In this lesson, we will look at solving word problems of the form ax+b=c, using both algebra tiles and pencil and paper.

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Now let’s look at a problem involving fractions.

#### Solving Two-Step Equations in the form $ax+b=c$.

Jia received an $\$80$ gift card for a candy shop. She used it to buy some jellybeans that cost $\$9.50$ per pound and a candy dispenser for $\$20$. After buying all of this, she had $\$31.50$ left on her card. How many pounds of jellybeans did she buy?

Before we can start solving this problem we have to write an equation.

So lets start by looking at the key information.

First we need to define our variable, so lets use $p$ to represent the number of pounds of jellybeans she bought. That would make the amount Jia spent on the candy was $ \$9.50p$. She also bought a candy dispenser for $ \$20$, so we would need to add that to the equation, leaving us with:

$ \$9.50p+\$20$ = the amount of money she spent.

We know that Jia started with $\$80$ and had $\$31.50$ left on the gift card, so the amount she spent would be the different between those two amounts.

$ \$9.50p+\$20=\$80-\$31.50$.

Next we should collect like terms by subtracting the numbers on the right hand side of the equation.

$ \$9.50p+\$20=\$48.50$.

Remember we want to get the variable by itself, so we will need to subtract $ \$20$ from each side of the equation (what ever we do on one side of the scale we have to do on the other side to keep the equation balanced)

\[ \begin{align*}

\$9.50p+\$20 \color{blue}{-\$20} &=\$48.50 \color{blue}{-\$20} \\

\$9.50p&=\$28.50 \end{align*}\]

To find the value of $p$ , we need to divide each side by $ \$9.50$.

\[ \begin{align*}

\frac{\$9.50p}{\color{blue}{\$9.50}}&=\frac{\$28.50}{\color{blue}{\$9.50}} \\

p&=3 \end{align*}\]

So Jia bought $3$ pounds of jellybeans.