Answer :
note:xy means x times y and x(y) means the same thing
so
first we get rid of square root then
make the equation equal to zero becaues if
xy=0 then x or/and y=0
squareroot(y-1)+3=y
isolate the squareroot
subtrac 3 from boht sides
squareroot(y-1)=y-3
square both sides (since they are equal, you should be able to square both sides and still make it true)
(squareroot(y-1))^2=(y-3)^2
(y-1)=(y-3)(y-3)
y-1=y^2-6y+9
subtrac y from both sides
-1=y^2-7y+9
add 1 to both sides
0=y^2-7y+10
find what two number multiply to make 10 and add to get -7
the answer is -2 and -5
0=(y-5)(y-2)
therfore
y-5=0
and/or
y-2=0
therefor
y=5 or/and 2 might work
let's try out 2
square root(2-1)+3=2
square root(1)+3=2
1+3=2
false
so 2 doesn't work
let's try 5
squareroot(5-1)+3=5
squareroot(4)+3=5
2+3=5
5=5
true
y=5
so
first we get rid of square root then
make the equation equal to zero becaues if
xy=0 then x or/and y=0
squareroot(y-1)+3=y
isolate the squareroot
subtrac 3 from boht sides
squareroot(y-1)=y-3
square both sides (since they are equal, you should be able to square both sides and still make it true)
(squareroot(y-1))^2=(y-3)^2
(y-1)=(y-3)(y-3)
y-1=y^2-6y+9
subtrac y from both sides
-1=y^2-7y+9
add 1 to both sides
0=y^2-7y+10
find what two number multiply to make 10 and add to get -7
the answer is -2 and -5
0=(y-5)(y-2)
therfore
y-5=0
and/or
y-2=0
therefor
y=5 or/and 2 might work
let's try out 2
square root(2-1)+3=2
square root(1)+3=2
1+3=2
false
so 2 doesn't work
let's try 5
squareroot(5-1)+3=5
squareroot(4)+3=5
2+3=5
5=5
true
y=5
