8.5: Solve Equations with Variables and Constants on Both Sides (Part 2)
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Solve Equations Using a General Strategy
Each of the first few sections of this chapter has dealt with solving one specific form of a linear equation. It’s time now to lay out an overall strategy that can be used to solve any linear equation. We call this the general strategy. Some equations won’t require all the steps to solve, but many will. Simplifying each side of the equation as much as possible first makes the rest of the steps easier.
HOW TO: USE A GENERAL STRATEGY FOR SOLVING LINEAR EQUATIONS
Step 1. Simplify each side of the equation as much as possible. Use the Distributive Property to remove any parentheses. Combine like terms.
Step 2. Collect all the variable terms to one side of the equation. Use the Addition or Subtraction Property of Equality.
Step 3. Collect all the constant terms to the other side of the equation. Use the Addition or Subtraction Property of Equality.
Step 4. Make the coefficient of the variable term to equal to 1. Use the Multiplication or Division Property of Equality. State the solution to the equation.
Step 5. Check the solution. Substitute the solution into the original equation to make sure the result is a true statement.
Example \(\PageIndex{11}\):
Solve: 3(x + 2) = 18.
Solution
Simplify each side of the equation as much as possible. Use the Distributive Property.  $$3x + 6 = 18 \tag{8.3.46}$$ 
Collect all variable terms on one side of the equation—all x's are already on the left side.  
Collect constant terms on the other side of the equation. Subtract 6 from each side.  $$3x + 6 \textcolor{red}{6} = 18 \textcolor{red}{6} \tag{8.3.47}$$ 
Simplify.  $$3x = 12 \tag{8.3.48}$$ 
Make the coefficient of the variable term equal to 1. Divide each side by 3.  $$\dfrac{3x}{\textcolor{red}{3}} = \dfrac{12}{\textcolor{red}{3}} \tag{8.3.49}$$ 
Simplify.  $$x = 4 \tag{8.3.50}$$ 
Check: Let x = 4.  $$\begin{split} 3(x + 2) &= 18 \\ 3(\textcolor{red}{4} + 2 &\stackrel{?}{=} 18 \\ 3(6) &\stackrel{?}{=} 18 \\ 18 &\stackrel{?}{=} 18\; \checkmark \end{split}$$ 
Exercise \(\PageIndex{21}\):
Solve: 5(x + 3) = 35.
 Answer

x = 4
Exercise \(\PageIndex{22}\):
Solve: 6(y − 4) = −18.
 Answer

y = 1
Example \(\PageIndex{12}\):
Solve: −(x + 5) = 7.
Solution
Simplify each side of the equation as much as possible by distributing. The only x term is on the left side, so all variable terms are on the left side of the equation.  $$x  5 = 7 \tag{8.3.51}$$ 
Add 5 to both sides to get all constant terms on the right side of the equation.  $$x  5 \textcolor{red}{+5} = 7 \textcolor{red}{+5} \tag{8.3.52}$$ 
Simplify.  $$x = 12 \tag{8.3.53}$$ 
Make the coefficient of the variable term equal to 1 by multiplying both sides by 1.  $$\textcolor{red}{1} (x) = \textcolor{red}{1} (12) \tag{8.3.54}$$ 
Simplify.  $$x = 12 \tag{8.3.55}$$ 
Check: Let x = −12.  $$\begin{split} (x + 5) &= 7 \\ (\textcolor{red}{12} + 5) &\stackrel{?}{=} 7 \\ (7) &\stackrel{?}{=} 7 \\ 7 &= 7\; \checkmark \end{split}$$ 
Exercise \(\PageIndex{23}\):
Solve: −(y + 8) = −2.
 Answer

y = 6
Exercise \(\PageIndex{24}\):
Solve: −(z + 4) = −12.
 Answer

z = 8
Example \(\PageIndex{13}\):
Solve: 4(x − 2) + 5 = −3.
Solution
Simplify each side of the equation as much as possible. Distribute.  $$4x  8 + 5 = 3 \tag{8.3.56}$$ 
Combine like terms.  $$4x  3 = 3 \tag{8.3.57}$$ 
The only x is on the left side, so all variable terms are on one side of the equation.  
Add 3 to both sides to get all constant terms on the other side of the equation.  $$4x  3 \textcolor{red}{+3} = 3 \textcolor{red}{+3} \tag{8.3.58}$$ 
Simplify.  $$4x = 0 \tag{8.3.59}$$ 
Make the coefficient of the variable term equal to 1 by dividing both sides by 4.  $$\dfrac{4x}{\textcolor{red}{4}} = \dfrac{0}{\textcolor{red}{4}} \tag{8.3.60}$$ 
Simplify.  $$x = 0 \tag{8.3.61}$$ 
Check: Let x = 0.  $$\begin{split} 4(x  2) + 5 &= 3 \\ 4(\textcolor{red}{0}  2) + 5 &\stackrel{?}{=} 3 \\ 4(2) + 5 &\stackrel{?}{=} 3 \\ 8 + 5 &\stackrel{?}{=} 3 \\ 3 &= 3\; \checkmark \end{split}$$ 
Exercise \(\PageIndex{25}\):
Solve: 2(a − 4) + 3 = −1.
 Answer

a = 2
Exercise \(\PageIndex{26}\):
Solve: 7(n − 3) − 8 = −15.
 Answer

n = 2
Example \(\PageIndex{14}\):
Solve: 8 − 2(3y + 5) = 0.
Solution
Be careful when distributing the negative.
Simplify—use the Distributive Property.  $$8  6y  10 = 0 \tag{8.3.62}$$ 
Combine like terms.  $$6y  2 = 0 \tag{8.3.63}$$ 
Add 2 to both sides to collect constants on the right.  $$6y  2 \textcolor{red}{+2} = 0 \textcolor{red}{+2} \tag{8.3.64}$$ 
Simplify.  $$y =  \dfrac{1}{3} \tag{8.3.65}$$ 
Divide both sides by −6.  $$\dfrac{6y}{\textcolor{red}{6}} = \dfrac{2}{\textcolor{red}{6}} \tag{8.3.66}$$ 
Simplify.  $$y =  \dfrac{1}{3} \tag{8.3.67}$$ 
Check: Let y = \(− \dfrac{1}{3}\).  \[\begin{split} 8  2(3y + 5) &= 0 \\ 8  2 \Bigg[ 3 \left(\textcolor{red}{ \dfrac{1}{3}}\right) + 5 \Bigg] &= 0 \\ 8  2(1 + 5) &\stackrel{?}{=} 0 \\ 8  2(4) &\stackrel{?}{=} 0 \\ 8  8 &\stackrel{?}{=} 0 \\ 0 &= 0; \checkmark \end{split}$$ 
Exercise \(\PageIndex{27}\):
Solve: 12 − 3(4j + 3) = −17.
 Answer

\(j = \frac{5}{3}\)
Exercise \(\PageIndex{28}\):
Solve: −6 − 8(k − 2) = −10.
 Answer

\(k = \frac{5}{2}\)
Example \(\PageIndex{15}\):
Solve: 3(x − 2) − 5 = 4(2x + 1) + 5.
Solution
Distribute.  $$3x  6  5 = 8x + 4 + 5 \tag{8.3.68}$$ 
Combine like terms.  $$3x  11 = 8x + 9 \tag{8.3.69}$$ 
Subtract 3x to get all the variables on the right since 8 > 3.  $$3x \textcolor{red}{3x}  11 = 8x \textcolor{red}{3x} + 9 \tag{8.3.70}$$ 
Simplify.  $$11 = 5x + 9 \tag{8.3.71}$$ 
Subtract 9 to get the constants on the left.  $$11 \textcolor{red}{9} = 5x + 9 \textcolor{red}{9} \tag{8.3.72}$$ 
Simplify.  $$20 = 5x \tag{8.3.73}$$ 
Divide by 5.  $$\dfrac{20}{\textcolor{red}{5}} = \dfrac{5x}{\textcolor{red}{5}} \tag{8.3.74}$$ 
Simplify.  $$4 = x \tag{8.3.75}$$ 
Check: Substitute: −4 = x.  \[\begin{split} 3(x  2)  5 &= 4(2x + 1) + 5 \\ 3(\textcolor{red}{4}  2)  5 &\stackrel{?}{=} 4[2(\textcolor{red}{4}) + 1] + 5 \\ 3(6)  5 &\stackrel{?}{=} 4(8 + 1) + 5 \\ 18  5 &\stackrel{?}{=} 4(7) + 5 \\ 23 &\stackrel{?}{=} 28 + 5 \\ 23 &= 23\; \checkmark \end{split}$$ 
Exercise \(\PageIndex{29}\):
Solve: 6(p − 3) − 7 = 5(4p + 3) − 12.
 Answer

p = 2
Exercise \(\PageIndex{30}\):
Solve: 8(q + 1) − 5 = 3(2q − 4) − 1.
 Answer

q = 8
Example \(\PageIndex{16}\):
Solve: \(\dfrac{1}{2}\)(6x − 2) = 5 − x.
Solution
Distribute.  $$3x  1 = 5  x \tag{8.3.76}$$ 
Add x to get all the variables on the left.  $$3x  1 \textcolor{red}{+x} = 5  x \textcolor{red}{+x} \tag{8.3.77}$$ 
Simplify.  $$4x  1 = 5 \tag{8.3.78}$$ 
Add 1 to get constants on the right.  $$4x  1 \textcolor{red}{+1} = 5 \textcolor{red}{+1} \tag{8.3.79}$$ 
Simplify.  $$4x = 6 \tag{8.3.80}$$ 
Divide by 4.  $$\dfrac{4x}{\textcolor{red}{4}} = \dfrac{6}{\textcolor{red}{4}} \tag{8.3.81}$$ 
Simplify.  $$x = \dfrac{3}{2} \tag{8.3.82}$$ 
Check: Let x = \(\dfrac{3}{2}\).  $$\begin{split} \dfrac{1}{2} (6x  2) &= 5  x \\ \dfrac{1}{2} \left(6 \cdot \textcolor{red}{\dfrac{3}{2}}  2 \right) &\stackrel{?}{=} 5  \textcolor{red}{\dfrac{3}{2}} \\ \dfrac{1}{2} (9  2) &\stackrel{?}{=} \dfrac{10}{2}  \dfrac{3}{2} \\ \dfrac{1}{2} (7) &\stackrel{?}{=} \dfrac{7}{2} \\ \dfrac{7}{2} &= \dfrac{7}{2}\; \checkmark \end{split}$$ 
Exercise \(\PageIndex{31}\):
Solve: \(\dfrac{1}{3}\)(6u + 3) = 7 − u.
 Answer

u = 2
Exercise \(\PageIndex{32}\):
Solve: \(\dfrac{2}{3}\)(9x − 12) = 8 + 2x.
 Answer

x = 4
In many applications, we will have to solve equations with decimals. The same general strategy will work for these equations.
Example \(\PageIndex{17}\):
Solve: 0.24(100x + 5) = 0.4(30x + 15).
Solution
Distribute.  $$24x + 1.2 = 12x + 6 \tag{8.3.83}$$ 
Subtract 12x to get all the x s to the left.  $$24x + 1.2 \textcolor{red}{12x} = 12x + 6 \textcolor{red}{12x} \tag{8.3.84}$$ 
Simplify.  $$12x + 1.2 = 6 \tag{8.3.85}$$ 
Subtract 1.2 to get the constants to the right.  $$12x + 1.2 \textcolor{red}{1.2} = 6 \textcolor{red}{1.2} \tag{8.3.86}$$ 
Simplify.  $$12x = 4.8 \tag{8.3.87}$$ 
Divide.  $$\dfrac{12x}{\textcolor{red}{12}} = \dfrac{4.8}{\textcolor{red}{12}} \tag{8.3.88}$$ 
Simplify.  $$x = 0.4 \tag{8.3.89}$$ 
Check: Let x = 0.4.  \[\begin{split} 0.24(100x + 5) &= 0.4(30x + 15) \\ 0.24[100(\textcolor{red}{0.4}) + 5] &\stackrel{?}{=} 0.4[30(\textcolor{red}{0.4}) + 15] \\ 0.24(40 + 5) &\stackrel{?}{=} 0.4(12 + 15) \\ 0.24(45) &\stackrel{?}{=} 0.4(27) \\ 10.8 &= 10.8\; \checkmark \end{split}$$ 
Exercise \(\PageIndex{33}\):
Solve: 0.55(100n + 8) = 0.6(85n + 14).
 Answer

n = 1
Exercise \(\PageIndex{34}\):
Solve: 0.15(40m − 120) = 0.5(60m + 12).
n = 1
Practice Makes Perfect
Solve an Equation with Constants on Both Sides
In the following exercises, solve the equation for the variable.
 6x − 2 = 40
 7x − 8 = 34
 11w + 6 = 93
 14y + 7 = 91
 3a + 8 = −46
 4m + 9 = −23
 −50 = 7n − 1
 −47 = 6b + 1
 25 = −9y + 7
 29 = −8x − 3
 −12p − 3 = 15
 −14q − 15 = 13
Solve an Equation with Variables on Both Sides
In the following exercises, solve the equation for the variable.
 8z = 7z − 7
 9k = 8k − 11
 4x + 36 = 10x
 6x + 27 = 9x
 c = −3c − 20
 b = −4b − 15
 5q = 44 − 6q
 7z = 39 − 6z
 3y + \(\dfrac{1}{2}\) = 2y
 8x + \(\dfrac{3}{4}\) = 7x
 −12a − 8 = −16a
 −15r − 8 = −11r
Solve an Equation with Variables and Constants on Both Sides
In the following exercises, solve the equations for the variable.
 6x − 15 = 5x + 3
 4x − 17 = 3x + 2
 26 + 8d = 9d + 11
 21 + 6 f = 7 f + 14
 3p − 1 = 5p − 33
 8q − 5 = 5q − 20
 4a + 5 = − a − 40
 9c + 7 = −2c − 37
 8y − 30 = −2y + 30
 12x − 17 = −3x + 13
 2z − 4 = 23 − z
 3y − 4 = 12 − y
 \(\dfrac{5}{4}\)c − 3 = \(\dfrac{1}{4}\)c − 16
 \(\dfrac{4}{3}\)m − 7 = \(\dfrac{1}{3}\)m − 13
 8 − \(\dfrac{2}{5}\)q = \(\dfrac{3}{5}\)q + 6
 11 − \(\dfrac{1}{4}\)a = \(\dfrac{3}{4}\)a + 4
 \(\dfrac{4}{3}\)n + 9 = \(\dfrac{1}{3}\)n − 9
 \(\dfrac{5}{4}\)a + 15 = \(\dfrac{3}{4}\)a − 5
 \(\dfrac{1}{4}\)y + 7 = \(\dfrac{3}{4}\)y − 3
 \(\dfrac{3}{5}\)p + 2 = \(\dfrac{4}{5}\)p − 1
 14n + 8.25 = 9n + 19.60
 13z + 6.45 = 8z + 23.75
 2.4w − 100 = 0.8w + 28
 2.7w − 80 = 1.2w + 10
 5.6r + 13.1 = 3.5r + 57.2
 6.6x − 18.9 = 3.4x + 54.7
Solve an Equation Using the General Strategy
In the following exercises, solve the linear equation using the general strategy.
 5(x + 3) = 75
 4(y + 7) = 64
 8 = 4(x − 3)
 9 = 3(x − 3)
 20(y − 8) = −60
 14(y − 6) = −42
 −4(2n + 1) = 16
 −7(3n + 4) = 14
 3(10 + 5r) = 0
 8(3 + 3p) = 0
 \(\dfrac{2}{3}\)(9c − 3) = 22
 \(\dfrac{3}{5}\)(10x − 5) = 27
 5(1.2u − 4.8) = −12
 4(2.5v − 0.6) = 7.6
 0.2(30n + 50) = 28
 0.5(16m + 34) = −15
 −(w − 6) = 24
 −(t − 8) = 17
 9(3a + 5) + 9 = 54
 8(6b − 7) + 23 = 63
 10 + 3(z + 4) = 19
 13 + 2(m − 4) = 17
 7 + 5(4 − q) = 12
 −9 + 6(5 − k) = 12
 15 − (3r + 8) = 28
 18 − (9r + 7) = −16
 11 − 4(y − 8) = 43
 18 − 2(y − 3) = 32
 9(p − 1) = 6(2p − 1)
 3(4n − 1) − 2 = 8n + 3
 9(2m − 3) − 8 = 4m + 7
 5(x − 4) − 4x = 14
 8(x − 4) − 7x = 14
 5 + 6(3s − 5) = −3 + 2(8s − 1)
 −12 + 8(x − 5) = −4 + 3(5x − 2)
 4(x − 1) − 8 = 6(3x − 2) − 7
 7(2x − 5) = 8(4x − 1) − 9
Everyday Math
 Making a fence Jovani has a fence around the rectangular garden in his backyard. The perimeter of the fence is 150 feet. The length is 15 feet more than the width. Find the width, w, by solving the equation 150 = 2(w + 15) + 2w.
 Concert tickets At a school concert, the total value of tickets sold was $1,506. Student tickets sold for $6 and adult tickets sold for $9. The number of adult tickets sold was 5 less than 3 times the number of student tickets. Find the number of student tickets sold, s, by solving the equation 6s + 9(3s − 5) = 1506.
 Coins Rhonda has $1.90 in nickels and dimes. The number of dimes is one less than twice the number of nickels. Find the number of nickels, n, by solving the equation 0.05n + 0.10(2n − 1) = 1.90.
 Fencing Micah has 74 feet of fencing to make a rectangular dog pen in his yard. He wants the length to be 25 feet more than the width. Find the length, L, by solving the equation 2L + 2(L − 25) = 74.
Writing Exercises
203. When solving an equation with variables on both sides, why is it usually better to choose the side with the larger coefficient as the variable side? 204. Solve the equation 10x + 14 = −2x + 38, explaining all the steps of your solution. 205. What is the first step you take when solving the equation 3 − 7(y − 4) = 38? Explain why this is your first step. 206. Solve the equation 1 4 (8x + 20) = 3x − 4 explaining all the steps of your solution as in the examples in this section. 207. Using your own words, list the steps in the General Strategy for Solving Linear Equations. 208. Explain why you should simplify both sides of an equation as much as possible before collecting the variable terms to one side and the constant terms to the other side.
Self Check
(a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
(b) What does this checklist tell you about your mastery of this section? What steps will you take to improve?
Contributors and Attributions
Lynn Marecek (Santa Ana College) and MaryAnne AnthonySmith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at http://cnx.org/contents/fd53eae1fa2...49835c3c@5.191."