Solve y-1=sqrt{-y+7}

$$y - 1 = \sqrt { - y + 7 }$$

Solve for y

$$y=3$$

Solution Steps

Square both sides of the equation.

$$\left(y-1\right)^{2}=\left(\sqrt{-y+7}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-1\right)^{2}.

$$y^{2}-2y+1=\left(\sqrt{-y+7}\right)^{2}$$

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

$$y^{2}-2y+1=-y+7$$

Add y to both sides.

$$y^{2}-2y+1+y=7$$

Combine -2y and y to get -y.

$$y^{2}-y+1=7$$

Subtract 7 from both sides.

$$y^{2}-y+1-7=0$$

Subtract 7 from 1 to get -6.

$$y^{2}-y-6=0$$

To solve the equation, factor y^{2}-y-6 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.

$$a+b=-1 ab=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

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

$$\left(y-3\right)\left(y+2\right)$$

To find equation solutions, solve y-3=0 and y+2=0.

$$y=3 y=-2$$

Substitute 3 for y in the equation y-1=\sqrt{-y+7}.

$$3-1=\sqrt{-3+7}$$

Simplify. The value y=3 satisfies the equation.

$$2=2$$

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

$$-2-1=\sqrt{-\left(-2\right)+7}$$

Simplify. The value y=-2 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-3=3$$

Equation y-1=\sqrt{7-y} has a unique solution.

$$y=3$$