Solve sqrt{y+1}+sqrt{2y-5}=3

$$\sqrt { y + 1 } + \sqrt { 2 y - 5 } = 3$$

Solve for y

$$y=3$$

Solution Steps

Subtract \sqrt{2y-5} from both sides of the equation.

$$\sqrt{y+1}=3-\sqrt{2y-5}$$

Square both sides of the equation.

$$\left(\sqrt{y+1}\right)^{2}=\left(3-\sqrt{2y-5}\right)^{2}$$

Calculate \sqrt{y+1} to the power of 2 and get y+1.

$$y+1=\left(3-\sqrt{2y-5}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-\sqrt{2y-5}\right)^{2}.

$$y+1=9-6\sqrt{2y-5}+\left(\sqrt{2y-5}\right)^{2}$$

Calculate \sqrt{2y-5} to the power of 2 and get 2y-5.

$$y+1=9-6\sqrt{2y-5}+2y-5$$

Subtract 5 from 9 to get 4.

$$y+1=4-6\sqrt{2y-5}+2y$$

Subtract 4+2y from both sides of the equation.

$$y+1-\left(4+2y\right)=-6\sqrt{2y-5}$$

To find the opposite of 4+2y, find the opposite of each term.

$$y+1-4-2y=-6\sqrt{2y-5}$$

Subtract 4 from 1 to get -3.

$$y-3-2y=-6\sqrt{2y-5}$$

Combine y and -2y to get -y.

$$-y-3=-6\sqrt{2y-5}$$

Square both sides of the equation.

$$\left(-y-3\right)^{2}=\left(-6\sqrt{2y-5}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-y-3\right)^{2}.

$$y^{2}+6y+9=\left(-6\sqrt{2y-5}\right)^{2}$$

Expand \left(-6\sqrt{2y-5}\right)^{2}.

$$y^{2}+6y+9=\left(-6\right)^{2}\left(\sqrt{2y-5}\right)^{2}$$

Calculate -6 to the power of 2 and get 36.

$$y^{2}+6y+9=36\left(\sqrt{2y-5}\right)^{2}$$

Calculate \sqrt{2y-5} to the power of 2 and get 2y-5.

$$y^{2}+6y+9=36\left(2y-5\right)$$

Use the distributive property to multiply 36 by 2y-5.

$$y^{2}+6y+9=72y-180$$

Subtract 72y from both sides.

$$y^{2}+6y+9-72y=-180$$

Combine 6y and -72y to get -66y.

$$y^{2}-66y+9=-180$$

Add 180 to both sides.

$$y^{2}-66y+9+180=0$$

Add 9 and 180 to get 189.

$$y^{2}-66y+189=0$$

To solve the equation, factor y^{2}-66y+189 using formula y^{2}+\left(a+b\right)y+ab=\left(y+a\right)\left(y+b\right). To find a and b, set up a system to be solved.

$$a+b=-66 ab=189$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 189.

$$-1,-189 -3,-63 -7,-27 -9,-21$$

Calculate the sum for each pair.

$$-1-189=-190 -3-63=-66 -7-27=-34 -9-21=-30$$

The solution is the pair that gives sum -66.

$$a=-63 b=-3$$

Rewrite factored expression \left(y+a\right)\left(y+b\right) using the obtained values.

$$\left(y-63\right)\left(y-3\right)$$

To find equation solutions, solve y-63=0 and y-3=0.

$$y=63 y=3$$

Substitute 63 for y in the equation \sqrt{y+1}+\sqrt{2y-5}=3.

$$\sqrt{63+1}+\sqrt{2\times 63-5}=3$$

Simplify. The value y=63 does not satisfy the equation.

$$19=3$$

Substitute 3 for y in the equation \sqrt{y+1}+\sqrt{2y-5}=3.

$$\sqrt{3+1}+\sqrt{2\times 3-5}=3$$

Simplify. The value y=3 satisfies the equation.

$$3=3$$

Equation \sqrt{y+1}=-\sqrt{2y-5}+3 has a unique solution.

$$y=3$$