Equations with Variables on Both Sides 2

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Question 1 of 5

1. Question
Solve

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

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

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

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

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

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

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

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

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

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

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

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

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

$2$ $2$ $n$ $n$ $\div 2$ $divide2$ $=$ $=$ $8$ $8$ $\div 2$ $divide2$

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

Check our work

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

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

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

$9$ $9$ $=$ $=$ $9$ $9$

Since the equation is true, the answer is correct.

$n = 4$ $n=4$
Question 2 of 5

2. Question
Solve for $x$ $x$

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

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

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

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

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

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

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

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

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

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

$9 x$ $9x$ $\div 9$ $divide9$ $=$ $=$ $9$ $9$ $\div 9$ $divide9$

$9$ $9$ $x$ $x$ $\div 9$ $divide9$ $=$ $=$ $9$ $9$ $\div 9$ $divide9$ $\times 9 \div 9$ $times9divide9$ cancels out

$x$ $x$ $=$ $=$ $1$ $1$

$x = 1$ $x=1$
Question 3 of 5

3. Question
Solve for $x$ $x$

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

Add $8 x$ $8x$ to both sides of the equation

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

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

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

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

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

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

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

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

$12 x$ $12x$ $\div 12$ $divide12$ $=$ $=$ $- 12$ $-12$ $\div 12$ $divide12$

$12$ $12$ $x$ $x$ $\div 12$ $divide12$ $=$ $=$ $- 12$ $-12$ $\div 12$ $divide12$ $\times 12 \div 12$ $times12divide12$ cancels out

$x$ $x$ $=$ $=$ $- 1$ $-1$

$x = - 1$ $x=-1$

Question 4 of 5

Solve for

p

$p$

p - 5 + 3 p = - 9

$p-5+3p=-9$

$p =$ $p=$ (-1)

Question 5 of 5

Solve for

r

$r$

- 12 = - 2 r - 4 r

$-12=-2r-4r$

$r =$ $r=$ (2)

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

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

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

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

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

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

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

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

$9 x$ $9x$ $\div 9$ $divide9$	$=$ $=$	$9$ $9$ $\div 9$ $divide9$
$9$ $9$ $x$ $x$ $\div 9$ $divide9$	$=$ $=$	$9$ $9$ $\div 9$ $divide9$	$\times 9 \div 9$ $times9divide9$ cancels out
$x$ $x$	$=$ $=$	$1$ $1$

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

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

$12 x$ $12x$ $\div 12$ $divide12$	$=$ $=$	$- 12$ $-12$ $\div 12$ $divide12$
$12$ $12$ $x$ $x$ $\div 12$ $divide12$	$=$ $=$	$- 12$ $-12$ $\div 12$ $divide12$	$\times 12 \div 12$ $times12divide12$ cancels out
$x$ $x$	$=$ $=$	$- 1$ $-1$

$- 12$ $-12$	$=$ $=$	$- 2 r - 4 r$ $-2r-4r$
$- 12$ $-12$	$=$ $=$	$- 6 r$ $-6r$

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