 2, divide both sides of the last equation by 2.
\[ \begin{aligned} \frac{2x}{2} = \frac{-18}{2} ~ & \textcolor{red}{ \text{ Divide both sides by 2.}} \\ x = -9 ~ & \textcolor{red}{ \text{ Simplify: } -18/2 = =-9.} \end{aligned}\nonumber \]
Check
Substitute −9 for x in the original equation.
\[ \begin{aligned} 5x = 3x - 18 ~ & \textcolor{red}{ \text{ Original equation.}} \\ 5(-9) = 3(-9)-18 ~ & \textcolor{red}{ \text{ Substitute } -9 \text{ for } x.} \\ -45 = -27-18 ~ & \textcolor{red}{ \text{ Multiply first on both sides.}} \\ -45 = -45 ~ & \textcolor{red}{ \text{ Subtract on the right: } -27 - 18 = -45.} \end{aligned}\nonumber \]
Because the last line of the check is a true statement, −9 is a solution of the original equation.
Exercise
Solve for x:
4x − 3 = x
- Answer
-
x = 1
Example 4
Solve for x:
5x = 3 + 6x
Solution
To isolate the variables on one side of the equation, subtract 6x from both sides of the equation and simplify.
\[ \begin{aligned} 5x = 3+6x ~ & \textcolor{red}{ \text{ Original equation.}} \\ 5x - 6x = 3+6x - 6x ~ & \textcolor{red}{ \text{ Subtract } 6x \text{ from both sides.}} \\ -x=3 ~ & \textcolor{red}{ \text{ Combine like terms: } 5x - 6x = -x} \\ ~ & \textcolor{red}{ \text{ and } 6x - 6x = 0.} \end{aligned}\nonumber \]
Note that the variable is now isolated on the left-hand side of the equation.
There are a couple of ways we can finish this solution. Remember, −x is the same as (−1)x, so we could undo the effects of multiplying by −1 by dividing both sides of the equation by −1. Multiplying both sides of the equation by −1 will work equally well. But perhaps the easiest way to proceed is to simply negate both sides of the equation.
\[ \begin{aligned} -(-x) = -3 ~ & \textcolor{red}{ \text{ Negate both sides.}} \\ x = -3 ~ & \textcolor{red}{ \text{ Simplify: } -(-x) = x.} \end{aligned}\nonumber \]
Check
Substitute −3 for x in the original equation.
\[ \begin{aligned} 5x = 3+6x ~ & \textcolor{red}{ \text{ Original equation.}} \\ 5(-3) = 3+6(-3) ~ & \textcolor{red}{ \text{ Substitute } -3 \text{ for } x.} \\ -15 = 3-18 ~ & \textcolor{red}{ \text{ Multiply first on both sides.}} \\ -15 = -15 ~ & \textcolor{red}{ \text{ Subtract on the right: } 3 -8 = -15.} \end{aligned}\nonumber \]
Because the last line of the check is a true statement, −3 is a solution of the original equation.
Exercise
Solve for x:
7x = 18 + 9x
- Answer
-
x = −9
Dealing with −x.
If your equation has the form
−x = c,
where c is some integer, note that this is equivalent to the equation (−1)x = c. Therefore, dividing both sides by −1 will produce a solution for x. Multiplying both sides by −1 works equally well. However, perhaps the easiest thing to do is negate each side, producing
−(−x) = −c, which is equivalent to x = −c.
Example 5
Solve for x:
6x − 5 = 12x + 19
Solution
To isolate the variables on one side of the equation, subtract 12x from both sides of the equation and simplify.
\[ \begin{aligned} 6x-5 = 12x+19 ~ & \textcolor{red}{ \text{ Original equation.}} \\ 6x-5-12x = 12x + 19 - 12x ~ & \textcolor{red}{ \text{ Subtract } 12x \text{ from both sides.}} \\ -6x - 5 = 19 ~ & \textcolor{red}{ \text{ Combine like terms: } 6x-12x=-6x} \\ ~ & \textcolor{red}{ \text{ and } 12x-12x = 0.} \end{aligned}\nonumber \]
Note that the variable is now isolated on the left-hand side of the equation. Next, to “undo” subtracting 5, add 5 to both sides of the equation.
\[ \begin{aligned} -6x -5+5 = 19+5 ~ & \textcolor{red}{ \text{ Add 5 to both sides.}} \\ -6x = 24 ~ & \textcolor{red}{ \text{ Simplify: } -5+5=0 \text{ and } 19+5 = 24.} \end{aligned}\nonumber \]
Finally, to “undo” multiplying by −6, divide both sides of the equation by −6.
\[ \begin{aligned} \frac{-6x}{-6} = \frac{24}{-6} ~ & \textcolor{red}{ \text{ Divide both sides by } -6.} \\ x = -4 ~ & \textcolor{red}{ \text{ Simplify: } 24/(-6)=-4.} \end{aligned}\nonumber \]
Check. Substitute −4 for x in the original equation
\[ \begin{aligned} 6x - 5 = 12x+19 ~ & \textcolor{red}{ \text{ Original equation.}} \\ 6(-4)-5 = 12(-4)+19 ~ & \textcolor{red}{ \text{ Substitute } -4 \text{ for } x.} \\ -24-5 = -48 + 19 ~ & \textcolor{red}{ \text{ Multiply first on both sides.}} \\ -29 = -29 ~ & \textcolor{red}{ \text{ Simplify each side: } -24 - 5 = -29, -48 + 19 = -29.} \end{aligned}\nonumber \]
Because the last line of the check is a true statement, −4 is a solution of the original equation.
Exercise
Solve for x:
2x + 3 = 18 − 3x
- Answer
-
x = 3
Example 6
Solve for x:
2(3x + 2) − 3(4 − x) = x + 8
Solution
Use the distributive property to remove parentheses on the left-hand side of the equation.
\[ \begin{aligned} 2(3x+2)-3(4-x)=x+8 ~ & \textcolor{red}{ \text{ Original equation.}} \\ 6x+4 - 12 + 3x = x + 8 ~ & \textcolor{red}{ \text{ Use the distributive property.}} \\ 9x - 8 = x + 8 ~ & \textcolor{red}{ \text{ Combine like terms: } 6x + 3x = 9x} \\ ~ & \textcolor{red}{ \text{ and } 4-12 = -8.} \end{aligned}\nonumber \]
Isolate the variables on the left by subtracting x from both sides of the equation
\[ \begin{aligned} 9x-8 -x = x + 8 - x ~ & \textcolor{red}{ \text{ Subtract } x \text{ from both sides.}} \\ 8x - 8 = 8 ~ & \textcolor{red}{ \text{ Combine like terms: } 9x-x = 8x} \\ ~ & \textcolor{red}{ \text{ and } x-x=0.} \end{aligned}\nonumber \]
Note that the variable is now isolated on the left-hand side of the equation. Next, to “undo” subtracting 8, add 8 to both sides of the equation.
\[ \begin{aligned} 8x-8+8=8+8 ~ & \textcolor{red}{ \text{ Add 8 to both sides.}} \\ 8x = 16 ~ & \textcolor{red}{ \text{ Simplify: } -8+8=0 \text{ and } 8+8 = 16.} \end{aligned}\nonumber \]
Finally, to “undo” multiplying by 8, divide both sides of the equation by 8.
\[ \begin{aligned} \frac{8x}{8} = \frac{16}{8} ~ & \textcolor{red}{ \text{ Divide both sides by 8.}} \\ x = 2 ~ & \textcolor{red}{ \text{ Simplify: } 16/8=2.} \end{aligned}\nonumber \]
Check. Substitute 2 for x in the original equation.
\[ \begin{aligned} \end{aligned}\nonumber \]
Because the last line of the check is a true statement, 2 is a solution of the original equation.
Exercise
Solve for x:
3(2x − 4) − 2(5 − x) = 18
- Answer
-
x = 5
Exercises
Solve the equation.
1. −9x + x = −8
3. −4 = 3x − 4x
5. 27x + 51 = −84
7. 9 = 5x + 9 − 6x
11. 41 = 28x + 97
13. 8x − 8 − 9x = −3
15. −85x + 85 = 0
17. −6x = −5x − 9
19. 6x − 7 = 5x
21. 4x − 3 = 5x − 1
23. −3x +5 = 3x − 1
31. 6x = 4x − 4
33. −8x + 2 = −6x + 6
35. 1 − (x − 2) = −3
37. −7x + 6(x + 8) = −2
39. 8(−6x − 1) = −8
43. 2 − 9(x − 5) = −16
45. 7x + 2(x + 9) = −9
49. 8 + 2(x − 5) = −4
51. 9x − 2(x + 5) = −10
53. 4(−7x + 5) + 8 = 3(−9x − 1) − 2
55. −8(−2x − 6) = 7(5x − 1) − 2
57. 2(2x − 9) + 5 = −7(−x − 8)
61. 2(−2x − 3) = 3(−x + 2)
63. −5(−9x + 7) + 7 = −(−9x − 8)
Answers
1. 1
3. 4
5. −5
7. 0
11. −2
13. −5
15. 1
17. 9
19. 7
21. −2
23. 1
31. −2
33. −2
35. 6
37. 50
39. 0
43. 7
45. −3
49. −1
51. 0
53. 33
55. 3
57. −23
61. −12
63. 1