Solve sqrt{2y+7}=4-y

$$\sqrt { 2 y + 7 } = 4 - y$$

Solve for y

$$y=1$$

Solution Steps

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-y\right)^{2}.

$$2y+7=16-8y+y^{2}$$

Subtract 16 from both sides.

$$2y+7-16=-8y+y^{2}$$

Subtract 16 from 7 to get -9.

$$2y-9=-8y+y^{2}$$

Add 8y to both sides.

$$2y-9+8y=y^{2}$$

Combine 2y and 8y to get 10y.

$$10y-9=y^{2}$$

Subtract y^{2} from both sides.

$$10y-9-y^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-y^{2}+10y-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -y^{2}+ay+by-9. To find a and b, set up a system to be solved.

$$a+b=10 ab=-\left(-9\right)=9$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 10.

$$a=9 b=1$$

Rewrite -y^{2}+10y-9 as \left(-y^{2}+9y\right)+\left(y-9\right).

$$\left(-y^{2}+9y\right)+\left(y-9\right)$$

Factor out -y in -y^{2}+9y.

$$-y\left(y-9\right)+y-9$$

Factor out common term y-9 by using distributive property.

$$\left(y-9\right)\left(-y+1\right)$$

To find equation solutions, solve y-9=0 and -y+1=0.

$$y=9 y=1$$

Substitute 9 for y in the equation \sqrt{2y+7}=4-y.

$$\sqrt{2\times 9+7}=4-9$$

Simplify. The value y=9 does not satisfy the equation because the left and the right hand side have opposite signs.

$$5=-5$$

Substitute 1 for y in the equation \sqrt{2y+7}=4-y.

$$\sqrt{2\times 1+7}=4-1$$

Simplify. The value y=1 satisfies the equation.

$$3=3$$

Equation \sqrt{2y+7}=4-y has a unique solution.

$$y=1$$