Solve for y square root of 2y-5 = square root of y-1

$\sqrt{2 y - 5} = \sqrt{y - 1}$

Step 1

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

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

Step 2

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

Rewrite ${\sqrt{y - 1}}^{2}$ as $y - 1$ .

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

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

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

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

Combine $\frac{1}{2}$ and $2$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$2 y - 5 = y - 1$

Step 3

Solve for $y$ .

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

Subtract $y$ from both sides of the equation.

$2 y - 5 - y = - 1$

Subtract $y$ from $2 y$ .

$y - 5 = - 1$

Move all terms not containing $y$ to the right side of the equation.

Add $5$ to both sides of the equation.

$y = - 1 + 5$

Add $- 1$ and $5$ .

$y = 4$

Solve for y square root of 2y-5 = square root of y-1

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