Solve for y
y=4
y=-2
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\sqrt{10y+24}=y+1+3
Subtract -3 from both sides of the equation.
\sqrt{10y+24}=y+4
Add 1 and 3 to get 4.
\left(\sqrt{10y+24}\right)^{2}=\left(y+4\right)^{2}
Square both sides of the equation.
10y+24=\left(y+4\right)^{2}
Calculate \sqrt{10y+24} to the power of 2 and get 10y+24.
10y+24=y^{2}+8y+16
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+4\right)^{2}.
10y+24-y^{2}=8y+16
Subtract y^{2} from both sides.
10y+24-y^{2}-8y=16
Subtract 8y from both sides.
2y+24-y^{2}=16
Combine 10y and -8y to get 2y.
2y+24-y^{2}-16=0
Subtract 16 from both sides.
2y+8-y^{2}=0
Subtract 16 from 24 to get 8.
-y^{2}+2y+8=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=2 ab=-8=-8
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -y^{2}+ay+by+8. To find a and b, set up a system to be solved.
-1,8 -2,4
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 -8.
-1+8=7 -2+4=2
Calculate the sum for each pair.
a=4 b=-2
The solution is the pair that gives sum 2.
\left(-y^{2}+4y\right)+\left(-2y+8\right)
Rewrite -y^{2}+2y+8 as \left(-y^{2}+4y\right)+\left(-2y+8\right).
-y\left(y-4\right)-2\left(y-4\right)
Factor out -y in the first and -2 in the second group.
\left(y-4\right)\left(-y-2\right)
Factor out common term y-4 by using distributive property.
y=4 y=-2
To find equation solutions, solve y-4=0 and -y-2=0.
\sqrt{10\times 4+24}-3=4+1
Substitute 4 for y in the equation \sqrt{10y+24}-3=y+1.
5=5
Simplify. The value y=4 satisfies the equation.
\sqrt{10\left(-2\right)+24}-3=-2+1
Substitute -2 for y in the equation \sqrt{10y+24}-3=y+1.
-1=-1
Simplify. The value y=-2 satisfies the equation.
y=4 y=-2
List all solutions of \sqrt{10y+24}=y+4.
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