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## College Algebra

### Unit 1: Lesson 1

Solving equations with one unknown

# Multi-step equations review

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.
\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:
\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.
\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}
a, equals, start color #11accd, minus, 2, end color #11accd
Check our work:
\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}

### 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.
\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}
e, equals, start color #7854ab, 3, end color #7854ab
Check our work:
\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}

## Practice

Problem 1
Solve for b.
4, b, plus, 5, equals, 1, plus, 5, b
b, equals

Want more practice? Check out these exercises:

## Want to join the conversation?

• how did you get 5/3b?
• So as we can see the question starts off as:
2/3b + 5 = 20-b
first, add +b to both sides of the equation
We got 5/3 by adding fractions:
2/3+1/1.
Step 1: Multiply the denominators (x/3)
Step 2: Cross multiply the numerators and denominators (2x1 and 3x1)
Step 3: Add the two products together (2x1=2, 3x1=3 therefore, add 2+3). WITHOUT touching the denominator!
Step 4: 5/3b + 5 = 20. Subtract 5 from both sides of the equation to cancel out 5.
Step 5. divide 5/3 to 15. Keep change Flip
Keep the fraction change the division sign to multiplication and flip the second fraction (example 2/3 to 3/2). So, 5/3 to 3/5 and multiply both sides of the equation, lastly, your answer is 4.
• What do I do if the variable is equal to 0? How do I check my answer?
• To check your answer, just plug 0 in for the variable.

For example, let's say you solved this equation:
9x - 1 = 5x - 1
9x - 5x = 1 - 1
4x = 0
x = 0
It looks like the variable is equal to 0.

Now, double check your answer with the original equation by replacing all the x's with 0's:
9x - 1 = 5x - 1
9(0) - 1 = 5(0) - 1
0 - 1 = 0 - 1
-1 = -1
The equation is true, so 0 is valid.

Hope this helps!
• These comments be getting answers years later. Like bro they already progressed into High School or College, they most likely know after then.
• This type of math is a little confusing because sometimes when u have a -3x you divide or subtract at least i think so i don't know why but my teacher said that and can some one explain it please?
• You divide. you can think of -3x as -3 times x, so using the opposite of multiplication(division), you can get rid of it.
• how do you solve an equation like 0.5(5-7x) = 8-(4x+6)
• 1) Use distributive property to remove the parentheses. Distribute the 0.5 and the "-"
2) On the right side, you will have 2 terms that are like terms. Combine them.
3) Move all the "x" terms to the same side of the equation
4) Move the constants to the opposite side from the x's
5) Then divide by the coefficient of "x" (the number in front of x)

See if you can solve the equation. Comment back if you get stuck with what you have done so far.
• If I have one cookie and my friend eats my cookie is he really my friend?
• It depends on a lot of circumstances; have you already eaten the rest of the bag and are offering him the last one?
If you really wanted the cookie, did you give it away, or did they just take it?
Why is there only one cookie?
I do not feel we have enough information to answer this question.
• Why is peter so silly