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Algebra 1

Course: Algebra 1 > Unit 2

Lesson 2: Linear equations with parentheses
Math
Algebra 1
Solving equations & inequalities
Linear equations with parentheses
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Multi-step equations review

Google Classroom
To solve an equation we find the value of the variable that makes the equation true. For more complicated, fancier equations, this process can take several steps.
When solving an equation, our goal is to find the value of the variable that makes the equation true.

Example 1: Two-step equation

Solve for x.
3, x, plus, 7, equals, 13
We need to manipulate the equation to get x by itself.
3x+7=133x+7−7=13−73x=63x3=63x=2\begin{aligned} 3x+7&=13 \\\\ 3x+7\redD{-7}&=13\redD{-7} \\\\ 3x&=6 \\\\ \dfrac{3x}{\redD{3}}&=\dfrac{6}{\redD{3}} \\\\ x&=2 \end{aligned}
We call this a two-step equation because it took two steps to solve. The first step was to subtract 7 from both sides, and the second step was to divide both sides by 3. Want an explanation of why we do the same thing to both sides of the equation? Check out this video.
We check the solution by plugging start color #e84d39, 2, end color #e84d39 back into the original equation:
3x+7=133⋅2+7=?136+7=?1313=13       Yes!\begin{aligned} 3x+7&=13 \\\\ 3\cdot \redD 2 + 7 &\stackrel?= 13 \\\\ 6+7 &\stackrel?= 13 \\\\ 13 &= 13 ~~~~~~~\text{Yes!} \end{aligned}

Example 2: Variables on both sides

Solve for a.
5, plus, 14, a, equals, 9, a, minus, 5
We need to manipulate the equation to get a by itself.
5+14a=9a−55+14a−9a=9a−5−9a5+5a=−55+5a−5=−5−55a=−105a5=−105a=−2\begin{aligned} 5 + 14a &= 9a - 5 \\\\ 5 + 14a \blueD{- 9a} &= 9a - 5 \blueD{- 9a} \\\\ 5 + 5a &= -5 \\\\ 5 + 5a \blueD{-5} &= -5 \blueD{- 5}\\\\ 5a &= -10\\\\ \dfrac{5a}{\blueD5} &= \dfrac{-10}{\blueD5} \\\\ a &= \blueD{-2} \end{aligned}
The answer:
a, equals, start color #11accd, minus, 2, end color #11accd
Check our work:
5+14a=9a−55+14(−2)=?9(−2)−55+(−28)=?−18−5−23=−23       Yes!\begin{aligned} 5 + 14a &= 9a - 5 \\\\ 5 + 14(\blueD{-2}) &\stackrel?= 9(\blueD{-2}) - 5 \\\\ 5 + (-28) &\stackrel?= -18 - 5 \\\\ -23 &= -23 ~~~~~~~\text{Yes!} \end{aligned}
Want to learn more about solving equations with variables on both sides? Check out this video.

Example 3: Distributive property

Solve for e.
7, left parenthesis, 2, e, minus, 1, right parenthesis, minus, 11, equals, 6, plus, 6, e
We need to manipulate the equation to get e by itself.
7(2e−1)−11=6+6e14e−7−11=6+6e14e−18=6+6e14e−18−6e=6+6e−6e8e−18=68e−18+18=6+188e=248e8=248e=3\begin{aligned} 7(2e-1)-11 &= 6+6e \\\\ 14e-7 -11&= 6+6e\\\\ 14e-18 &= 6+6e\\\\ 14e-18\purpleD{-6e} &= 6+6e\purpleD{-6e} \\\\ 8e-18&=6\\\\ 8e-18\purpleD{+18} &=6 \purpleD{+18} \\\\ 8e &=24\\\\ \dfrac{8e}{\purpleD{8}}&= \dfrac{24}{\purpleD{8}}\\\\ e &= \purpleD{3} \end{aligned}
The answer:
e, equals, start color #7854ab, 3, end color #7854ab
Check our work:
7(2e−1)−11=6+6e7(2(3)−1)−11=?6+6(3)7(6−1)−11=?6+187(5)−11=?2435−11=?2424=24       Yes!\begin{aligned} 7(2e-1)-11 &= 6+6e \\\\ 7(2(\purpleD{3})-1) -11&\stackrel?= 6+6(\purpleD{3}) \\\\ 7(6-1)-11 &\stackrel?= 6+18 \\\\ 7(5)-11&\stackrel?=24 \\\\ 35-11&\stackrel?=24 \\\\ 24 &=24 ~~~~~~~\text{Yes!} \end{aligned}
Want to learn more about solving equations with the distributive property? Check out this video.

Practice

Problem 1
  • Current
Solve for b.
4, b, plus, 5, equals, 1, plus, 5, b
b, equals
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

Want more practice? Check out these exercises:

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