Solve for y square root of 6y+15-4=y

$\sqrt{6 y + 15} - 4 = y$

Step 1

Add $4$ to both sides of the equation.

$\sqrt{6 y + 15} = y + 4$

Step 2

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

${\sqrt{6 y + 15}}^{2} = {(y + 4)}^{2}$

Step 3

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{6 y + 15}$ as ${(6 y + 15)}^{\frac{1}{2}}$ .

${({(6 y + 15)}^{\frac{1}{2}})}^{2} = {(y + 4)}^{2}$

Simplify the left side.

Simplify ${({(6 y + 15)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(6 y + 15)}^{\frac{1}{2}})}^{2}$ .

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

${(6 y + 15)}^{\frac{1}{2} \cdot 2} = {(y + 4)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(6 y + 15)}^{\frac{1}{2} \cdot 2} = {(y + 4)}^{2}$

Rewrite the expression.

${(6 y + 15)}^{1} = {(y + 4)}^{2}$

Simplify.

$6 y + 15 = {(y + 4)}^{2}$

Simplify the right side.

Simplify ${(y + 4)}^{2}$ .

Rewrite ${(y + 4)}^{2}$ as $(y + 4) (y + 4)$ .

$6 y + 15 = (y + 4) (y + 4)$

Expand $(y + 4) (y + 4)$ using the FOIL Method.

Apply the distributive property.

$6 y + 15 = y (y + 4) + 4 (y + 4)$

Apply the distributive property.

$6 y + 15 = y \cdot y + y \cdot 4 + 4 (y + 4)$

Apply the distributive property.

$6 y + 15 = y \cdot y + y \cdot 4 + 4 y + 4 \cdot 4$

Simplify and combine like terms.

Simplify each term.

Multiply $y$ by $y$ .

$6 y + 15 = y^{2} + y \cdot 4 + 4 y + 4 \cdot 4$

Move $4$ to the left of $y$ .

$6 y + 15 = y^{2} + 4 \cdot y + 4 y + 4 \cdot 4$

Multiply $4$ by $4$ .

$6 y + 15 = y^{2} + 4 y + 4 y + 16$

Add $4 y$ and $4 y$ .

$6 y + 15 = y^{2} + 8 y + 16$

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.

$y^{2} + 8 y + 16 = 6 y + 15$

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

Subtract $6 y$ from both sides of the equation.

$y^{2} + 8 y + 16 - 6 y = 15$

Subtract $6 y$ from $8 y$ .

$y^{2} + 2 y + 16 = 15$

Subtract $15$ from both sides of the equation.

$y^{2} + 2 y + 16 - 15 = 0$

Subtract $15$ from $16$ .

$y^{2} + 2 y + 1 = 0$

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

$y^{2} + 2 y + 1^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 y = 2 \cdot y \cdot 1$

Rewrite the polynomial.

$y^{2} + 2 \cdot y \cdot 1 + 1^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = y$ and $b = 1$ .

${(y + 1)}^{2} = 0$

Set the $y + 1$ equal to $0$ .

$y + 1 = 0$

Subtract $1$ from both sides of the equation.

$y = - 1$

Solve for y square root of 6y+15-4=y

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