Equations with Variables on Both Sides (Fractions) 2

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Question 1 of 5

Solve

$(x-1)/5=(x+5)/2$

$x=$ (-9)

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Inverse Operations

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

Distributive Property

$a$

$(b+c)=$ $a$

$b+$ $a$

$c$

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

First, remove the fractions by finding the Least Common Denominator

$(LCD)$ of the denominators

$5$ and

$2$ .

Multiples of

$5$ :

$5$ $10$

$15 20 25$

Multiples of

$2$ :

$2 4 6 8$ $10$

The

$LCD$ of

$5$ and

$2$ is $10$

Multiply the

$LCD$ to both sides of the equation to remove the fractions.

$\frac{\color{#00880A}{x}-1}{5}\color{#D800AD}{\times10}$	$=$	$\frac{\color{#00880A}{x}+5}{2}\color{#D800AD}{\times10}$

$\frac{10(\color{#00880A}{x}-1)}{5}$	$=$	$\frac{10(\color{#00880A}{x}+5)}{2}$

$2$ $($ $x$ $-1)$	$=$	$5$ $($ $x$ $+5)$	Divide $10$ by each denominator
$2$ $($ $x$ $)-$ $2$ $(1)$	$=$	$5$ $($ $x$ $)+$ $5$ $(5)$	Distribute the constants inside the parentheses
$2$ $x$ $-2$	$=$	$5$ $x$ $+25$	Simplify

Next, move

$5x$ to the other side by subtracting

$5x$ from both sides of the equation.

$2$ $x$ $-2$	$=$	$5$ $x$ $+25$
$2$ $x$ $-2$ $-5x$	$=$	$5$ $x$ $+25$ $-5x$
$-3$ $x$ $-2$	$=$	$25$	$5x-5x$ cancels out

Then, move

$-2$ to the other side by adding

$2$ to both sides of the equation.

$-3$ $x$ $-2$	$=$	$25$
$-3$ $x$ $-2$ $+2$	$=$	$25$ $+2$
$-3$ $x$	$=$	$27$	$-2+2$ cancels out

Finally, remove

$-3$ by dividing both sides of the equation by

$-3$ .

$-3$ $x$	$=$	$27$
$-3$ $x$ $divide-3$	$=$	$27$ $divide-3$
$x$	$=$	$-9$	$-3divide-3$ cancels out

Check our work

To confirm our answer, substitute

$x=-9$ to the original equation.

$(x-1)/5$	$=$	$(x+5)/2$

$(-9-1)/5$	$=$	$(-9+5)/2$	Substitute $x=-9$

$(-10)/5$	$=$	$(-4)/2$

$-2$	$=$	$-2$

Since the equation is true, the answer is correct.

$x=-9$

Question 2 of 5

Solve for

$a$

$(3a)/5 + a/2 = 4$

$a=$ (40/11)

Question 3 of 5

Solve

$3+2/3 y=4+1/2 y$

$y=$ (6)

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Inverse Operations

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

Distributive Property

$a$

$(b+c)=$ $a$

$b+$ $a$

$c$

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

First, remove the fractions by finding the Least Common Denominator

$(LCD)$ of the denominators

$3$ and

$2$ .

Multiples of

$3$ :

$3$ $6$

$9 12 15$

Multiples of

$2$ :

$2 4$ $6$

$8 10$

The

$LCD$ of

$3$ and

$2$ is $6$

Multiply the

$LCD$ to both sides of the equation to remove the fractions. Use the Distributive Property.

$\left(3+\frac{2}{3}\color{#00880A}{y}\right)\color{#D800AD}{\times6}$	$=$	$\left(4+\frac{1}{2}\color{#00880A}{y}\right)\color{#D800AD}{\times6}$

$\color{#007DDC}{6}\left(3+\frac{2}{3}\color{#00880A}{y}\right)$	$=$	$\color{#007DDC}{6}\left(4+\frac{1}{2}\color{#00880A}{y}\right)$

$\color{#007DDC}{6}(3)+\color{#007DDC}{6}\left(\frac{2}{3}\color{#00880A}{y}\right)$	$=$	$\color{#007DDC}{6}(4)+\color{#007DDC}{6}\left(\frac{1}{2}\color{#00880A}{y}\right)$

$18+\frac{12}{3}\color{#00880A}{y}$	$=$	$24+\frac{6}{2}\color{#00880A}{y}$

$18+4$ $y$	$=$	$24+3$ $y$	Simplify

Next, move

$3y$ to the other side by subtracting

$3y$ from both sides of the equation.

$18+4$ $y$	$=$	$24+3$ $y$
$18+4$ $y$ $-3y$	$=$	$24+3$ $y$ $-3y$
$18+$ $y$	$=$	$24$	$3y-3y$ cancels out

Finally, move

$18$ to the other side by subtracting

$18$ from both sides of the equation.

$18+$ $y$	$=$	$24$
$18+$ $y$ $-18$	$=$	$24$ $-18$
$y$	$=$	$6$	$18-18$ cancels out

Check our work

To confirm our answer, substitute

$y=6$ to the original equation.

$3+2/3 y$	$=$	$4+1/2 y$

$3+2/3 (6)$	$=$	$4+1/2 (6)$	Substitute $y=6$

$3+12/3$	$=$	$4+6/2$

$3+4$	$=$	$4+3$
$7$	$=$	$7$

Since the equation is true, the answer is correct.

$y=6$

Question 4 of 5

Solve

$1/4 m-2=1/3 m+4$

$m=$ (-72)

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Inverse Operations

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

Distributive Property

$a$

$(b+c)=$ $a$

$b+$ $a$

$c$

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

First, remove the fractions by finding the Least Common Denominator

$(LCD)$ of the denominators

$4$ and

$3$ .

Multiples of

$4$ :

$4 8$ $12$

$16 20$

Multiples of

$3$ :

$3 6 9$ $12$

$15$

The

$LCD$ of

$4$ and

$3$ is $12$

Multiply the

$LCD$ to both sides of the equation to remove the fractions. Use the Distributive Property.

$\left(\frac{1}{4}\color{#00880A}{m}-2\right)\color{#D800AD}{\times12}$	$=$	$\left(\frac{1}{3}\color{#00880A}{m}+4\right)\color{#D800AD}{\times12}$

$\color{#007DDC}{12}\left(\frac{1}{4}\color{#00880A}{m}-2\right)$	$=$	$\color{#007DDC}{12}\left(\frac{1}{3}\color{#00880A}{m}+4\right)$

$\color{#007DDC}{12}\left(\frac{1}{4}\color{#00880A}{m}\right)+\color{#007DDC}{12}(-2)$	$=$	$\color{#007DDC}{12}\left(\frac{1}{3}\color{#00880A}{m}\right)+\color{#007DDC}{12}(4)$

$\frac{12}{4}\color{#00880A}{m}-24$	$=$	$\frac{12}{3}\color{#00880A}{m}+48$

$3$ $m$ $-24$	$=$	$4$ $m$ $+48$	Simplify

Next, move

$3m$ to the other side by subtracting

$3m$ from both sides of the equation.

$3$ $m$ $-24$	$=$	$4$ $m$ $+48$
$3$ $m$ $-24$ $-3m$	$=$	$4$ $m$ $+48$ $-3m$
$-24$	$=$	$m$ $+48$	$3m-3m$ cancels out

Finally, move

$48$ to the other side by subtracting

$48$ from both sides of the equation.

$-24$	$=$	$m$ $+48$
$-24$ $-48$	$=$	$m$ $+48$ $-48$
$-72$	$=$	$m$	$48-48$ cancels out
$m$	$=$	$-72$

Check our work

To confirm our answer, substitute

$m=-72$ to the original equation.

$1/4 m-2$	$=$	$1/3 m+4$

$1/4 (-72)-2$	$=$	$1/3 (-72)+4$	Substitute $m=-72$

$(-72)/4 -2$	$=$	$(-72)/3 +4$

$-18-2$	$=$	$-24+4$
$-20$	$=$	$-20$

Since the equation is true, the answer is correct.

$m=-72$

Question 5 of 5

5. Question
Solve

$(6x-3)/5=(5x+3)/6$
- $x=$ (3)
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English
Chapters
Chapters

Inverse Operations

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

Cross Product

Distributive Property

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

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

First, remove the fractions by using the Cross Product.

$\frac{6\color{#00880A}{x}-3}{5}$ $=$ $\frac{5\color{#00880A}{x}+3}{6}$

$6(6$ $x$ $-3)$ $=$ $5(5$ $x$ $+3)$

Expand both sides of the equation by using the Distributive Property.

$6$ $(6$ $x$ $-3)$ $=$ $5$ $(5$ $x$ $+3)$

$6$ $(6$ $x$ $)-$ $6$ $(3)$ $=$ $5$ $(5$ $x$ $)+$ $5$ $(3)$

$36$ $x$ $-18$ $=$ $25$ $x$ $+15$

Next, move $-18$ to the other side by adding $18$ to both sides of the equation.

$36$ $x$ $-18$ $=$ $25$ $x$ $+15$

$36$ $x$ $-18$ $+18$ $=$ $25$ $x$ $+15$ $+18$

$36$ $x$ $=$ $25$ $x$ $+33$ $-18+18$ cancels out

Then, move $25x$ to the other side by subtracting $25x$ from both sides of the equation.

$36$ $x$ $=$ $25$ $x$ $+33$

$36$ $x$ $-25x$ $=$ $25$ $x$ $+33$ $-25x$

$11$ $x$ $=$ $33$ $25x-25x$ cancels out

Finally, remove $11$ by dividing both sides of the equation by $11$ .

$11$ $x$ $=$ $33$

$11$ $x$ $-:11$ $=$ $33$ $-:11$

$x$ $=$ $3$ $11divide11$ cancels out

Check our work

To confirm our answer, substitute $x=3$ to the original equation.

$(6x-3)/5$ $=$ $(5x+3)/6$

$(6(3)-3)/5$ $=$ $(5(3)+3)/6$ Substitute $x=3$

$(18-3)/5$ $=$ $(15+3)/6$

$15/5$ $=$ $18/6$

$3$ $=$ $3$

Since the equation is true, the answer is correct.

$x=3$

$frac{6a+5a}{10}$	$=$	$4$

$frac{11a}{10}$	$=$	$4$

$frac{11a}{10}$ $times10$	$=$	$4$ $times10$	Multiply both sides by $10$

$frac{10(11a)}{10}$	$=$	$4$ $times10$

$11a$	$=$	$40$	The coefficient $frac{10}{10}$ cancels out

$11a$ $divide11$	$=$	$40$ $divide11$	Divide both sides by $11$
$11$ $a$ $divide11$	$=$	$40$ $divide11$	$times11divide11$ cancels out

$a$	$=$	$40/11$

Equations with Variables on Both Sides (Fractions) 2

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Quiz summary

Information

1. Question

Hint

Excellent!

Inverse Operations

Distributive Property

2. Question

Hint

Great Work!

3. Question

Hint

Good Job!

Inverse Operations

Distributive Property

4. Question

Hint

Excellent!

Inverse Operations

Distributive Property

5. Question

Hint

Well Done!

Inverse Operations

Cross Product

Distributive Property

Quizzes

$frac{3a}{5}+frac{a}{2}$	$=$	$4$

$frac{3a}{5}$ $timesfrac{2}{2}$ $+frac{a}{2}$ $timesfrac{5}{5}$	$=$	$4$

$frac{6a}{10}+frac{5a}{10}$	$=$	$4$

$\frac{6\color{#00880A}{x}-3}{5}$	$=$	$\frac{5\color{#00880A}{x}+3}{6}$

$6(6$ $x$ $-3)$	$=$	$5(5$ $x$ $+3)$

$6$ $(6$ $x$ $-3)$	$=$	$5$ $(5$ $x$ $+3)$
$6$ $(6$ $x$ $)-$ $6$ $(3)$	$=$	$5$ $(5$ $x$ $)+$ $5$ $(3)$
$36$ $x$ $-18$	$=$	$25$ $x$ $+15$

$36$ $x$ $-18$	$=$	$25$ $x$ $+15$
$36$ $x$ $-18$ $+18$	$=$	$25$ $x$ $+15$ $+18$
$36$ $x$	$=$	$25$ $x$ $+33$	$-18+18$ cancels out

$36$ $x$	$=$	$25$ $x$ $+33$
$36$ $x$ $-25x$	$=$	$25$ $x$ $+33$ $-25x$
$11$ $x$	$=$	$33$	$25x-25x$ cancels out

$11$ $x$	$=$	$33$
$11$ $x$ $-:11$	$=$	$33$ $-:11$
$x$	$=$	$3$	$11divide11$ cancels out

$(6x-3)/5$	$=$	$(5x+3)/6$

$(6(3)-3)/5$	$=$	$(5(3)+3)/6$	Substitute $x=3$

$(18-3)/5$	$=$	$(15+3)/6$

$15/5$	$=$	$18/6$

$3$	$=$	$3$