Equations with Variables on Both Sides (Fractions) 1

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Question 1 of 5

1. Question
Solve

$x + 6 = \frac{5}{7} x$ $x+6=5/7 x$
- $x =$ $x=$ (-21)
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Inverse Operations

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

Distributive Property

$a$ $a$ $(b + c) =$ $(b+c)=$ $a$ $a$ $b +$ $b+$ $a$ $a$ $c$ $c$

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

Start by removing the fraction by multiplying both sides of the equation by $7$ $7$ .

$x$ $x$ $+ 6$ $+6$ $=$ $=$ $\frac{5}{7}$ $5/7$ $x$ $x$

$($ $($ $x$ $x$ $+ 6)$ $+6)$ $\times 7$ $times7$ $=$ $=$ $\frac{5}{7}$ $5/7$ $x$ $x$ $\times 7$ $times7$

$7 ($ $7($ $x$ $x$ $+ 6)$ $+6)$ $=$ $=$ $5$ $5$ $x$ $x$ $\frac{1}{7} \times 7$ $1/7times7$ cancels out

Next, expand the left side by using the Distributive Property.

$7$ $7$ $($ $($ $x$ $x$ $+ 6)$ $+6)$ $=$ $=$ $5$ $5$ $x$ $x$

$7$ $7$ $x$ $x$ $+$ $+$ $7$ $7$ $(6)$ $(6)$ $=$ $=$ $5$ $5$ $x$ $x$

$7$ $7$ $x$ $x$ $+ 42$ $+42$ $=$ $=$ $5$ $5$ $x$ $x$

Next, move $7 x$ $7x$ to the other side by subtracting $7 x$ $7x$ from both sides of the equation.

$7$ $7$ $x$ $x$ $+ 42$ $+42$ $=$ $=$ $5$ $5$ $x$ $x$

$7$ $7$ $x$ $x$ $+ 42$ $+42$ $- 7 x$ $-7x$ $=$ $=$ $5$ $5$ $x$ $x$ $- 7 x$ $-7x$

$42$ $42$ $=$ $=$ $- 2$ $-2$ $x$ $x$ $7 x - 7 x$ $7x-7x$ cancels out

Finally, remove $- 2$ $-2$ by dividing both sides of the equation by $- 2$ $-2$ .

$42$ $42$ $=$ $=$ $- 2$ $-2$ $x$ $x$

$42$ $42$ $\div - 2$ $divide-2$ $=$ $=$ $- 2$ $-2$ $x$ $x$ $\div - 2$ $divide-2$

$- 21$ $-21$ $=$ $=$ $x$ $x$ $- 2 \div - 2$ $-2divide-2$ cancels out

$x$ $x$ $=$ $=$ $- 21$ $-21$

Check our work

To confirm our answer, substitute $x = - 21$ $x=-21$ to the original equation.

$x + 6$ $x+6$ $=$ $=$ $\frac{5}{7} x$ $5/7 x$

$- 21 + 6$ $-21+6$ $=$ $=$ $\frac{5}{7} (- 21)$ $5/7 (-21)$ Substitute $x = - 21$ $x=-21$

$- 15$ $-15$ $=$ $=$ $5 (- 3)$ $5(-3)$

$- 15$ $-15$ $=$ $=$ $- 15$ $-15$

Since the equation is true, the answer is correct.

$x = - 21$ $x=-21$

Question 2 of 5

Solve

\frac{4 u}{5} + 3 = u

$(4u)/5+3=u$

$u =$ $u=$ (15)

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Inverse Operations

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

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

Start by removing the fraction by multiplying both sides of the equation by

5

$5$ .

$\frac{4\color{#00880A}{u}}{5}+3$	$=$ $=$	$u$ $u$

$\left(\frac{4\color{#00880A}{u}}{5}+3\right)\color{#CC0000}{\times5}$	$=$ $=$	$u$ $u$ $\times 5$ $times5$

$\frac{4\color{#00880A}{u}}{5}\color{#CC0000}{(5)}+3\color{#CC0000}{(5)}$	$=$ $=$	$5$ $5$ $u$ $u$	Distribute $5$ $5$ to the parenthesis

$4$ $4$ $u$ $u$ $+ 15$ $+15$	$=$ $=$	$5$ $5$ $u$ $u$	$\frac{1}{5} \times 5$ $1/5times5$ cancels out

Next, move

4 u

$4u$ to the other side by subtracting

4 u

$4u$ from both sides of the equation.

$4$ $4$ $u$ $u$ $+ 15$ $+15$	$=$ $=$	$5$ $5$ $u$ $u$
$4$ $4$ $u$ $u$ $+ 15$ $+15$ $- 4 u$ $-4u$	$=$ $=$	$5$ $5$ $u$ $u$ $- 4 u$ $-4u$
$15$ $15$	$=$ $=$	$u$ $u$	$4 u - 4 u$ $4u-4u$ cancels out
$u$ $u$	$=$ $=$	$15$ $15$

Check our work

To confirm our answer, substitute

u = 15

$u=15$ to the original equation.

$\frac{4 u}{5} + 3$ $(4u)/5+3$	$=$ $=$	$u$ $u$

$\frac{4 (15)}{5} + 3$ $(4(15))/5+3$	$=$ $=$	$15$ $15$	Substitute $u = 15$ $u=15$

$\frac{60}{5} + 3$ $60/5+3$	$=$ $=$	$15$ $15$

$12 + 3$ $12+3$	$=$ $=$	$15$ $15$
$15$ $15$	$=$ $=$	$15$ $15$

Since the equation is true, the answer is correct.

u = 15

$u=15$

Question 3 of 5

Solve

7 x = \frac{5 x - 8}{3}

$7x=(5x-8)/3$

$x =$ $x=$ (-1/2)

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Inverse Operations

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

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

Start by removing the fraction by multiplying both sides of the equation by

3

$3$ .

$7$ $7$ $x$ $x$	$=$ $=$	$\frac{5\color{#00880A}{x}-8}{3}$

$7$ $7$ $x$ $x$ $\times 3$ $times3$	$=$ $=$	$\frac{5\color{#00880A}{x}-8}{3}\color{#CC0000}{\times3}$

$21$ $21$ $x$ $x$	$=$ $=$	$5$ $5$ $x$ $x$ $- 8$ $-8$	$\frac{1}{3} \times 3$ $1/3times3$ cancels out

Next, move

5 x

$5x$ to the other side by subtracting

5 x

$5x$ from both sides of the equation.

$21$ $21$ $x$ $x$	$=$ $=$	$5$ $5$ $x$ $x$ $- 8$ $-8$
$21$ $21$ $x$ $x$ $- 5 x$ $-5x$	$=$ $=$	$5$ $5$ $x$ $x$ $- 8$ $-8$ $- 5 x$ $-5x$
$16$ $16$ $x$ $x$	$=$ $=$	$- 8$ $-8$	$5 x - 5 x$ $5x-5x$ cancels out

Finally, remove

16

$16$ by dividing both sides of the equation by

16

$16$ .

$16$ $16$ $x$ $x$	$=$ $=$	$- 8$ $-8$
$16$ $16$ $x$ $x$ $\div 16$ $divide16$	$=$ $=$	$- 8$ $-8$ $\div 16$ $divide16$
$x$ $x$	$=$ $=$	$- \frac{1}{2}$ $-1/2$	$16 \div 16$ $16divide16$ cancels out

Check our work

To confirm our answer, substitute

x = - \frac{1}{2}

$x=-1/2$ to the original equation.

$7 x$ $7x$	$=$ $=$	$\frac{5 x - 8}{3}$ $(5x-8)/3$

$7 (- \frac{1}{2})$ $7(-1/2)$	$=$ $=$	$\frac{5 (- \frac{1}{2}) - 8}{3}$ $[5(-1/2)-8]/3$	Substitute $x = - \frac{1}{2}$ $x=-1/2$

$- \frac{7}{2}$ $-7/2$	$=$ $=$	$\frac{- \frac{5}{2} - 8}{3}$ $[-5/2-8]/3$

$- \frac{7}{2}$ $-7/2$	$=$ $=$	$\frac{- \frac{21}{2}}{3}$ $[-21/2]/3$

$- \frac{7}{2}$ $-7/2$	$=$ $=$	$- \frac{21}{2} \times \frac{1}{3}$ $-21/2 times 1/3$

$- \frac{7}{2}$ $-7/2$	$=$ $=$	$- \frac{21}{6}$ $-21/6$

$- \frac{7}{2}$ $-7/2$	$=$ $=$	$- \frac{7}{2}$ $-7/2$

Since the equation is true, the answer is correct.

x = - \frac{1}{2}

$x=-1/2$

Question 4 of 5

Solve for

t

$t$

3 t - 2 = \frac{3}{5} t + 1

$3t-2=3/5 t+1$

1. $1 \frac{1}{4}$ $1 1/4$
2. $\frac{1}{5}$ $1/5$
3. $1 \frac{1}{5}$ $1 1/5$
4. $\frac{2}{5}$ $2/5$

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Inverse Operations

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

Distributive Property

$a$ $a$

(b + c) =

$(b+c)=$ $a$ $a$

b +

$b+$ $a$ $a$

c

$c$

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

Start by removing the fraction by multiplying both sides of the equation by

5

$5$ , then using the Distributive Property.

$3$ $3$ $t$ $t$ $- 2$ $-2$	$=$ $=$	$\frac{3}{5}$ $3/5$ $t$ $t$ $+ 1$ $+1$

$(3$ $(3$ $t$ $t$ $- 2)$ $-2)$ $\times 5$ $times5$	$=$ $=$	$\left(\frac{3}{5}\color{#00880A}{t}+1\right)\color{#CC0000}{\times5}$

$5$ $5$ $(3$ $(3$ $t$ $t$ $- 2)$ $-2)$	$=$ $=$	$\color{#007DDC}{5}\left(\frac{3}{5}\color{#00880A}{t}+1\right)$

$5$ $5$ $(3$ $(3$ $t$ $t$ $) +$ $)+$ $5$ $5$ $(- 2)$ $(-2)$	$=$ $=$	$\color{#007DDC}{5}\left(\frac{3}{5}\color{#00880A}{t}\right)+\color{#007DDC}{5}(1)$

$15$ $15$ $t$ $t$ $- 10$ $-10$	$=$ $=$	$3$ $3$ $t$ $t$ $+ 5$ $+5$	$\frac{1}{5} \times 5$ $1/5times5$ cancels out

Next, move

- 10

$-10$ to the other side by adding

10

$10$ to both sides of the equation.

$15$ $15$ $t$ $t$ $- 10$ $-10$	$=$ $=$	$3$ $3$ $t$ $t$ $+ 5$ $+5$
$15$ $15$ $t$ $t$ $- 10$ $-10$ $+ 10$ $+10$	$=$ $=$	$3$ $3$ $t$ $t$ $+ 5$ $+5$ $+ 10$ $+10$
$15$ $15$ $t$ $t$	$=$ $=$	$3$ $3$ $t$ $t$ $+ 15$ $+15$	$- 10 + 10$ $-10+10$ cancels out

Now, move

3 t

$3t$ to the other side by subtracting

3 t

$3t$ from both sides of the equation.

$15$ $15$ $t$ $t$	$=$ $=$	$3$ $3$ $t$ $t$ $+ 15$ $+15$
$15$ $15$ $t$ $t$ $- 3 t$ $-3t$	$=$ $=$	$3$ $3$ $t$ $t$ $+ 15$ $+15$ $- 3 t$ $-3t$
$12$ $12$ $t$ $t$	$=$ $=$	$15$ $15$	$3 t - 3 t$ $3t-3t$ cancels out

Finally, remove

12

$12$ by dividing both sides of the equation by

12

$12$ .

$12$ $12$ $t$ $t$	$=$ $=$	$15$ $15$
$12$ $12$ $t$ $t$ $\div 12$ $divide12$	$=$ $=$	$15$ $15$ $\div 12$ $divide12$

$t$ $t$	$=$ $=$	$\frac{15}{12}$ $15/12$	$12 \div 12$ $12divide12$ cancels out

$t$ $t$	$=$ $=$	$\frac{5}{4}$ $5/4$	Simplify the fraction

$t$ $t$	$=$ $=$	$1 \frac{1}{4}$ $1 1/4$

Check our work

To confirm our answer, substitute

t = 1 \frac{1}{4}

$t=1 1/4$ or

t = \frac{5}{4}

$t=5/4$ to the original equation.

$3 t - 2$ $3t-2$	$=$ $=$	$\frac{3}{5} t + 1$ $3/5 t+1$

$3 (\frac{5}{4}) - 2$ $3(5/4)-2$	$=$ $=$	$\frac{3}{5} (\frac{5}{4}) + 1$ $3/5 (5/4)+1$	Substitute $x = \frac{5}{4}$ $x=5/4$

$\frac{15}{4} - 2$ $15/4-2$	$=$ $=$	$\frac{15}{20} + 1$ $15/20+1$

$\frac{15 - 8}{4}$ $(15-8)/4$	$=$ $=$	$\frac{15 + 20}{20}$ $(15+20)/20$

$\frac{7}{4}$ $7/4$	$=$ $=$	$\frac{35}{20}$ $35/20$

$\frac{7}{4}$ $7/4$	$=$ $=$	$\frac{7}{4}$ $7/4$

Since the equation is true, the answer is correct.

x = 1 \frac{1}{4}

$x=1 1/4$

Question 5 of 5

Solve for

x

$x$

\frac{x}{4} - \frac{x}{5} = 12

$x/4-x/5=12$

$x =$ $x=$ (240)

$\frac{5 x - 4 x}{20}$ $frac{5x-4x}{20}$	$=$ $=$	$12$ $12$

$\frac{x}{20}$ $frac{x}{20}$	$=$	$12$

$frac{x}{20}$ $times20$	$=$	$12$ $times20$	Multiply both sides by $20$

$frac{20x}{20}$	$=$	$12$ $times20$

$x$	$=$	$240$	The coefficient $frac{20}{20}$ cancels out

Equations with Variables on Both Sides (Fractions) 1

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Quiz summary

Information

1. Question

Keep Going!

Inverse Operations

Distributive Property

2. Question

Hint

Great Work!

Inverse Operations

3. Question

Hint

Fantastic!

Inverse Operations

4. Question

Hint

Correct!

Inverse Operations

Distributive Property

5. Question

Hint

Keep Going!

Quizzes

$x$ $x$ $+ 6$ $+6$	$=$ $=$	$\frac{5}{7}$ $5/7$ $x$ $x$

$($ $($ $x$ $x$ $+ 6)$ $+6)$ $\times 7$ $times7$	$=$ $=$	$\frac{5}{7}$ $5/7$ $x$ $x$ $\times 7$ $times7$

$7 ($ $7($ $x$ $x$ $+ 6)$ $+6)$	$=$ $=$	$5$ $5$ $x$ $x$	$\frac{1}{7} \times 7$ $1/7times7$ cancels out

$7$ $7$ $($ $($ $x$ $x$ $+ 6)$ $+6)$	$=$ $=$	$5$ $5$ $x$ $x$
$7$ $7$ $x$ $x$ $+$ $+$ $7$ $7$ $(6)$ $(6)$	$=$ $=$	$5$ $5$ $x$ $x$
$7$ $7$ $x$ $x$ $+ 42$ $+42$	$=$ $=$	$5$ $5$ $x$ $x$

$42$ $42$	$=$ $=$	$- 2$ $-2$ $x$ $x$
$42$ $42$ $\div - 2$ $divide-2$	$=$ $=$	$- 2$ $-2$ $x$ $x$ $\div - 2$ $divide-2$
$- 21$ $-21$	$=$ $=$	$x$ $x$	$- 2 \div - 2$ $-2divide-2$ cancels out
$x$ $x$	$=$ $=$	$- 21$ $-21$

$x + 6$ $x+6$	$=$ $=$	$\frac{5}{7} x$ $5/7 x$

$- 21 + 6$ $-21+6$	$=$ $=$	$\frac{5}{7} (- 21)$ $5/7 (-21)$	Substitute $x = - 21$ $x=-21$

$- 15$ $-15$	$=$ $=$	$5 (- 3)$ $5(-3)$
$- 15$ $-15$	$=$ $=$	$- 15$ $-15$

$\frac{x}{4} - \frac{x}{5}$ $frac{x}{4}-frac{x}{5}$	$=$ $=$	$12$ $12$

$\frac{x}{4}$ $frac{x}{4}$ $\times \frac{5}{5}$ $timesfrac{5}{5}$ $- \frac{x}{5}$ $-frac{x}{5}$ $\times \frac{4}{4}$ $timesfrac{4}{4}$	$=$ $=$	$12$ $12$

$\frac{5 x}{20} - \frac{4 x}{20}$ $frac{5x}{20}-frac{4x}{20}$	$=$ $=$	$12$ $12$