Solve (3)/(y+8)-2=(4)/(y-2) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\frac{ 3 }{ y+8 } -2= \frac{ 4 }{ y-2 }$$

Solve for y

$y=-\frac{1}{2}=-0.5$
$y=-6$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $y$ cannot be equal to any of the values $-8,2$ since division by zero is not defined. Multiply both sides of the equation by $\left(y-2\right)\left(y+8\right)$, the least common multiple of $y+8,y-2$.

$$\left(y-2\right)\times 3+\left(y-2\right)\left(y+8\right)\left(-2\right)=\left(y+8\right)\times 4$$

Use the distributive property to multiply $y-2$ by $3$.

$$3y-6+\left(y-2\right)\left(y+8\right)\left(-2\right)=\left(y+8\right)\times 4$$

Use the distributive property to multiply $y-2$ by $y+8$ and combine like terms.

$$3y-6+\left(y^{2}+6y-16\right)\left(-2\right)=\left(y+8\right)\times 4$$

Use the distributive property to multiply $y^{2}+6y-16$ by $-2$.

$$3y-6-2y^{2}-12y+32=\left(y+8\right)\times 4$$

Combine $3y$ and $-12y$ to get $-9y$.

$$-9y-6-2y^{2}+32=\left(y+8\right)\times 4$$

Add $-6$ and $32$ to get $26$.

$$-9y+26-2y^{2}=\left(y+8\right)\times 4$$

Use the distributive property to multiply $y+8$ by $4$.

$$-9y+26-2y^{2}=4y+32$$

Subtract $4y$ from both sides.

$$-9y+26-2y^{2}-4y=32$$

Combine $-9y$ and $-4y$ to get $-13y$.

$$-13y+26-2y^{2}=32$$

Subtract $32$ from both sides.

$$-13y+26-2y^{2}-32=0$$

Subtract $32$ from $26$ to get $-6$.

$$-13y-6-2y^{2}=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$-2y^{2}-13y-6=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-2$ for $a$, $-13$ for $b$, and $-6$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$y=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\left(-2\right)\left(-6\right)}}{2\left(-2\right)}$$

Square $-13$.

$$y=\frac{-\left(-13\right)±\sqrt{169-4\left(-2\right)\left(-6\right)}}{2\left(-2\right)}$$

Multiply $-4$ times $-2$.

$$y=\frac{-\left(-13\right)±\sqrt{169+8\left(-6\right)}}{2\left(-2\right)}$$

Multiply $8$ times $-6$.

$$y=\frac{-\left(-13\right)±\sqrt{169-48}}{2\left(-2\right)}$$

Add $169$ to $-48$.

$$y=\frac{-\left(-13\right)±\sqrt{121}}{2\left(-2\right)}$$

Take the square root of $121$.

$$y=\frac{-\left(-13\right)±11}{2\left(-2\right)}$$

The opposite of $-13$ is $13$.

$$y=\frac{13±11}{2\left(-2\right)}$$

Multiply $2$ times $-2$.

$$y=\frac{13±11}{-4}$$

Now solve the equation $y=\frac{13±11}{-4}$ when $±$ is plus. Add $13$ to $11$.

$$y=\frac{24}{-4}$$

Divide $24$ by $-4$.

$$y=-6$$

Now solve the equation $y=\frac{13±11}{-4}$ when $±$ is minus. Subtract $11$ from $13$.

$$y=\frac{2}{-4}$$

Reduce the fraction $\frac{2}{-4}$ to lowest terms by extracting and canceling out $2$.

$$y=-\frac{1}{2}$$

The equation is now solved.

$$y=-6$$ $$y=-\frac{1}{2}$$

Steps for Completing the Square

Variable $y$ cannot be equal to any of the values $-8,2$ since division by zero is not defined. Multiply both sides of the equation by $\left(y-2\right)\left(y+8\right)$, the least common multiple of $y+8,y-2$.

$$\left(y-2\right)\times 3+\left(y-2\right)\left(y+8\right)\left(-2\right)=\left(y+8\right)\times 4$$

Use the distributive property to multiply $y-2$ by $3$.

$$3y-6+\left(y-2\right)\left(y+8\right)\left(-2\right)=\left(y+8\right)\times 4$$

Use the distributive property to multiply $y-2$ by $y+8$ and combine like terms.

$$3y-6+\left(y^{2}+6y-16\right)\left(-2\right)=\left(y+8\right)\times 4$$

Use the distributive property to multiply $y^{2}+6y-16$ by $-2$.

$$3y-6-2y^{2}-12y+32=\left(y+8\right)\times 4$$

Combine $3y$ and $-12y$ to get $-9y$.

$$-9y-6-2y^{2}+32=\left(y+8\right)\times 4$$

Add $-6$ and $32$ to get $26$.

$$-9y+26-2y^{2}=\left(y+8\right)\times 4$$

Use the distributive property to multiply $y+8$ by $4$.

$$-9y+26-2y^{2}=4y+32$$

Subtract $4y$ from both sides.

$$-9y+26-2y^{2}-4y=32$$

Combine $-9y$ and $-4y$ to get $-13y$.

$$-13y+26-2y^{2}=32$$

Subtract $26$ from both sides.

$$-13y-2y^{2}=32-26$$

Subtract $26$ from $32$ to get $6$.

$$-13y-2y^{2}=6$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$-2y^{2}-13y=6$$

Divide both sides by $-2$.

$$\frac{-2y^{2}-13y}{-2}=\frac{6}{-2}$$

Dividing by $-2$ undoes the multiplication by $-2$.

$$y^{2}+\left(-\frac{13}{-2}\right)y=\frac{6}{-2}$$

Divide $-13$ by $-2$.

$$y^{2}+\frac{13}{2}y=\frac{6}{-2}$$

Divide $6$ by $-2$.

$$y^{2}+\frac{13}{2}y=-3$$

Divide $\frac{13}{2}$, the coefficient of the $x$ term, by $2$ to get $\frac{13}{4}$. Then add the square of $\frac{13}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+\frac{13}{2}y+\left(\frac{13}{4}\right)^{2}=-3+\left(\frac{13}{4}\right)^{2}$$

Square $\frac{13}{4}$ by squaring both the numerator and the denominator of the fraction.

$$y^{2}+\frac{13}{2}y+\frac{169}{16}=-3+\frac{169}{16}$$

Add $-3$ to $\frac{169}{16}$.

$$y^{2}+\frac{13}{2}y+\frac{169}{16}=\frac{121}{16}$$

Factor $y^{2}+\frac{13}{2}y+\frac{169}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(y+\frac{13}{4}\right)^{2}=\frac{121}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{13}{4}\right)^{2}}=\sqrt{\frac{121}{16}}$$

Simplify.

$$y+\frac{13}{4}=\frac{11}{4}$$ $$y+\frac{13}{4}=-\frac{11}{4}$$

Subtract $\frac{13}{4}$ from both sides of the equation.

$$y=-\frac{1}{2}$$ $$y=-6$$