Equations with Variables on Both Sides 2

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Question 1 of 5

1. Question
Solve

$n+5=13-n$
- $n=$ (4)
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Inverse Operations

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

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

Start by moving $5$ to the other side by subtracting $5$ from both sides of the equation.

$n$ $+5$ $=$ $13-$ $n$

$n$ $+5$ $-5$ $=$ $13-$ $n$ $-5$

$n$ $=$ $8-$ $n$ $5-5$ cancels out

Next, move $-n$ to the other side by adding $n$ to both sides of the equation.

$n$ $=$ $8-$ $n$

$n$ $+n$ $=$ $8-$ $n$ $+n$

$2$ $n$ $=$ $8$ $-n+n$ cancels out

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

$2$ $n$ $=$ $8$

$2$ $n$ $divide2$ $=$ $8$ $divide2$

$n$ $=$ $4$ $2-:2$ cancels out

Check our work

To confirm our answer, substitute $n=4$ to the original equation.

$n+5$ $=$ $13-n$

$4+5$ $=$ $13-4$ Substitute $n=4$

$9$ $=$ $9$

Since the equation is true, the answer is correct.

$n=4$
Question 2 of 5

2. Question
Solve for $x$

$10x-5=x+4$
- $x=$ (1)
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To solve for $x$ , get $x$ by itself

Subtract $x$ from both sides of the equation

$10x-5$ $-x$ $=$ $x+4$ $-x$

$10x-5$ $-x$ $=$ $x$ $+4$ $-x$ $x-x$ cancels out

$9x-5$ $=$ $4$

Next, add $5$ to both sides of the equation

$9x-5$ $+5$ $=$ $4$ $+5$

$9x$ $-5$ $+5$ $=$ $4$ $+5$ $-5+5$ cancels out

$9x$ $=$ $9$

Finally, divide both sides of the equation by $9$

$9x$ $divide9$ $=$ $9$ $divide9$

$9$ $x$ $divide9$ $=$ $9$ $divide9$ $times9divide9$ cancels out

$x$ $=$ $1$

$x=1$
Question 3 of 5

3. Question
Solve for $x$

$4x+5=-8x-7$
- $x=$ (-1)
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To solve for $x$ , get $x$ by itself

Add $8x$ to both sides of the equation

$4x+5$ $+8x$ $=$ $-8x-7$ $+8x$

$4x+5$ $+8x$ $=$ $-8x$ $-7$ $+8x$ $-8x+8x$ cancels out

$12x+5$ $=$ $-7$

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

$12x+5$ $-5$ $=$ $-7$ $-5$

$12x$ $+5$ $-5$ $=$ $-7$ $-5$ $5-5$ cancels out

$12x$ $=$ $-12$

Finally, divide both sides of the equation by $12$

$12x$ $divide12$ $=$ $-12$ $divide12$

$12$ $x$ $divide12$ $=$ $-12$ $divide12$ $times12divide12$ cancels out

$x$ $=$ $-1$

$x=-1$

Question 4 of 5

Solve for

$p$

$p-5+3p=-9$

$p=$ (-1)

Question 5 of 5

5. Question
Solve for $r$

$-12=-2r-4r$
- $r=$ (2)
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To solve for $r$ , get $r$ by itself

Combine like terms to simplify.

$-12$ $=$ $-2r-4r$

$-12$ $=$ $-6r$

Get the $r$ variable on one side of the equation

$-12$ $divide(-6)$ $=$ $-6r$ $divide(-6)$ Divide both sides by $-6$

$-12$ $divide(-6)$ $=$ $-6$ $r$ $divide(-6)$ $times(-6)divide(-6)$ cancels out

$2$ $=$ $r$

$r$ $=$ $2$

$r=2$

$p$ $-5$ $+3p$	$=$	$-9$
$(p+3p)$ $-5$	$=$	$-9$
$4p-5$	$=$	$-9$

$4p-5$ $+5$	$=$	$-9$ $+5$
$4p$ $-5$ $+5$	$=$	$-9$ $+5$	$-5+5$ cancels out
$4p$	$=$	$-4$
$4p$ $divide4$	$=$	$-4$ $divide4$	Divide both sides by $4$
$4$ $p$ $divide4$	$=$	$-4$ $divide4$	$times4divide4$ cancels out
$p$	$=$	$-1$

Equations with Variables on Both Sides 2

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Quiz summary

Information

1. Question

Hint

Excellent!

Inverse Operations

2. Question

Hint

Well Done!

3. Question

Hint

Nice Job!

4. Question

Keep Going!

5. Question

Correct!

Quizzes

$n$ $+5$	$=$	$13-$ $n$
$n$ $+5$ $-5$	$=$	$13-$ $n$ $-5$
$n$	$=$	$8-$ $n$	$5-5$ cancels out

$n$	$=$	$8-$ $n$
$n$ $+n$	$=$	$8-$ $n$ $+n$
$2$ $n$	$=$	$8$	$-n+n$ cancels out

$2$ $n$	$=$	$8$
$2$ $n$ $divide2$	$=$	$8$ $divide2$
$n$	$=$	$4$	$2-:2$ cancels out

$10x-5$ $-x$	$=$	$x+4$ $-x$
$10x-5$ $-x$	$=$	$x$ $+4$ $-x$	$x-x$ cancels out
$9x-5$	$=$	$4$

$9x-5$ $+5$	$=$	$4$ $+5$
$9x$ $-5$ $+5$	$=$	$4$ $+5$	$-5+5$ cancels out
$9x$	$=$	$9$

$9x$ $divide9$	$=$	$9$ $divide9$
$9$ $x$ $divide9$	$=$	$9$ $divide9$	$times9divide9$ cancels out
$x$	$=$	$1$

$4x+5$ $+8x$	$=$	$-8x-7$ $+8x$
$4x+5$ $+8x$	$=$	$-8x$ $-7$ $+8x$	$-8x+8x$ cancels out
$12x+5$	$=$	$-7$

$12x+5$ $-5$	$=$	$-7$ $-5$
$12x$ $+5$ $-5$	$=$	$-7$ $-5$	$5-5$ cancels out
$12x$	$=$	$-12$

$12x$ $divide12$	$=$	$-12$ $divide12$
$12$ $x$ $divide12$	$=$	$-12$ $divide12$	$times12divide12$ cancels out
$x$	$=$	$-1$

$-12$ $divide(-6)$	$=$	$-6r$ $divide(-6)$	Divide both sides by $-6$
$-12$ $divide(-6)$	$=$	$-6$ $r$ $divide(-6)$	$times(-6)divide(-6)$ cancels out
$2$	$=$	$r$
$r$	$=$	$2$

$n+5$	$=$	$13-n$
$4+5$	$=$	$13-4$	Substitute $n=4$
$9$	$=$	$9$