Evaluate 1/( square root of y+6)=-3/(y+2)

$\frac{1}{\sqrt{y + 6}} = - \frac{3}{y + 2}$

Step 1

Cross multiply.

Cross multiply by setting the product of the numerator of the right side and the denominator of the left side equal to the product of the numerator of the left side and the denominator of the right side.

$- 3 \cdot \sqrt{y + 6} = 1 \cdot (y + 2)$

Simplify the left side.

Multiply $- 3$ by $\sqrt{y + 6}$ .

$- 3 \sqrt{y + 6} = 1 \cdot (y + 2)$

Simplify the right side.

Multiply $y + 2$ by $1$ .

$- 3 \sqrt{y + 6} = y + 2$

Step 2

To remove the radical on the left side of the equation, square both sides of the equation.

${(- 3 \sqrt{y + 6})}^{2} = {(y + 2)}^{2}$

Step 3

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{y + 6}$ as ${(y + 6)}^{\frac{1}{2}}$ .

${(- 3 {(y + 6)}^{\frac{1}{2}})}^{2} = {(y + 2)}^{2}$

Simplify the left side.

Simplify ${(- 3 {(y + 6)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $- 3 {(y + 6)}^{\frac{1}{2}}$ .

${(- 3)}^{2} {({(y + 6)}^{\frac{1}{2}})}^{2} = {(y + 2)}^{2}$

Raise $- 3$ to the power of $2$ .

$9 {({(y + 6)}^{\frac{1}{2}})}^{2} = {(y + 2)}^{2}$

Multiply the exponents in ${({(y + 6)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$9 {(y + 6)}^{\frac{1}{2} \cdot 2} = {(y + 2)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$9 {(y + 6)}^{\frac{1}{2} \cdot 2} = {(y + 2)}^{2}$

Rewrite the expression.

$9 {(y + 6)}^{1} = {(y + 2)}^{2}$

Simplify.

$9 (y + 6) = {(y + 2)}^{2}$

Apply the distributive property.

$9 y + 9 \cdot 6 = {(y + 2)}^{2}$

Multiply $9$ by $6$ .

$9 y + 54 = {(y + 2)}^{2}$

Simplify the right side.

Simplify ${(y + 2)}^{2}$ .

Rewrite ${(y + 2)}^{2}$ as $(y + 2) (y + 2)$ .

$9 y + 54 = (y + 2) (y + 2)$

Expand $(y + 2) (y + 2)$ using the FOIL Method.

Apply the distributive property.

$9 y + 54 = y (y + 2) + 2 (y + 2)$

Apply the distributive property.

$9 y + 54 = y \cdot y + y \cdot 2 + 2 (y + 2)$

Apply the distributive property.

$9 y + 54 = y \cdot y + y \cdot 2 + 2 y + 2 \cdot 2$

Simplify and combine like terms.

Simplify each term.

Multiply $y$ by $y$ .

$9 y + 54 = y^{2} + y \cdot 2 + 2 y + 2 \cdot 2$

Move $2$ to the left of $y$ .

$9 y + 54 = y^{2} + 2 \cdot y + 2 y + 2 \cdot 2$

Multiply $2$ by $2$ .

$9 y + 54 = y^{2} + 2 y + 2 y + 4$

Add $2 y$ and $2 y$ .

$9 y + 54 = y^{2} + 4 y + 4$

Step 4

Solve for $y$ .

Since $y$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$y^{2} + 4 y + 4 = 9 y + 54$

Move all terms containing $y$ to the left side of the equation.

Subtract $9 y$ from both sides of the equation.

$y^{2} + 4 y + 4 - 9 y = 54$

Subtract $9 y$ from $4 y$ .

$y^{2} - 5 y + 4 = 54$

Subtract $54$ from both sides of the equation.

$y^{2} - 5 y + 4 - 54 = 0$

Subtract $54$ from $4$ .

$y^{2} - 5 y - 50 = 0$

Factor $y^{2} - 5 y - 50$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 50$ and whose sum is $- 5$ .

$- 10, 5$

Write the factored form using these integers.

$(y - 10) (y + 5) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$y - 10 = 0$

$y + 5 = 0$

Set $y - 10$ equal to $0$ and solve for $y$ .

Set $y - 10$ equal to $0$ .

$y - 10 = 0$

Add $10$ to both sides of the equation.

$y = 10$

Set $y + 5$ equal to $0$ and solve for $y$ .

Set $y + 5$ equal to $0$ .

$y + 5 = 0$

Subtract $5$ from both sides of the equation.

$y = - 5$

The final solution is all the values that make $(y - 10) (y + 5) = 0$ true.

$y = 10, - 5$

Step 5

Exclude the solutions that do not make $\frac{1}{\sqrt{y + 6}} = - \frac{3}{y + 2}$ true.

$y = - 5$

Evaluate 1/( square root of y+6)=-3/(y+2)

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