Reviewing Equations

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

Choose 2 answers:

A. $y+11=22$

B. $8=18−y$

C. $60=6y$

D. $\frac{y}{5}=4$

E. $10y=20$

2. Which value of $f$ makes $3f+5=32$ a true statement?

Choose 1 answer:

A. $f = 6$

B. $f = 8$

C. $f = 9$

D. $f = 12$

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

Choose 3 answers:

A. $13=9+m$

B. $7−m=3$

C. $15=11m$

D. $5=m+2$

E. $20÷m=5$

4. Which value of $h$ makes $\frac{8+h}{10}=1$ a true statement?

Choose 1 answer:

A. $h = 1$

B. $h = 2$

C. $h = 3$

D. $h = 4$

5. Select all the equations where $b = 11$ is a solution.

Choose 2 answers:

A. $2b=211$

B. $b+18=7$

C. $77=7b$

D. $9=b−2$

E. $11=33÷b$

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

Choose 1 answer:

A. $g = 11$

B. $g = 12$

C. $g = 13$

D. $g = 14$

7. Select all the equations where $c = 9$ is a solution.

Choose 2 answers:

A. $4-c=5$

B. $20=14+c$

C. $15=c-6$

D. $\frac{c}{3}= 3$

E. $36=4c$

1.

B. $8=18−y$

C. $60=6y$

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

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

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

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

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

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

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

• $8=18−y$
• $60=6y$

2.

C. $f = 9$

Let's substitute each $f$ value into the equation to see if it is a solution.

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

$\begin{array}{r} 3(f)+5=32 \\ 3(6)+5 \stackrel{?}{=} 32 \\ 18+5 \stackrel{?}{=} 32 \\ 23 \neq 32 \end{array}$

No, $f = 6$ does not make a true statement.

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

$\begin{array}{r} 3(f)+5=32 \\ 3(8)+5 \stackrel{?}{=} 32 \\ 24+5 \stackrel{?}{=} 32 \\ 29 \neq 32 \end{array}$

No, $f = 8$ does not make a true statement.

Let's substitute $f = 9$ and see if the equation is true.

$\begin{array}{r} 3(f)+5=32 \\ 3(9)+5 \stackrel{?}{=} 32 \\ 27+5 \stackrel{?}{=} 32 \\ 32 \stackrel{\checkmark}{=} 32 \end{array}$

Yes, $f = 9$ does make a true statement.

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

$\begin{array}{r} 3(f)+5=32 \\ 3(12)+5 \stackrel{?}{=} 32 \\ 36+5 \stackrel{?}{=} 32 \\ 41 \neq 32 \end{array}$

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

The $f$ value that makes $3f+5=32$ a true statement is $f = 9$.

3.

A. $13=9+m$

B. $7−m=3$

E. $20÷m=5$

For each equation, $m = 4$ is a solution if it makes the equation true.

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

1. Substitute $m = 4$ into the equation
2. Simplify.
3. Check if both sides of the equation have the same value.

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

$\begin{aligned} &13=9+m \\ &13 \stackrel{?}{=} 9+4 \\ &13 \stackrel{\checkmark}{=} 13 \end{aligned}$

Yes, $m = 4$ is a solution.

$m = 4$ is a solution for the following equations:

• $13=9+m$
• $7−m=3$
• $20÷m=5$

4.

B. $h = 2$

Let's substitute each $h$ value into the equation to see if it is a solution.

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

$\begin{aligned} \frac{8+h}{10} &=1 \\ \frac{8+1}{10} & \stackrel{?}{=} 1 \\ \frac{9}{10} & \stackrel{?}{=} 1 \\ \frac{9}{10} & \neq 1 \end{aligned}$

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

Let's substitute $h = 2$ and see if the equation is true.

$\begin{array}{r} \frac{8+h}{10}=1 \\ \frac{8+2}{10} \stackrel{?}{=} 1 \\ \frac{10}{10} \stackrel{?}{=} 1 \\ 1 \stackrel{\checkmark}{=} 1 \end{array}$

Yes, $h = 2$ does make a true statement.

Let's substitute $h = 3$ and see if the equation is true.

$\begin{array}{r} \frac{8+h}{10}=1 \\ \frac{8+3}{10} \stackrel{?}{=} 1 \\ \frac{11}{10} \stackrel{?}{=} 1 \\ \frac{11}{10} \neq 1 \end{array}$

No, $h = 3$ does not make a true statement.

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

$\begin{array}{r} \frac{8+h}{10}=1 \\ \frac{8+4}{10} \stackrel{?}{=} 1 \\ \frac{12}{10} \stackrel{?}{=} 1 \\ \frac{6}{5} \neq 1 \end{array}$

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

The $h$-value that makes $\frac{8+h}{10}=1$ a true statement is $h =2$.

5.

C. $77=7b$

D. $9=b−2$

For each equation, $b = 11$ is a solution if it makes the equation true.

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

1. Substitute $b = 11$ into the equation.
2. Simplify.
3. Check if both sides of the equation have the same value.

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

$\begin{array}{r} 2 b=211 \\ 2 \times 11 \stackrel{?}{=} 211 \\ 22 \neq 211 \end{array}$

No, $b = 11$ is not a solution.

$b = 11$ is a solution for the following equations:

• $77=7b$
• $9=b−2$

6.

A. $g = 11$

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 $g = 11$.

7.

D. $\frac{c}{3}= 3$

E. $36=4c$

For each equation, $c = 9$ is a solution if it makes the equation true.

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

1. Substitute $c = 9$ into the equation.
2. Simplify.
3. Check if both sides of the equation have the same value.

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

$\begin{gathered} 4-c=5 \\ 4-9 \stackrel{?}{=} 5 \\ -5 \neq 5 \end{gathered}$

No, $c = 9$ is not a solution.

$c = 9$ is a solution for the following equations:

• $\frac {c}{3} = 3$
• $36=4c$