Two Step Equations 3

Years

Year 8

Equations

Two Step Equations

Two Step Equations 3

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Question 1 of 5

1. Question
Solve

$-2y-13=-11$
- $y=$ (-1)
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Inverse Operations

When moving a term to the other side of an equation, the operation is inversed.

To solve for $y$ , it needs to be alone on one side.

Start by moving $13$ to the other side by adding $13$ to both sides of the equation.

$-2$ $y$ $-13$ $=$ $-11$

$-2$ $y$ $-13$ $+13$ $=$ $-11$ $+13$

$-2$ $y$ $=$ $2$ $-13+13$ cancels out

Next, remove $-2$ by dividing both sides of the equation by $-2$ .

$-2$ $y$ $=$ $2$

$-2$ $y$ $-:-2$ $=$ $2$ $-:-2$

$y$ $=$ $-1$ $-2-:-2$ cancels out

Check our work

To confirm our answer, substitute $y=-1$ to the original equation.

$-2y-13$ $=$ $-11$

$-2(-1)-13$ $=$ $-11$ Substitute $y=-1$

$2-13$ $=$ $-11$

$-11$ $=$ $-11$

Since the equation is true, the answer is correct.

$y=-1$
Question 2 of 5

2. Question
Solve for $r$

$-4r-1=31$
- $r=$ (-8)
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To solve for $r$ , get $r$ by itself

Add $1$ to both sides of the equation

$-4r-1$ $+1$ $=$ $31$ $+1$

$-4r$ $-1$ $+1$ $=$ $32$ $-1+1$ cancels out

$-4r$ $=$ $32$

Finally, divide both sides of the equation by $-4$

$-4r$ $divide -4$ $=$ $32$ $divide -4$

$-4$ $r$ $divide -4$ $=$ $-8$ $-4 divide -4$ cancels out

$r$ $=$ $-8$

$r=-8$
Question 3 of 5

3. Question
Solve

$x/3 +4=7$
- $x=$ (9)
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Inverse Operations

When moving a term to the other side of an equation, the operation is inversed.

To solve for $x$ , it needs to be alone on one side.

Start by moving $4$ to the other side by subtracting $4$ from both sides of the equation.

$\frac{\color{#00880A}{x}}{3}+4$ $=$ $7$

$\frac{\color{#00880A}{x}}{3}+4\color{#CC0000}{-4}$ $=$ $7$ $-4$

$\frac{\color{#00880A}{x}}{3}$ $=$ $3$ $4-4$ cancels out

Next, remove $1/3$ by multiplying both sides of the equation by $3$ .

$\frac{\color{#00880A}{x}}{3}$ $=$ $3$

$\frac{\color{#00880A}{x}}{3}\color{#CC0000}{\times3}$ $=$ $3$ $xx3$

$x$ $=$ $9$ $1/3 times3$ cancels out

Check our work

To confirm our answer, substitute $x=9$ to the original equation.

$x/3 +4$ $=$ $7$

$9/3 +4$ $=$ $7$ Substitute $x=9$

$3+4$ $=$ $7$

$7$ $=$ $7$

Since the equation is true, the answer is correct.

$x=9$
Question 4 of 5

4. Question
Solve

$x/7 +3=-2$
- $x=$ (-35)
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Inverse Operations

When moving a term to the other side of an equation, the operation is inversed.

To solve for $x$ , it needs to be alone on one side.

Start by moving $3$ to the other side by subtracting $3$ from both sides of the equation.

$\frac{\color{#00880A}{x}}{7}+3$ $=$ $-2$

$\frac{\color{#00880A}{x}}{7}+3\color{#CC0000}{-3}$ $=$ $-2$ $-3$

$\frac{\color{#00880A}{x}}{7}$ $=$ $-5$ $3-3$ cancels out

Next, remove $1/7$ by multiplying both sides of the equation by $7$ .

$\frac{\color{#00880A}{x}}{7}$ $=$ $-5$

$\frac{\color{#00880A}{x}}{7}\color{#CC0000}{\times7}$ $=$ $-5$ $xx7$

$x$ $=$ $-35$ $1/7 times7$ cancels out

Check our work

To confirm our answer, substitute $x=-35$ to the original equation.

$x/7 +3$ $=$ $-2$

$(-35)/7 +3$ $=$ $-2$ Substitute $x=-35$

$-5+3$ $=$ $-2$

$-2$ $=$ $-2$

Since the equation is true, the answer is correct.

$x=-35$
Question 5 of 5

5. Question
Solve

$t/5 -3=-6$
- $t=$ (-15)
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Inverse Operations

When moving a term to the other side of an equation, the operation is inversed.

To solve for $t$ , it needs to be alone on one side.

Start by moving $3$ to the other side by adding $3$ to both sides of the equation.

$\frac{\color{#00880A}{t}}{5}-3$ $=$ $-6$

$\frac{\color{#00880A}{t}}{5}-3\color{#CC0000}{+3}$ $=$ $-6$ $+3$

$\frac{\color{#00880A}{t}}{5}$ $=$ $-3$ $-3+3$ cancels out

Next, remove $1/5$ by multiplying both sides of the equation by $5$ .

$\frac{\color{#00880A}{t}}{5}$ $=$ $-3$

$\frac{\color{#00880A}{t}}{5}\color{#CC0000}{\times5}$ $=$ $-3$ $xx5$

$t$ $=$ $-15$ $1/5 times5$ cancels out

Check our work

To confirm our answer, substitute $t=-15$ to the original equation.

$t/5 -3$ $=$ $-6$

$(-15)/5 -3$ $=$ $-6$ Substitute $t=-15$

$-3-3$ $=$ $-6$

$-6$ $=$ $-6$

Since the equation is true, the answer is correct.

$t=-15$

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Quiz summary

Information

1. Question

Hint

Fantastic!

Inverse Operations

2. Question

Well Done!

3. Question

Hint

Good Job!

Inverse Operations

4. Question

Hint

Great Work!

Inverse Operations

5. Question

Hint

Excellent!

Inverse Operations

Quizzes

$-2$ $y$ $-13$	$=$	$-11$
$-2$ $y$ $-13$ $+13$	$=$	$-11$ $+13$
$-2$ $y$	$=$	$2$	$-13+13$ cancels out

$-2$ $y$	$=$	$2$
$-2$ $y$ $-:-2$	$=$	$2$ $-:-2$
$y$	$=$	$-1$	$-2-:-2$ cancels out

$-2y-13$	$=$	$-11$
$-2(-1)-13$	$=$	$-11$	Substitute $y=-1$
$2-13$	$=$	$-11$
$-11$	$=$	$-11$

$-4r-1$ $+1$	$=$	$31$ $+1$
$-4r$ $-1$ $+1$	$=$	$32$	$-1+1$ cancels out
$-4r$	$=$	$32$

$-4r$ $divide -4$	$=$	$32$ $divide -4$
$-4$ $r$ $divide -4$	$=$	$-8$	$-4 divide -4$ cancels out
$r$	$=$	$-8$

$\frac{\color{#00880A}{x}}{3}+4$	$=$	$7$

$\frac{\color{#00880A}{x}}{3}+4\color{#CC0000}{-4}$	$=$	$7$ $-4$

$\frac{\color{#00880A}{x}}{3}$	$=$	$3$	$4-4$ cancels out

$\frac{\color{#00880A}{x}}{3}$	$=$	$3$

$\frac{\color{#00880A}{x}}{3}\color{#CC0000}{\times3}$	$=$	$3$ $xx3$

$x$	$=$	$9$	$1/3 times3$ cancels out

$x/3 +4$	$=$	$7$

$9/3 +4$	$=$	$7$	Substitute $x=9$

$3+4$	$=$	$7$
$7$	$=$	$7$

$\frac{\color{#00880A}{x}}{7}+3$	$=$	$-2$

$\frac{\color{#00880A}{x}}{7}+3\color{#CC0000}{-3}$	$=$	$-2$ $-3$

$\frac{\color{#00880A}{x}}{7}$	$=$	$-5$	$3-3$ cancels out

$\frac{\color{#00880A}{x}}{7}$	$=$	$-5$

$\frac{\color{#00880A}{x}}{7}\color{#CC0000}{\times7}$	$=$	$-5$ $xx7$

$x$	$=$	$-35$	$1/7 times7$ cancels out

$\frac{\color{#00880A}{t}}{5}-3$	$=$	$-6$

$\frac{\color{#00880A}{t}}{5}-3\color{#CC0000}{+3}$	$=$	$-6$ $+3$

$\frac{\color{#00880A}{t}}{5}$	$=$	$-3$	$-3+3$ cancels out

$\frac{\color{#00880A}{t}}{5}$	$=$	$-3$

$\frac{\color{#00880A}{t}}{5}\color{#CC0000}{\times5}$	$=$	$-3$ $xx5$

$t$	$=$	$-15$	$1/5 times5$ cancels out

$t/5 -3$	$=$	$-6$

$(-15)/5 -3$	$=$	$-6$	Substitute $t=-15$

$-3-3$	$=$	$-6$
$-6$	$=$	$-6$

$x/7 +3$	$=$	$-2$

$(-35)/7 +3$	$=$	$-2$	Substitute $x=-35$

$-5+3$	$=$	$-2$
$-2$	$=$	$-2$