2.3: Equations with the Distributive Property
Morgan, Connor, and Jake 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 days food ration costs $7. If the class has only $500 for the survival kits, how many days can they go on their trip?
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Khan Academy Solving Equations with the Distributive Property
Guidance
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.
Let’s see what this looks like in math terms. Say you had \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*}).
When solving equations that require you using 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.
Example A
\begin{align*}2(3x+5)=-2\end{align*}
You can solve this problem using the balance method.
Your first step is to remove the brackets. To do this multiply the 2 by the numbers inside the brackets. 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*}
Example B
\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*}
Example C
\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*}
Concept Problem Revisited
Morgan, Connor, and Jake 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 days food ration costs $7. If the class has only $500 for the survival kits, how many days can they go on their trip?
First let’s write down what we know:
- Number of students going on the trip = 15 (Morgan, Connor, Jake + 12 others)
- Total money available = $500
- Each flashlight costs $10
- Each first aid kit costs $9
- Each days 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.
Vocabulary
- Distributive Property
- The distributive property is a mathematical way of grouping terms. It 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. For example, in the expression: \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*}.
- Variable
- A variable is an unknown quantity in a mathematical expression. It is represented by a letter. It is often referred to as the literal coefficient.
Guided Practice
1. Use the balance method to solve for the variable in the problem \begin{align*}6(-3x+2)=-6\end{align*}.
2. Use the algebra tiles to solve for the variable in the problem \begin{align*}2(r-2)=-6\end{align*}.
3. Determine the most efficient method to solve for the variable in the problem \begin{align*}2(j+1)+3=3(j-1)\end{align*}. Explain your choice of method for solving this problem.
Answers:
1. \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*}.
\begin{align*}\text{Check:} &\\ 6(-3x+2) & =-6\\ 6(-3({\color{red}1})+2) &= -6\\ 6(-3+2) &= -6\\ 6(-1) &= -6\\ -6 &= -6 \ \ \end{align*}
2. \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*}.
\begin{align*}\text{Check:} &\\ 2(r-2) &= -6\\ 2(({\color{red}-1})-2) &= -6\\ 2(-3) &= -6\\ -6 &= -6 \ \ \end{align*}
3. \begin{align*}2(j+1)+3=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 brackets, you must start with the distributive property and remove the brackets.
Therefore \begin{align*}j = 5\end{align*}.
\begin{align*}\text{Check:} &\\ 2(j+1) &= 3(j-1)\\ 2(({\color{red}5})+1) &= 3(({\color{red}5})-1)\\ 2(6) &= 3(4)\\ 12 &= 12 \ \ \end{align*}
Practice
Use the balance method to find the solution for the variable in each of the following problems.
- \begin{align*}5(4x+3)=75\end{align*}
- \begin{align*}3(s-4)=15\end{align*}
- \begin{align*}5(k-4)=10\end{align*}
- \begin{align*}43=4(t+6)-1\end{align*}
- \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 problems.
- \begin{align*}2(d-3)=4\end{align*}
- \begin{align*}5+2(x+7)=20\end{align*}
- \begin{align*}2(3x-4)=22\end{align*}
- \begin{align*}2(3x+2)-x=-6\end{align*}
- \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 problems. Remember to choose an efficient method to solve for the variable.
- \begin{align*}-6=-6(3x-8)\end{align*}
- \begin{align*}-2(x-2)=11\end{align*}
- \begin{align*}2+3(-2x+1)=20\end{align*}
- \begin{align*}3(x+2)-x=12\end{align*}
- \begin{align*}5(2-3x)=-8-6x\end{align*}
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Here you will learn to solve equations that require the distributive property.
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