Do you like candy? Take a look at this yummy dilemma.

Eight children were given some candy. Then six different children were given the same unknown amount of candy. Next, two children were that same unknown amount of candy plus three additional pieces of candy. The total number of pieces of candy given out was thirty -eight.

If this is the case, what is the unknown amount of candy?

Do you know how to solve this problem? Write an equation and then solve it for the unknown amount of candy. Pay attention and you will see this dilemma at the end of the Concept.

### Guidance

To solve some multi-step equations you will need to use the distributive property and combine like terms. When this happens, you will see that there is more than one term with the same variable or there is more than one number in the equation. You always want to combine everything that you can before moving on to solving the equation.

Let's apply this to the following situation.

*Solve for \begin{align*}m\end{align*}: \begin{align*}6(1 +2m) -3m = 24\end{align*}*

Apply the distributive property to the left side of the equation. Multiply each of the two numbers inside the parentheses by 6 and then add those products.

\begin{align*}6(1 + 2m) - 3m &= 24\\ (6 \times 1) + (6 \times 2m) - 3m &=24\\ 6 + 12m - 3m &= 24\end{align*}

Next, subtract the like terms—\begin{align*}12m\end{align*} and \begin{align*}3m\end{align*}—on the left side of the equation.

\begin{align*}6 + 12m - 3m &= 24\\ 6 + (12m - 3m) &= 24\\ 6 + 9m &= 24\end{align*}

Finally, solve as you would solve any two-step equation. Subtract 6 from both sides of the equation.

\begin{align*}6 + 9m &= 24\\ 6-6 + 9m &= 24-6\\ 0 + 9m &= 18\\ 9m &= 18\end{align*}

Now, divide both sides of the equation by 9.

\begin{align*}9m &= 18\\ \frac{9m}{9} &= \frac{18}{9}\\ 1m &= 2\\ m &= 2\\\end{align*}

**The value of \begin{align*}m\end{align*} is 2.**

Here is another one.

*Solve for \begin{align*}b\end{align*}: \begin{align*}-4(2 + 3b) + 5b = 13\end{align*}*

Apply the distributive property to the left side of the equation. Multiply each of the two numbers inside the parentheses by -4 and then add those products.

\begin{align*}-4(2 + 3b) + 5b &= 13\\ (-4 \times 2) + (-4 \times 3b) + 5b &= 13\\ -8 + (-12b) + 5b &= 13\end{align*}

Next, add the like terms on the left side of the equation. To add those like terms, \begin{align*}-12b\end{align*} and \begin{align*}5b\end{align*}, you will need to use what you know about adding integers. To review how to add and subtract integers, look back at Lesson 2.5.

\begin{align*}-8 + (-12b) + 5b &= 13\\ -8 + (-12b + 5b) &= 13\\ -8 + (-7b) &= 13\end{align*}

Finally, solve as you would solve any two-step equation. Since -8 is added to \begin{align*}(-7b)\end{align*}, you can subtract -8 from both sides of the equation to solve it.

\begin{align*}-8 + (-7b) &= 13\\ -8 - (-8) + (-7b) &= 13 - (-8)\\ (-8 + 8) + (-7b) &= 13 + 8\\ 0 + (-7b) &= 21\\ -7b &= 21\end{align*}

Now, divide both sides of the equation by -7.

\begin{align*}-7b &= 21\\ \frac{-7b}{-7} &= \frac{21}{-7}\\ 1b &= -3\\ b &= -3\end{align*}

**The value of \begin{align*}b\end{align*} is -3.**

#### Example A

\begin{align*}6(x+4)+3x - 2 = 54\end{align*}

**Solution: \begin{align*}x = 4\end{align*}**

#### Example B

\begin{align*}6y + 3(y-4) = 33\end{align*}

**Solution: \begin{align*}y = 5\end{align*}**

#### Example C

\begin{align*}5(a+3)+6(a+1)+8a=40\end{align*}

**Solution: \begin{align*}a = 1\end{align*}**

Now let's go back to the dilemma from the beginning of the Concept.

Eight children were given some candy. Then six different children were given the same unknown amount of candy. Next, two children were that same unknown amount of candy plus three additional pieces of candy. The total number of pieces of candy given out was thirty -eight.

First, write an equation. Break down the problem a piece at a time. We will use \begin{align*}c\end{align*} for the unknown amount of candy given.

\begin{align*}8c + 6c + 2(c + 3) = 38\end{align*}

This is our equation.

Now solve the equation by first getting rid of the parentheses.

\begin{align*}8c + 6c + 2c + 6 = 38\end{align*}

Next, combine like terms.

\begin{align*}16c + 6 = 38\end{align*}

Now subtract six from both sides of the equation.

\begin{align*}16c + 6 - 6 = 38 - 6\end{align*}

\begin{align*}16c = 32\end{align*}

\begin{align*}c = 2\end{align*}

**The unknown amount of candy was two pieces.**

### Vocabulary

- Like Terms
- terms that include a common variable.

- Commutative Property of Addition
- states that the order that you add different numbers does not change the sum.

- Associative Property of Addition
- states that you can change the groupings of numbers being added without changing the sum.

- Distributive Property
- states that you can multiply a term outside of a set of parentheses with the terms inside the parentheses to simplify the set of parentheses.

### Guided Practice

Here is one for you to try on your own.

Solve for \begin{align*}x\end{align*}.

\begin{align*}-5x+3(x+1)-4x=45\end{align*}

**Solution**

First, distribute the three and get rid of the parentheses.

\begin{align*}-5x+3x+3-4x=45\end{align*}

Now combine like terms.

\begin{align*}-2x+3-4x=45\end{align*}

Combine again because there are many terms to combine in this problem.

\begin{align*}-6x+3=45\end{align*}

Next subtract three from both sides of the equation.

\begin{align*}-6x+3-3=45-3\end{align*}

\begin{align*}-6x=42\end{align*}

\begin{align*}x = -7\end{align*}

**This is our answer.**

### Video Review

Khan Academy Solving Linear Equations 2

### Practice

Directions: Distribute and combine like terms and then solve each equation.

- \begin{align*}x+8(x+2)=52\end{align*}
- \begin{align*}2y+6(y+3)=34\end{align*}
- \begin{align*}4y+2(y-2)=8\end{align*}
- \begin{align*}9y+3(y-6)=30\end{align*}
- \begin{align*}6(x+2)-4x=30\end{align*}
- \begin{align*}3(y-1)+2(y+3)=13\end{align*}
- \begin{align*}4(a+3)-2(a+6)=20\end{align*}
- \begin{align*}6(x+2)-4x+6=36\end{align*}
- \begin{align*}-9(x+3)+4x=-2\end{align*}
- \begin{align*}-4(y+3)-2y=24\end{align*}
- \begin{align*}4(a+2)-9=11\end{align*}
- \begin{align*}-8(y+2)-16=16\end{align*}
- \begin{align*}5(a+4)-6a+1=12\end{align*}
- \begin{align*}x+3x+2x+3(x+1)=30\end{align*}
- \begin{align*}2x+4x+6x-2(x+3)=34\end{align*}