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\sqrt{y-1}=y-3
Subtract 3 from both sides of the equation.
\left(\sqrt{y-1}\right)^{2}=\left(y-3\right)^{2}
Square both sides of the equation.
y-1=\left(y-3\right)^{2}
Calculate \sqrt{y-1} to the power of 2 and get y-1.
y-1=y^{2}-6y+9
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-3\right)^{2}.
y-1-y^{2}=-6y+9
Subtract y^{2} from both sides.
y-1-y^{2}+6y=9
Add 6y to both sides.
7y-1-y^{2}=9
Combine y and 6y to get 7y.
7y-1-y^{2}-9=0
Subtract 9 from both sides.
7y-10-y^{2}=0
Subtract 9 from -1 to get -10.
-y^{2}+7y-10=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=7 ab=-\left(-10\right)=10
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -y^{2}+ay+by-10. To find a and b, set up a system to be solved.
1,10 2,5
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 10.
1+10=11 2+5=7
Calculate the sum for each pair.
a=5 b=2
The solution is the pair that gives sum 7.
\left(-y^{2}+5y\right)+\left(2y-10\right)
Rewrite -y^{2}+7y-10 as \left(-y^{2}+5y\right)+\left(2y-10\right).
-y\left(y-5\right)+2\left(y-5\right)
Factor out -y in the first and 2 in the second group.
\left(y-5\right)\left(-y+2\right)
Factor out common term y-5 by using distributive property.
y=5 y=2
To find equation solutions, solve y-5=0 and -y+2=0.
\sqrt{5-1}+3=5
Substitute 5 for y in the equation \sqrt{y-1}+3=y.
5=5
Simplify. The value y=5 satisfies the equation.
\sqrt{2-1}+3=2
Substitute 2 for y in the equation \sqrt{y-1}+3=y.
4=2
Simplify. The value y=2 does not satisfy the equation.
y=5
Equation \sqrt{y-1}=y-3 has a unique solution.