Let's Practice Solutions to Equations

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 ‍\(x=6\).