# Unit 3 Family Materials

Writing and Solving Equations

### Representing Situations of the Form $px+q=r$ and $p(x+q)=r$

In this unit, your student will be representing situations with diagrams and equations. There are two main categories of situations with associated diagrams and equations.

Here is an example of the first type: A standard deck of playing cards has four suits. In each suit, there are 3 face cards and $$x$$ other cards. There are 52 total cards in the deck. A diagram we might use to represent this situation is:

and its associated equation could be $$52=4(3+x)$$. There are 4 groups of cards, each group contains $$x+3$$ cards, and there are 52 cards in all.

Here is an example of the second type: A chef makes 52 pints of spaghetti sauce. She reserves 3 pints to take home to her family, and divides the remaining sauce equally into 4 containers. A diagram we might use to represent this situation is:

and its associated equation could be $$52=4x+3$$. From the 52 pints of sauce, 3 were set aside, and each of 4 containers holds $$x$$ pints of sauce.

1. Draw a diagram to represent the equation $$3x+6=39$$
2. Draw a diagram to represent the equation $$39=3(y+6)$$
3. Decide which story goes with which equation-diagram pair:
• Three friends went cherry picking and each picked the same amount of cherries, in pounds. Before they left the cherry farm, someone gave them an additional 6 pounds of cherries. Altogether, they had 39 pounds of cherries.
• One of the friends made three cherry tarts. She put the same number of cherries in each tart, and then added 6 more cherries to each tart. Altogether, the three tarts contained 39 cherries.

Solution:

Diagram A represents $$3x+6=39$$ and the story about cherry picking. Diagram B represents $$3(y+6)=39$$ and the story about making cherry tarts.

### Solving Equations of the Form $px+q=r$ and $p(x+q)=r$ and Problems That Lead to Those Equations

Your student is studying efficient methods to solve equations and working to understand why these methods work. Sometimes to solve an equation, we can just think of a number that would make the equation true. For example, the solution to $$12-c=10$$ is 2, because we know that $$12-2=10$$. For more complicated equations that may include decimals, fractions, and negative numbers, the solution may not be so obvious.

An important method for solving equations is doing the same thing to each side. For example, let's show how we might solve $$\text-4(x-1)=20$$ by doing the same thing to each side.

\begin{align} \text-4(x-1) &= 24 \\ \text-\tfrac14 \boldcdot \text-4(x-1) &= \text-\tfrac14 \boldcdot 24 & \text{multiply each side by }\text-\tfrac14 \\ x-1 &= \text-6 \\ x-1+1 &= \text- 6 + 1 & \text{ add 1 to each side} \\ x &= \text-5 \\ \end{align}

Another helpful tool for solving equations is to apply the distributive property. In the example above, instead of multiplying each side by $$\text-\frac14$$, you could apply the distributive property to $$\text-4(x-1)$$ and replace it with $$\text-4x+4$$. Your solution would look like this:

\begin{align} \text-4(x-1) &= 24 \\ \text-4x+4 &= 24 & \text{apply the distributive property} \\ \text-4x+4-4 &= 24-4 &\text{subtract 4 from each side} \\ \text-4x &= 20 \\ \text-4x \div \text-4 &= 20\div\text-4 & \text{divide each side by }\text-4 \\ x &= \text-5 \\ \end{align}

Elena picks a number, adds 45 to it, and then multiplies by $$\frac12$$. The result is 29. Elena says that you can find her number by solving the equation $$29=\frac12(x+45)$$.
\begin{align} 29 &= \frac12(x+45) \\ 2 \boldcdot 29 &= 2 \boldcdot \frac12 (x+45) & \text{multiply each side by }2 \\ 58 &= x+45 \\ 58-45 &= x+45-45 & \text{subtract 45 from each side} \\ 13 &=x \\ \end{align}