Let's Practice Solutions to Equations

1. Which value of ‍$$g$$ makes ‍$$26 = 7(g-9)+12$$ a true statement?

   Choose 1 answer:
   
   1. $$g = 11$$
   2. $$g = 12$$
   3. $$g = 13$$

   4. $$g = 14$$

2. Select all the equations where ‍$$z=5$$ is a solution.

   Choose 3 answers:

   1. $$8=z+3$$
   2. $$z-1=6$$
   3. $$20=40 \div z$$
   4. $$12z=60$$

   5. $$12+z=17$$

3. Which value of ‍$$v$$ makes ‍$$12=12+\dfrac{v-8}{2}$$ a true statement?

   Choose 1 answer:
   
   1. $$v = 8$$
   2. $$v = 10$$
   3. $$v = 12$$

   4. $$v= 14$$

4. Select all the equations where ‍$$a=8$$ is a solution.

   Choose 3 answers:

   1. $$a+11=15$$
   2. $$1=8 \div a$$
   3. $$15+a=23$$
   4. $$142=7a$$

   5. $$a-5=3$$

5. Which value of ‍$$w$$ makes ‍$$14=11+\dfrac{w}{8}\cdot6$$ a true statement?

   Choose 1 answer:
   
   1. $$w = 1$$
   2. $$w = 4$$
   3. $$w = 16$$

   4. $$w = 24$$

6. Select all the equations where ‍$$y=10$$ is a solution.

   Choose 2 answers:

   1. $$y+11=22$$
   2. $$8=18-y$$
   3. $$60=6y$$
   4. $$\frac{y}{5} =4$$

   5. $$10y=20$$

7. Which value of ‍$$x$$ makes ‍$$7+5(x-3)=22$$ a true statement?

   Choose 1 answer:
   
   1. $$x = 4$$
   2. $$x = 5$$
   3. $$x = 6$$

   4. $$x= 7$$

Source: Khan Academy, https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-equations-and-inequalities/cc-6th-intro-equations/a/introduction-to-equations
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License.

1. Let's substitute each ‍g-value into the equation to see if it is a solution.

Let's substitute ‍$$g=11$$ and see if the equation is true.

$$
\begin{aligned}
26&=7({g}-9)+12 \\\\
26&\stackrel{?}=7({11}-9)+12 \\\\
26&\stackrel{?}=7(2)+12 \\\\
26&\stackrel{?}=14+12 \\\\
26&\stackrel{\checkmark}=26
\end{aligned}$$

Yes, ‍$$g=11$$ does make a true statement.

Let's substitute ‍$$g=12$$ and see if the equation is true.

$$ \begin{aligned}
26&=7({g}-9)+12 \\\\
26&\stackrel{?}=7({12}-9)+12 \\\\
26&\stackrel{?}=7(3)+12 \\\\
26&\stackrel{?}=21+12 \\\\
26&\neq 33
\end{aligned}$$

No, ‍ $$g=12$$ does not make a true statement.

Let's substitute ‍$$g=13$$ and see if the equation is true.

$$\begin{aligned}
26&=7({g}-9)+12 \\\\
26&\stackrel{?}=7({13}-9)+12 \\\\
26&\stackrel{?}=7(4)+12 \\\\
26&\stackrel{?}=28+12 \\\\
26&\neq 40
\end{aligned}$$

No, ‍$$g=13$$ does not make a true statement.

Let's substitute ‍$$g=14$$ and see if the equation is true.

$$\begin{aligned}
26&=7({g}-9)+12 \\\\
26&\stackrel{?}=7({14}-9)+12 \\\\
26&\stackrel{?}=7(5)+12 \\\\
26&\stackrel{?}=35+12 \\\\
26&\neq 47
\end{aligned}$$

No, ‍$$g=14$$ does not make a true statement.

The ‍g-value that makes ‍$$26=7(g-9)+12$$ a true statement is ‍$$g=11$$.

2.

• $$8=z+3$$
• $$12z=60$$
• $$12+z=17$$

For each equation, ‍$$z=5$$ is a solution if it makes the equation true.

To test the each equation, we can follow these steps:

• Substitute ‍$$z=5$$ into the equation.
• Simplify.
• Check if both sides of the equation have the same value.

For example, here's how we could test the first equation:

$$\begin{aligned}
8&={z}+3 \\\\
8&\stackrel{?}={5}+3 \\\\
8&\stackrel{\checkmark}=8
\end{aligned}$$

Yes, $$z=5$$ is a solution.

$$z=5$$ is a solution for the following equations:

• $$8=z+3$$
• $$12z=60$$
• $$12+z=17$$

3. Let's substitute each ‍v-value into the equation to see if it is a solution.

Let's substitute ‍$$v=8$$ and see if the equation is true.

$$ \begin{aligned}
12&=12+\dfrac{{v}-8}{2} \\\\
12&\stackrel{?}=12+\dfrac{{8}-8}{2} \\\\
12&\stackrel{?}=12+\dfrac{0}{2} \\\\
12&\stackrel{?}=12+0 \\\\
12&\stackrel{\checkmark}{=}12
\end{aligned}$$

Yes, ‍$$v=8$$ does make a true statement.

Let's substitute ‍$$v=10$$ and see if the equation is true.

$$\begin{aligned}
12&=12+\dfrac{{v}-8}{2} \\\\
12&\stackrel{?}=12+\dfrac{{10}-8}{2} \\\\
12&\stackrel{?}=12+\dfrac{2}{2} \\\\
12&\stackrel{?}=12+1\\\\
12&\neq 13
\end{aligned}$$

No, ‍ $$v=10$$ does not make a true statement.

Let's substitute ‍$$v=12$$ and see if the equation is true.

$$\begin{aligned}
12&=12+\dfrac{{v}-8}{2} \\\\
12&\stackrel{?}=12+\dfrac{{12}-8}{2} \\\\
12&\stackrel{?}=12+\dfrac{4}{2} \\\\
12&\stackrel{?}=12+2\\\\
12&\neq 14
\end{aligned}$$

No, ‍$$v=12$$ does not make a true statement.

Let's substitute ‍$$v=14$$ and see if the equation is true.

$$\begin{aligned}
12&=12+\dfrac{{v}-8}{2} \\\\
12&\stackrel{?}=12+\dfrac{{14}-8}{2} \\\\
12&\stackrel{?}=12+\dfrac{6}{2} \\\\
12&\stackrel{?}=12+3\\\\
12&\neq 15
\end{aligned}$$

No, ‍$$v=14$$does not make a true statement.

The ‍v-value that makes ‍$$12=12+\dfrac{v-8}{2}$$ a true statement is ‍$$v=8$$.

4.

• $$1=8 \div a$$
• $$15+a=23$$
• $$a-5=3$$

For each equation, ‍$$a=8$$ is a solution if it makes the equation true.

To test the each equation, we can follow these steps:

• Substitute ‍$$a=8$$ into the equation.
• Simplify.
• Check if both sides of the equation have the same value.

For example, here's how we could test the first equation:

$$\begin{aligned}  {a}+11 &= 15 \\\\
{8}+11 &\stackrel{?}= 15 \\\\
19&\neq 15\end{aligned}$$

No, $$a=8$$ is not a solution

$$a=8$$ is a solution for the following equations:

• $$1=8 \div a$$
• $$15+a=23$$
• $$a-5=3$$

5. Let's substitute each ‍w-value into the equation to see if it is a solution.

Let's substitute ‍$$w=1$$ and see if the equation is true.

$$\begin{aligned}
14&=11+\dfrac{{w}}{8}\cdot 6 \\\\
14&\stackrel{?}=11+\dfrac{{1}}{8}\cdot 6 \\\\
14&\stackrel{?}=11+\dfrac{6}{8} \\\\
14&\stackrel{?}=11+\dfrac{3}{4} \\\\
14&\neq11\dfrac{3}{4}
\end{aligned}$$

No, ‍$$w=1$$ does not make a true statement.

Let's substitute ‍$$w=4$$ and see if the equation is true.

$$\begin{aligned}
14&=11+\dfrac{{w}}{8}\cdot 6 \\\\
14&\stackrel{?}=11+\dfrac{{4}}{8}\cdot 6 \\\\
14&\stackrel{?}=11+\dfrac{1}{2}\cdot 6 \\\\
14&\stackrel{?}=11+3 \\\\
14&\stackrel\checkmark =14
\end{aligned}$$

Yes, ‍$$w=4$$ does make a true statement.

Let's substitute ‍$$w=16$$ and see if the equation is true.

$$\begin{aligned}
14&=11+\dfrac{{w}}{8}\cdot 6 \\\\
14&\stackrel{?}=11+\dfrac{{16}}{8}\cdot 6 \\\\
14&\stackrel{?}=11+2\cdot 6 \\\\
14&\stackrel{?}=11+12 \\\\
14&\neq 23
\end{aligned}$$

No, ‍$$w=16$$ does not make a true statement.

Let's substitute ‍$$w=24$$ and see if the equation is true.

$$\begin{aligned}
14&=11+\dfrac{{w}}{8}\cdot 6 \\\\
14&\stackrel{?}=11+\dfrac{{24}}{8}\cdot 6 \\\\
14&\stackrel{?}=11+3\cdot 6 \\\\
14&\stackrel{?}=11+18 \\\\
14&\neq 29
\end{aligned}$$

No, ‍$$w=24$$ does not make a true statement.

The ‍w-value that makes ‍$$14=11+\dfrac{w}{8}\cdot6$$ a true statement is ‍$$w=4$$.

6.

• $$8=18-y$$
• $$60=6y$$

For each equation, ‍$$y=10$$ is a solution if it makes the equation true.

To test the each equation, we can follow these steps:

• Substitute ‍$$y=10$$ into the equation.
• Simplify.
• Check if both sides of the equation have the same value.

For example, here's how we could test the first equation:

$$\begin{aligned}
{y}+11&=22 \\\\
{10}+11&\stackrel{?}=22 \\\\
21&\neq 22
\end{aligned}$$

No, $$y=10$$ is not a solution

$$y=10$$ is a solution for the following equations:

• $$8=18-y$$
• $$60=6y$$

7. Let's substitute each ‍x-value into the equation to see if it is a solution.

Let's substitute ‍$$x=4$$ and see if the equation is true.

$$\begin{aligned}  7+5(x-3)&=22 \\\\
7+5(4-3) &\stackrel{\large?}{=}  22\\\\
7+5(1) &\stackrel{\large?}{=}  22\\\\
7+5 &\stackrel{\large?}{=}  22\\\\
12 &\neq 22\end{aligned}$$

No, ‍$$x=4$$ does not make a true statement.

Let's substitute ‍$$x=5$$ and see if the equation is true.

$$\begin{aligned}  7+5(x-3)&=22 \\\\
7+5(5-3) &\stackrel{\large?}{=}  22\\\\
7+5(2) &\stackrel{\large?}{=}  22\\\\
7+10 &\stackrel{\large?}{=}  22\\\\
17 &\neq 22\end{aligned}$$

No, ‍$$x=5$$ does not make a true statement.

Let's substitute ‍$$x=6$$ and see if the equation is true.

$$\begin{aligned}  7+5(x-3)&=22 \\\\
7+5(6-3) &\stackrel{\large?}{=}  22\\\\
7+5(3) &\stackrel{\large?}{=}  22\\\\
7+15 &\stackrel{\large?}{=}  22\\\\
22 &\stackrel{\checkmark}{=}  22\end{aligned}$$

Yes, ‍$$x=6$$ does make a true statement.

Let's substitute ‍$$x=7$$ and see if the equation is true.

$$\begin{aligned}  7+5(x-3)&=22 \\\\
7+5(7-3) &\stackrel{\large?}{=}  22\\\\
7+5(4) &\stackrel{\large?}{=}  22\\\\
7+20 &\stackrel{\large?}{=}  22\\\\
27 &\neq 22\end{aligned}$$

No, ‍$$x=7$$ does not make a true statement.

The ‍x-value that makes ‍$$7+5(x-3)=22$$ a true statement is ‍$$x=6$$.