Solve by Factoring y+1 = square root of 5y-1

$y + 1 = \sqrt{5 y - 1}$

Step 1

Subtract $\sqrt{5 y - 1}$ from both sides of the equation.

$y + 1 - \sqrt{5 y - 1} = 0$

Step 2

Move all terms not containing $- \sqrt{5 y - 1}$ to the right side of the equation.

Subtract $y$ from both sides of the equation.

$1 - \sqrt{5 y - 1} = - y$

Subtract $1$ from both sides of the equation.

$- \sqrt{5 y - 1} = - y - 1$

Step 3

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

${(- \sqrt{5 y - 1})}^{2} = {(- y - 1)}^{2}$

Step 4

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5 y - 1}$ as ${(5 y - 1)}^{\frac{1}{2}}$ .

${(- {(5 y - 1)}^{\frac{1}{2}})}^{2} = {(- y - 1)}^{2}$

Simplify the left side.

Simplify ${(- {(5 y - 1)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $- {(5 y - 1)}^{\frac{1}{2}}$ .

${(- 1)}^{2} {({(5 y - 1)}^{\frac{1}{2}})}^{2} = {(- y - 1)}^{2}$

Raise $- 1$ to the power of $2$ .

$1 {({(5 y - 1)}^{\frac{1}{2}})}^{2} = {(- y - 1)}^{2}$

Multiply ${({(5 y - 1)}^{\frac{1}{2}})}^{2}$ by $1$ .

${({(5 y - 1)}^{\frac{1}{2}})}^{2} = {(- y - 1)}^{2}$

Multiply the exponents in ${({(5 y - 1)}^{\frac{1}{2}})}^{2}$ .

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

${(5 y - 1)}^{\frac{1}{2} \cdot 2} = {(- y - 1)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(5 y - 1)}^{\frac{1}{2} \cdot 2} = {(- y - 1)}^{2}$

Rewrite the expression.

${(5 y - 1)}^{1} = {(- y - 1)}^{2}$

Simplify.

$5 y - 1 = {(- y - 1)}^{2}$

Simplify the right side.

Simplify ${(- y - 1)}^{2}$ .

Rewrite ${(- y - 1)}^{2}$ as $(- y - 1) (- y - 1)$ .

$5 y - 1 = (- y - 1) (- y - 1)$

Expand $(- y - 1) (- y - 1)$ using the FOIL Method.

Apply the distributive property.

$5 y - 1 = - y (- y - 1) - 1 (- y - 1)$

Apply the distributive property.

$5 y - 1 = - y (- y) - y \cdot - 1 - 1 (- y - 1)$

Apply the distributive property.

$5 y - 1 = - y (- y) - y \cdot - 1 - 1 (- y) - 1 \cdot - 1$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$5 y - 1 = - 1 \cdot - 1 y \cdot y - y \cdot - 1 - 1 (- y) - 1 \cdot - 1$

Multiply $y$ by $y$ by adding the exponents.

Move $y$ .

$5 y - 1 = - 1 \cdot - 1 (y \cdot y) - y \cdot - 1 - 1 (- y) - 1 \cdot - 1$

Multiply $y$ by $y$ .

$5 y - 1 = - 1 \cdot - 1 y^{2} - y \cdot - 1 - 1 (- y) - 1 \cdot - 1$

Multiply $- 1$ by $- 1$ .

$5 y - 1 = 1 y^{2} - y \cdot - 1 - 1 (- y) - 1 \cdot - 1$

Multiply $y^{2}$ by $1$ .

$5 y - 1 = y^{2} - y \cdot - 1 - 1 (- y) - 1 \cdot - 1$

Multiply $- y \cdot - 1$ .

Multiply $- 1$ by $- 1$ .

$5 y - 1 = y^{2} + 1 y - 1 (- y) - 1 \cdot - 1$

Multiply $y$ by $1$ .

$5 y - 1 = y^{2} + y - 1 (- y) - 1 \cdot - 1$

Multiply $- 1 (- y)$ .

Multiply $- 1$ by $- 1$ .

$5 y - 1 = y^{2} + y + 1 y - 1 \cdot - 1$

Multiply $y$ by $1$ .

$5 y - 1 = y^{2} + y + y - 1 \cdot - 1$

Multiply $- 1$ by $- 1$ .

$5 y - 1 = y^{2} + y + y + 1$

Add $y$ and $y$ .

$5 y - 1 = y^{2} + 2 y + 1$

Step 5

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.

$y^{2} + 2 y + 1 = 5 y - 1$

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

Subtract $5 y$ from both sides of the equation.

$y^{2} + 2 y + 1 - 5 y = - 1$

Subtract $5 y$ from $2 y$ .

$y^{2} - 3 y + 1 = - 1$

Add $1$ to both sides of the equation.

$y^{2} - 3 y + 1 + 1 = 0$

Add $1$ and $1$ .

$y^{2} - 3 y + 2 = 0$

Factor $y^{2} - 3 y + 2$ 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 $2$ and whose sum is $- 3$ .

$- 2, - 1$

Write the factored form using these integers.

$(y - 2) (y - 1) = 0$

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

$y - 2 = 0$

$y - 1 = 0$

Set $y - 2$ equal to $0$ and solve for $y$ .

Set $y - 2$ equal to $0$ .

$y - 2 = 0$

Add $2$ to both sides of the equation.

$y = 2$

Set $y - 1$ equal to $0$ and solve for $y$ .

Set $y - 1$ equal to $0$ .

$y - 1 = 0$

Add $1$ to both sides of the equation.

$y = 1$

The final solution is all the values that make $(y - 2) (y - 1) = 0$ true.

$y = 2, 1$

Solve by Factoring y+1 = square root of 5y-1

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