Equations with Variables on Both Sides 1

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Question 1 of 5

1. Question
Solve

$3a+4=2a+10$
- $a=$ (6)
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Inverse Operations

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

Get $a$ alone to the left side and all constants to the right.

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

$3$ $a$ $+4$ $=$ $2$ $a$ $+10$

$3$ $a$ $+4$ $-4$ $=$ $2$ $a$ $+10$ $-4$

$3$ $a$ $=$ $2$ $a$ $+6$ $4-4$ cancels out

Next, move $2a$ to the other side by subtracting $2a$ from both sides of the equation.

$3$ $a$ $=$ $2$ $a$ $+6$

$3$ $a$ $-2a$ $=$ $2$ $a$ $+6$ $-2a$

$a$ $=$ $6$ $2a-2a$ cancels out

Check our work

To confirm our answer, substitute $a=6$ to the original equation.

$3a+4$ $=$ $2a+10$

$3(6)+4$ $=$ $2(6)+10$ Substitute $a=6$

$18+4$ $=$ $12+10$

$22$ $=$ $22$

Since the equation is true, the answer is correct.

$a=6$
Question 2 of 5

2. Question
Solve

$4m-3=24+m$
- $m=$ (9)
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Inverse Operations

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

Get $m$ alone to the left side and all constants to the right.

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

$4$ $m$ $-3$ $=$ $24+$ $m$

$4$ $m$ $-3$ $+3$ $=$ $24+$ $m$ $+3$

$4$ $m$ $=$ $27+$ $m$ $-3+3$ cancels out

Next, move $m$ to the other side by subtracting $m$ from both sides of the equation.

$4$ $m$ $=$ $27+$ $m$

$4$ $m$ $-m$ $=$ $27+$ $m$ $-m$

$3$ $m$ $=$ $27$ $m-m$ cancels out

Finally, remove $3$ by dividing $3$ from both sides.

$3$ $m$ $=$ $27$

$3$ $m$ $-:3$ $=$ $27$ $-:3$

$m$ $=$ $9$ $3-:3$ cancels out

Check our work

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

$4m-3$ $=$ $24+m$

$4(9)-3$ $=$ $24+9$ Substitute $m=9$

$36-3$ $=$ $33$

$33$ $=$ $33$

Since the equation is true, the answer is correct.

$m=9$
Question 3 of 5

3. Question
Solve

$8y-2=6y+28$
- $y=$ (15)
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Inverse Operations

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

Get $y$ alone to the left side and all constants to the right.

Start by moving $-2$ to the other side by adding $2$ to both sides of the equation.

$8$ $y$ $-2$ $=$ $6$ $y$ $+28$

$8$ $y$ $-2$ $+2$ $=$ $6$ $y$ $+28$ $+2$

$8$ $y$ $=$ $6$ $y$ $+30$ $-2+2$ cancels out

Next, move $6y$ to the other side by subtracting $6y$ from both sides of the equation.

$8$ $y$ $=$ $6$ $y$ $+30$

$8$ $y$ $-6y$ $=$ $6$ $y$ $+30$ $-6y$

$2$ $y$ $=$ $30$ $6y-6y$ cancels out

Finally, remove $2$ by dividing $2$ from both sides.

$2$ $y$ $=$ $30$

$2$ $y$ $divide2$ $=$ $30$ $divide2$

$y$ $=$ $15$ $2-:2$ cancels out

Check our work

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

$8y-2$ $=$ $6y+28$

$8(15)-2$ $=$ $6(15)+28$ Substitute $y=15$

$120-2$ $=$ $90+28$

$118$ $=$ $118$

Since the equation is true, the answer is correct.

$y=15$
Question 4 of 5

4. Question
Solve for $x$

$3x+2=2x+9$
- $x=$ (7)
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To solve for $x$ , get $x$ by itself

Subtract $2x$ from both sides of the equation

$3x+2$ $-2x$ $=$ $2x+9$ $-2x$

$3x+2$ $-2x$ $=$ $2x$ $+9$ $-2x$ $2x-2x$ cancels out

$x+2$ $=$ $9$

Finally, subtract $2$ from both sides of the equation

$x+2$ $-2$ $=$ $9$ $-2$

$x$ $+2$ $-2$ $=$ $9$ $-2$ $2-2$ cancels out

$x$ $=$ $7$

$x=7$
Question 5 of 5

5. Question
Solve for $x$

$3x+5=2x+7$
- $x=$ (2)
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To solve for $x$ , get $x$ by itself

Subtract $2x$ from both sides of the equation

$3x+5$ $-2x$ $=$ $2x+7$ $-2x$

$3x+5$ $-2x$ $=$ $2x$ $+7$ $-2x$ $2x-2x$ cancels out

$x+5$ $=$ $7$

Finally, subtract $5$ from both sides of the equation

$x+5$ $-5$ $=$ $7$ $-5$

$x$ $+5$ $-5$ $=$ $7$ $-5$ $5-5$ cancels out

$x$ $=$ $2$

$x=2$

Equations with Variables on Both Sides 1

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Quiz summary

Information

1. Question

Hint

Fantastic!

Inverse Operations

2. Question

Hint

Good Job!

Inverse Operations

3. Question

Hint

Correct!

Inverse Operations

4. Question

Hint

Excellent!

5. Question

Hint

Fantastic!

Quizzes

$3$ $a$ $+4$	$=$	$2$ $a$ $+10$
$3$ $a$ $+4$ $-4$	$=$	$2$ $a$ $+10$ $-4$
$3$ $a$	$=$	$2$ $a$ $+6$	$4-4$ cancels out

$3$ $a$	$=$	$2$ $a$ $+6$
$3$ $a$ $-2a$	$=$	$2$ $a$ $+6$ $-2a$
$a$	$=$	$6$	$2a-2a$ cancels out

$3a+4$	$=$	$2a+10$
$3(6)+4$	$=$	$2(6)+10$	Substitute $a=6$
$18+4$	$=$	$12+10$
$22$	$=$	$22$

$4$ $m$ $-3$	$=$	$24+$ $m$
$4$ $m$ $-3$ $+3$	$=$	$24+$ $m$ $+3$
$4$ $m$	$=$	$27+$ $m$	$-3+3$ cancels out

$4$ $m$	$=$	$27+$ $m$
$4$ $m$ $-m$	$=$	$27+$ $m$ $-m$
$3$ $m$	$=$	$27$	$m-m$ cancels out

$3$ $m$	$=$	$27$
$3$ $m$ $-:3$	$=$	$27$ $-:3$
$m$	$=$	$9$	$3-:3$ cancels out

$4m-3$	$=$	$24+m$
$4(9)-3$	$=$	$24+9$	Substitute $m=9$
$36-3$	$=$	$33$
$33$	$=$	$33$

$8$ $y$ $-2$	$=$	$6$ $y$ $+28$
$8$ $y$ $-2$ $+2$	$=$	$6$ $y$ $+28$ $+2$
$8$ $y$	$=$	$6$ $y$ $+30$	$-2+2$ cancels out

$8$ $y$	$=$	$6$ $y$ $+30$
$8$ $y$ $-6y$	$=$	$6$ $y$ $+30$ $-6y$
$2$ $y$	$=$	$30$	$6y-6y$ cancels out

$2$ $y$	$=$	$30$
$2$ $y$ $divide2$	$=$	$30$ $divide2$
$y$	$=$	$15$	$2-:2$ cancels out

$8y-2$	$=$	$6y+28$
$8(15)-2$	$=$	$6(15)+28$	Substitute $y=15$
$120-2$	$=$	$90+28$
$118$	$=$	$118$

$3x+2$ $-2x$	$=$	$2x+9$ $-2x$
$3x+2$ $-2x$	$=$	$2x$ $+9$ $-2x$	$2x-2x$ cancels out
$x+2$	$=$	$9$

$x+2$ $-2$	$=$	$9$ $-2$
$x$ $+2$ $-2$	$=$	$9$ $-2$	$2-2$ cancels out
$x$	$=$	$7$

$3x+5$ $-2x$	$=$	$2x+7$ $-2x$
$3x+5$ $-2x$	$=$	$2x$ $+7$ $-2x$	$2x-2x$ cancels out
$x+5$	$=$	$7$

$x+5$ $-5$	$=$	$7$ $-5$
$x$ $+5$ $-5$	$=$	$7$ $-5$	$5-5$ cancels out
$x$	$=$	$2$