(√2y+7)+4=y Where 2y +7 is all under the same sr sign?

Answer 1

Subtract 4 from both sides

√(2y+7) = y-4

Square both sides

2y+7 = y^2-8y+16

Subtract 2y+7 from both sides

y^2-10y+9=0

Factorise (y-9)(y-1)=0

y=1,9

Answer 2

First, isolate the square root: √(2y+7)=y-3 Now square both sides: 2y+7=y²-6y+9 Render this into standard form: 0=y²-8y+2 Complete the square by adding 14: 14=y²-8y+16 Factor: 14=(y-4)² Take the square roots: y-4=±√14 Add 4: y=4±√14 Note that it is not necessarily the case that both of these solutions are correct, since one of the transformations we did (squaring both sides) was not invertible. Therefore, we test: y=4-√14 -- This cannot be a solution, because y-3=1-√14 is negative, but in the original problem, y-3 would have to equal the _positive_ square root of 2y+7. y=4+√14 -- In this case, y-3, which is 1+√14, is positive, so if its square is equal to 2y+7, then it is indeed the positive square root of 2y+7. So we check: (1+√14)²=15+2√14, and 2y+7 = 15+2√14, so this holds. Therefore, this is a solution. Thus your unique solution is y=4+√14

Answer 3

First get the radical by itself:
sqrt(2y+7) = y-4
Then square both sides:
2y+7 = y^2 + 16 - 8y

Combine terms and apply the quadratic formula

Answer 4

(√2y+7)+4=y
subtract 4 from each side
(√2y+7=y-4
Square both sides
2y+7=y^2-8y+16
subtract (2y+7) fromeach side
0=y^2-10y+9
(y-1)*(Y-9)=y^2-10y+9
y=1 or y=9

Answer 5

sqrt(2y+7) = y - 4

square both sides

2y + 7 = y^2 - 8y + 16

y^2 - 10y + 9 = 0

solve this for y and you're done!

Answer 6

(√2y+7)+4=y
(√2y+7) = y-4

square both sides

2y+7 = (y-4)(y-4)
2y+7 = y^2-8y+16
0 = y^2-10y+9
0 = (y-1)(y-9)

Set each set equal to zero:

(y-1)=0
y=1

OR

(y-9)=0
y=9

Hope you were able to follow each step.