Two Step Equations 4

Years

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 $y = 50$ $y=50$

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

Since the equation is true, the answer is correct.

$y=50$

Question 2 of 5

2. Question
Solve for $x$

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

Subtract $2$ from both sides of the equation

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

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

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

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

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

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

$x$ $=$ $42$ The coefficient $6/6$ cancels out

$x=42$
Question 3 of 5

3. Question
Solve for $x$

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

Add $2$ to both sides of the equation

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

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

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

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

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

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

$x$ $=$ $35$ The coefficient $7/7$ cancels out

$x=35$
Question 4 of 5

4. Question
Solve for $a$

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

Subtract $4$ from both sides of the equation

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

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

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

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

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

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

$a$ $=$ $80$ The coefficient $10/10$ cancels out

$a=80$
Question 5 of 5

5. Question
Solve for $v$

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

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

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

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

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

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

$x$	$=$	$42$	The coefficient $6/6$ cancels out

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

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

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

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

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

$x$	$=$	$35$	The coefficient $7/7$ cancels out

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

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

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

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

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

$a$	$=$	$80$	The coefficient $10/10$ 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