Two Step Equations 4

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Year 8

Equations

Two Step Equations

Two Step Equations 4

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Question 1 of 5

Solve

\frac{y}{- 5} + 6 = - 4

$y/(-5) +6=-4$

$y =$ $y=$ (50)

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Inverse Operations

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

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

Start by moving

6

$6$ to the other side by subtracting

6

$6$ from both sides of the equation.

$\frac{\color{#00880A}{y}}{-5}+6$	$=$ $=$	$- 4$ $-4$

$\frac{\color{#00880A}{y}}{-5}+6\color{#CC0000}{-6}$	$=$ $=$	$- 4$ $-4$ $- 6$ $-6$

$\frac{\color{#00880A}{y}}{-5}$	$=$ $=$	$- 10$ $-10$	$6 - 6$ $6-6$ cancels out

Next, remove

\frac{1}{- 5}

$1/(-5)$ by multiplying both sides of the equation by

- 5

$-5$ .

$\frac{\color{#00880A}{y}}{-5}$	$=$ $=$	$- 10$ $-10$

$\frac{\color{#00880A}{y}}{-5}\color{#CC0000}{\times-5}$	$=$ $=$	$- 10$ $-10$ $\times - 5$ $xx-5$

$y$ $y$	$=$ $=$	$50$ $50$	$\frac{1}{- 5} \times - 5$ $1/(-5) times-5$ cancels out

Check our work

To confirm our answer, substitute

y = 50

$y=50$ to the original equation.

$\frac{y}{- 5} + 6$ $y/(-5) +6$	$=$ $=$	$- 4$ $-4$

$\frac{50}{- 5} + 6$ $50/(-5) +6$	$=$ $=$	$- 4$ $-4$	Substitute $y = 50$ $y=50$

$- 10 + 6$ $-10+6$	$=$ $=$	$- 4$ $-4$
$- 4$ $-4$	$=$ $=$	$- 4$ $-4$

Since the equation is true, the answer is correct.

y = 50

$y=50$

Question 2 of 5

2. Question
Solve for $x$ $x$

$\frac{x}{6} + 2 = 9$ $x/6+2=9$
- $x =$ $x=$ (42)
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To solve for $x$ $x$ , get $x$ $x$ by itself

Subtract $2$ $2$ from both sides of the equation

$\frac{x}{6} + 2$ $frac{x}{6}+2$ $- 2$ $-2$ $=$ $=$ $9$ $9$ $- 2$ $-2$

$\frac{x}{6}$ $frac{x}{6}$ $+ 2$ $+2$ $- 2$ $-2$ $=$ $=$ $9$ $9$ $- 2$ $-2$ $2 - 2$ $2-2$ cancels out

$\frac{x}{6}$ $frac{x}{6}$ $=$ $=$ $7$ $7$

Finally, multiply both sides of the equation by $6$ $6$

$\frac{x}{6}$ $frac{x}{6}$ $\times 6$ $times6$ $=$ $=$ $7$ $7$ $\times 6$ $times6$

$\frac{6 x}{6}$ $frac{6x}{6}$ $=$ $=$ $7$ $7$ $\times 6$ $times6$

$x$ $x$ $=$ $=$ $42$ $42$ The coefficient $\frac{6}{6}$ $6/6$ cancels out

$x = 42$ $x=42$
Question 3 of 5

3. Question
Solve for $x$ $x$

$\frac{x}{7} - 2 = 3$ $x/7-2=3$
- $x =$ $x=$ (35)
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To solve for $x$ $x$ , get $x$ $x$ by itself

Add $2$ $2$ to both sides of the equation

$\frac{x}{7} - 2$ $frac{x}{7}-2$ $+ 2$ $+2$ $=$ $=$ $3$ $3$ $+ 2$ $+2$

$\frac{x}{7}$ $frac{x}{7}$ $- 2$ $-2$ $+ 2$ $+2$ $=$ $=$ $3$ $3$ $+ 2$ $+2$ $- 2 + 2$ $-2+2$ cancels out

$\frac{x}{7}$ $frac{x}{7}$ $=$ $=$ $5$ $5$

Finally, multiply both sides of the equation by $7$ $7$

$\frac{x}{7}$ $frac{x}{7}$ $\times 7$ $times7$ $=$ $=$ $5$ $5$ $\times 7$ $times7$ Multiply both sides by $7$ $7$

$\frac{7 x}{7}$ $frac{7x}{7}$ $=$ $=$ $5$ $5$ $\times 7$ $times7$

$x$ $x$ $=$ $=$ $35$ $35$ The coefficient $\frac{7}{7}$ $7/7$ cancels out

$x = 35$ $x=35$
Question 4 of 5

4. Question
Solve for $a$ $a$

$\frac{a}{10} + 4 = 12$ $a/10+4=12$
- $a =$ $a=$ (80)
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To solve for $a$ $a$ , get $a$ $a$ by itself

Subtract $4$ $4$ from both sides of the equation

$\frac{a}{10} + 4$ $frac{a}{10}+4$ $- 4$ $-4$ $=$ $=$ $12$ $12$ $- 4$ $-4$

$\frac{a}{10}$ $frac{a}{10}$ $+ 4$ $+4$ $- 4$ $-4$ $=$ $=$ $12$ $12$ $- 4$ $-4$ $4 - 4$ $4-4$ cancels out

$\frac{a}{10}$ $frac{a}{10}$ $=$ $=$ $8$ $8$

Finally, multiply both sides of the equation by $10$ $10$

$\frac{a}{10}$ $frac{a}{10}$ $\times 10$ $times10$ $=$ $=$ $8$ $8$ $\times 10$ $times10$

$\frac{10 a}{10}$ $frac{10a}{10}$ $=$ $=$ $8$ $8$ $\times 10$ $times10$

$a$ $a$ $=$ $=$ $80$ $80$ The coefficient $\frac{10}{10}$ $10/10$ cancels out

$a = 80$ $a=80$
Question 5 of 5

5. Question
Solve for $v$ $v$

$\frac{v}{7} + 1 = 3$ $v/7 + 1=3$
- $v=$ (14)
Correct

Fantastic!

Incorrect

To solve for $v$ , get $v$ by itself

Subtract $1$ from both sides of the equation

$v/7 + 1$ $-1$ $=$ $3$ $-1$

$v/7$ $+1$ $-1$ $=$ $3$ $-1$ $1-1$ cancels out

$v/7$ $=$ $2$

Finally, multiply both sides of the equation by $7$

$v/7$ $xx 7$ $=$ $2$ $xx 7$

$(7v)/7$ $=$ $2$ $xx 7$

$v$ $=$ $14$ The coefficient $7/7$ cancels out

$v=14$

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Quiz summary

Information

1. Question

Hint

Well Done!

Inverse Operations

2. Question

Hint

Excellent!

3. Question

Hint

Nice Job!

4. Question

Hint

Well Done!

5. Question

Fantastic!

Quizzes

$\frac{x}{6} + 2$ $frac{x}{6}+2$ $- 2$ $-2$	$=$ $=$	$9$ $9$ $- 2$ $-2$

$\frac{x}{6}$ $frac{x}{6}$ $+ 2$ $+2$ $- 2$ $-2$	$=$ $=$	$9$ $9$ $- 2$ $-2$	$2 - 2$ $2-2$ cancels out

$\frac{x}{6}$ $frac{x}{6}$	$=$ $=$	$7$ $7$

$\frac{x}{6}$ $frac{x}{6}$ $\times 6$ $times6$	$=$ $=$	$7$ $7$ $\times 6$ $times6$

$\frac{6 x}{6}$ $frac{6x}{6}$	$=$ $=$	$7$ $7$ $\times 6$ $times6$

$x$ $x$	$=$ $=$	$42$ $42$	The coefficient $\frac{6}{6}$ $6/6$ cancels out

$\frac{x}{7} - 2$ $frac{x}{7}-2$ $+ 2$ $+2$	$=$ $=$	$3$ $3$ $+ 2$ $+2$

$\frac{x}{7}$ $frac{x}{7}$ $- 2$ $-2$ $+ 2$ $+2$	$=$ $=$	$3$ $3$ $+ 2$ $+2$	$- 2 + 2$ $-2+2$ cancels out

$\frac{x}{7}$ $frac{x}{7}$	$=$ $=$	$5$ $5$

$\frac{x}{7}$ $frac{x}{7}$ $\times 7$ $times7$	$=$ $=$	$5$ $5$ $\times 7$ $times7$	Multiply both sides by $7$ $7$

$\frac{7 x}{7}$ $frac{7x}{7}$	$=$ $=$	$5$ $5$ $\times 7$ $times7$

$x$ $x$	$=$ $=$	$35$ $35$	The coefficient $\frac{7}{7}$ $7/7$ cancels out

$\frac{a}{10} + 4$ $frac{a}{10}+4$ $- 4$ $-4$	$=$ $=$	$12$ $12$ $- 4$ $-4$

$\frac{a}{10}$ $frac{a}{10}$ $+ 4$ $+4$ $- 4$ $-4$	$=$ $=$	$12$ $12$ $- 4$ $-4$	$4 - 4$ $4-4$ cancels out

$\frac{a}{10}$ $frac{a}{10}$	$=$ $=$	$8$ $8$

$\frac{a}{10}$ $frac{a}{10}$ $\times 10$ $times10$	$=$ $=$	$8$ $8$ $\times 10$ $times10$

$\frac{10 a}{10}$ $frac{10a}{10}$	$=$ $=$	$8$ $8$ $\times 10$ $times10$

$a$ $a$	$=$ $=$	$80$ $80$	The coefficient $\frac{10}{10}$ $10/10$ cancels out

$v/7 + 1$ $-1$	$=$	$3$ $-1$

$v/7$ $+1$ $-1$	$=$	$3$ $-1$	$1-1$ cancels out

$v/7$	$=$	$2$

$v/7$ $xx 7$	$=$	$2$ $xx 7$

$(7v)/7$	$=$	$2$ $xx 7$

$v$	$=$	$14$	The coefficient $7/7$ cancels out