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

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

Solve for y

$$y=10$$

Solution Steps

Subtract 5 from both sides of the equation.

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

Square both sides of the equation.

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

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

$$y^{2}-10y+25=\left(\sqrt{3y-5}\right)^{2}$$

Calculate \sqrt{3y-5} to the power of 2 and get 3y-5.

$$y^{2}-10y+25=3y-5$$

Subtract 3y from both sides.

$$y^{2}-10y+25-3y=-5$$

Combine -10y and -3y to get -13y.

$$y^{2}-13y+25=-5$$

Add 5 to both sides.

$$y^{2}-13y+25+5=0$$

Add 25 and 5 to get 30.

$$y^{2}-13y+30=0$$

To solve the equation, factor y^{2}-13y+30 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.

$$a+b=-13 ab=30$$

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 30.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Calculate the sum for each pair.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

The solution is the pair that gives sum -13.

$$a=-10 b=-3$$

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

$$\left(y-10\right)\left(y-3\right)$$

To find equation solutions, solve y-10=0 and y-3=0.

$$y=10 y=3$$

Substitute 10 for y in the equation y=5+\sqrt{3y-5}.

$$10=5+\sqrt{3\times 10-5}$$

Simplify. The value y=10 satisfies the equation.

$$10=10$$

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

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

Simplify. The value y=3 does not satisfy the equation.

$$3=7$$

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

$$y=10$$