Distributive Property for Multi-Step Equations

a(x + b) = c

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Equations with the Distributive Property

Morgan, Connor, and Ryan are going on a class field trip with 12 of their class mates. Each student is required to have a survival kit with a flashlight, a first aid kit, and enough food rations for the trip. A flashlight costs $10. A first aid kit costs$9. Each day's food ration costs $7. If the class has only$1000 for the survival kits, how many days can they go on their trip?

Distributive Property

The distributive property is a mathematical way of grouping terms. The distributive property states that the product of a number and a sum is equal to the sum of the individual products of the number and the addends.

Consider \begin{align*}{\color{red}3}({\color{blue}x + 5})\end{align*}. The distributive property states that the product of a number (\begin{align*}{\color{red}3}\end{align*}) and a sum \begin{align*}({\color{blue}x + 5})\end{align*} is equal to the sum of the individual products of the number \begin{align*}({\color{red}3})\end{align*} and the addends (\begin{align*}{\color{blue}x}\end{align*} and \begin{align*}{\color{blue}5}\end{align*}). So, \begin{align*}3(x+5)=3x+15\end{align*}.

When solving equations that require you to use the distributive property, you simply have one more step to follow. You still have the same goal of trying to get the variables on one side and the constants on the other side. When there are parentheses in the equation, your first step in solving the equation will be to use the distributive property to remove them.

Let's apply the distributive property to solve the following equations:

1. \begin{align*}2(3x+5)=-2\end{align*}

You can solve this problem using the balance method.

Your first step is to remove the parentheses. To do this multiply the 2 by the numbers inside the parentheses. Therefore, multiply 2 by \begin{align*}3x\end{align*} and 2 by 5.

Next, isolate the \begin{align*}x\end{align*} variable by subtracting 10 from both sides.

Simplifying you get:

Now divide by 6 to solve for the \begin{align*}x\end{align*} variable.

To finally solve for the variable, simplify each side.

Therefore \begin{align*}x = -2\end{align*}.

\begin{align*}\text{Check:}&\\ 2(3x+5) &= -2\\ 2(3({\color{red}-2})+5) &= -2\\ 2(-6+5) &= -2\\ 2(-1) &= -2\\ -2 &= -2 \ \ \end{align*}

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

You can solve this problem using algebra tiles.

Since all of our variables (\begin{align*}x\end{align*} tiles) are on the same side, you only need to subtract the 3 groups of 4 from both sides of the equal sign to isolate the variable.

Simplifying the right side of the equation and rearranging the tiles leave the following:

Therefore \begin{align*}y = -2\end{align*}.

\begin{align*}\text{Check:} &\\ 3(4+3y) &= -6\\ 3(4+3({\color{red}-2})) &= -6\\ 3(4-6) &= -6\\ 3(-2) &= -6\\ -6 &= -6 \ \ \end{align*}

1. \begin{align*}6(4x+1)=-18\end{align*}

Therefore \begin{align*}x = -1\end{align*}.

\begin{align*}\text{Check:} &\\ 6(4x+1) &= -18\\ 6(4({\color{red}-1})+1) &= -18\\ 6(-4+1) &= -18\\ 6(-3) &= -18\\ -18 &= -18 \ \ \end{align*}

Examples

Example 1

Earlier, you were given a problem about Morgan, Connor, and Ryan going on a class field trip with 12 of their class mates. Each student is required to have a survival kit with a flashlight, a first aid kit, and enough food rations for the trip. A flashlight costs $10. A first aid kit costs$9. Each day's food ration costs $7. If the class has only$1000 for the survival kits, how many days can they go on their trip?

Write down what you know:

• Number of students going on the trip = 15 (Morgan, Connor, Ryan + 12 others)
• Total money available = $500 • Each flashlight costs$10
• Each first aid kit costs $9 • Each day's food ration costs$7

One survival kit contains 1 flashlight + 1 first aid kit + \begin{align*}x\end{align*} day’s rations. Therefore, the cost of each survival kit is: \begin{align*}\10 + \9 + \7x = \19 + \7x\end{align*}.

For the 15 students, the total cost of the survival kits would be:

\begin{align*}15 (\19+\7x)=\285+\105x\end{align*}

Since the class has \$1000 to spend, you can calculate how many days they can go on their trip.

\begin{align*}\1000 &= \285+\105x\\ \1000 {\color{red}- \285} &= \285 {\color{red}-\285}+\105x\\ \715 &= \105x\\ \frac{\715}{\105} &= \frac{\105x}{\105}\\ 6.81 &= x\end{align*}

Since the class does not have enough money to buy 7 days of food rations for each student, they will buy six days of food rations and the class will go on a class trip for six days.

Example 2

Use the balance method to solve for the variable in the equation .\begin{align*}6(-3x+2)=-6\end{align*}

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

You can begin by using the distributive property on the left side of the equation.

Next subtract 12 from both sides of the equation to get the variable term alone.

Simplifying, you get:

You next have to isolate the variable. To do this, divide both sides of the equation by –18.

Finally, simplifying leaves you with:

Therefore \begin{align*}x = 1\end{align*}.

Example 3

Use algebra tiles to solve for the variable in the equation \begin{align*}2(r-2)=-6\end{align*}.

\begin{align*}2(r-2)=-6\end{align*}

First, you need to add 4 to both sides to get the variables alone on the one side of the equal sign.

Simplifying leaves you with:

Therefore \begin{align*}r = -1\end{align*}.

Example 4

Determine the most efficient method to solve for the variable in the equation \begin{align*}2(j+1)=3(j-1)\end{align*}. Explain your choice of method for solving this problem.

\begin{align*}2(j+1)=3(j-1)\end{align*}.

You could choose either method for this problem. Below you will see the balance method used to solve the problem. Work through the steps to see if you can follow them. Remember since there are parentheses, you must start with the distributive property and remove the parentheses.

Therefore \begin{align*}j = 5\end{align*}.

Review

Use the balance method to find the solution for the variable in each of the following equations.

1. \begin{align*}5(4x+3)=75\end{align*}
2. \begin{align*}3(s-4)=15\end{align*}
3. \begin{align*}5(k-4)=10\end{align*}
4. \begin{align*}43=4(t+6)-1\end{align*}
5. \begin{align*}6(x+4)=3(5x+2)\end{align*}

Use algebra tiles to find the solution for the variable in each of the following equations.

1. \begin{align*}2(d-3)=4\end{align*}
2. \begin{align*}5+2(x+7)=20\end{align*}
3. \begin{align*}2(3x-4)=22\end{align*}
4. \begin{align*}2(3x+2)-x=-6\end{align*}
5. \begin{align*}2(x+4)-x=9\end{align*}

Use the methods that you have learned for solving equations with variables to solve for the variables in each of the following equations. Remember to choose an efficient method to solve for the variable.

1. \begin{align*}-6=-6(3x-8)\end{align*}
2. \begin{align*}-2(x-2)=11\end{align*}
3. \begin{align*}2+3(-2x+1)=20\end{align*}
4. \begin{align*}3(x+2)-x=12\end{align*}
5. \begin{align*}5(2-3x)=-8-6x\end{align*}

To see the Review answers, open this PDF file and look for section 2.3.

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Color Highlighted Text Notes

Vocabulary Language: English

TermDefinition
distributive property The distributive property states that the product of an expression and a sum is equal to the sum of the products of the expression and each term in the sum. For example, $a(b + c) = ab + ac$.
factor Factors are the numbers being multiplied to equal a product. To factor means to rewrite a mathematical expression as a product of factors.
Variable A variable is a symbol used to represent an unknown or changing quantity. The most common variables are a, b, x, y, m, and n.

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