Solve sqrt{8y+4}=sqrt{7y+7}

$$\sqrt { 8 y + 4 } = \sqrt { 7 y + 7 }$$

Solve for y

$$y=3$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{8y+4}\right)^{2}=\left(\sqrt{7y+7}\right)^{2}$$

Calculate \sqrt{8y+4} to the power of 2 and get 8y+4.

$$8y+4=\left(\sqrt{7y+7}\right)^{2}$$

Calculate \sqrt{7y+7} to the power of 2 and get 7y+7.

$$8y+4=7y+7$$

Subtract 7y from both sides.

$$8y+4-7y=7$$

Combine 8y and -7y to get y.

$$y+4=7$$

Subtract 4 from both sides.

$$y=7-4$$

Subtract 4 from 7 to get 3.

$$y=3$$

Substitute 3 for y in the equation \sqrt{8y+4}=\sqrt{7y+7}.

$$\sqrt{8\times 3+4}=\sqrt{7\times 3+7}$$

Simplify. The value y=3 satisfies the equation.

$$2\times 7^{\frac{1}{2}}=2\times 7^{\frac{1}{2}}$$

Equation \sqrt{8y+4}=\sqrt{7y+7} has a unique solution.

$$y=3$$