Solve sqrt{3y+51}+7=y+6 | Microsoft Math Solver

Solve for y

Solution Steps

Graph

Quiz

Algebra

5 problems similar to:

Similar Problems from Web Search

https://socratic.org/questions/how-do-you-simplify-root3y-5-root3y

https://www.tiger-algebra.com/drill/sqrt(3y_4)=-2y_12/

https://socratic.org/questions/how-do-you-solve-sqrt-3y-2-sqrt-y-5

Copied to clipboard

\sqrt{3y+51}=y+6-7

Subtract 7 from both sides of the equation.

\sqrt{3y+51}=y-1

Subtract 7 from 6 to get -1.

\left(\sqrt{3y+51}\right)^{2}=\left(y-1\right)^{2}

Square both sides of the equation.

3y+51=\left(y-1\right)^{2}

Calculate \sqrt{3y+51} to the power of 2 and get 3y+51.

3y+51=y^{2}-2y+1

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

3y+51-y^{2}=-2y+1

Subtract y^{2} from both sides.

3y+51-y^{2}+2y=1

Add 2y to both sides.

5y+51-y^{2}=1

Combine 3y and 2y to get 5y.

5y+51-y^{2}-1=0

Subtract 1 from both sides.

5y+50-y^{2}=0

Subtract 1 from 51 to get 50.

-y^{2}+5y+50=0

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

a+b=5 ab=-50=-50

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -y^{2}+ay+by+50. To find a and b, set up a system to be solved.

-1,50 -2,25 -5,10

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -50.

-1+50=49 -2+25=23 -5+10=5

Calculate the sum for each pair.

a=10 b=-5

The solution is the pair that gives sum 5.

\left(-y^{2}+10y\right)+\left(-5y+50\right)

Rewrite -y^{2}+5y+50 as \left(-y^{2}+10y\right)+\left(-5y+50\right).

-y\left(y-10\right)-5\left(y-10\right)

Factor out -y in the first and -5 in the second group.

\left(y-10\right)\left(-y-5\right)

Factor out common term y-10 by using distributive property.

y=10 y=-5

To find equation solutions, solve y-10=0 and -y-5=0.