Could you check these two problems?

3sqrt (y+2) = 3 is y = 25
Answers were: true or false
I said true because I did it like this
Square both sides=y + 2 = 3^3
y+2 = 27
y = 25

and sqrt (2y-6) = 3-y for y
answes are:
no solution
5
3,5
3
I said 3 because I did it like this:
8y -15 + y^2 = 0
y^2 + 8y -15 =0
(y-5) (y+3) = 0
Y = 3,5 but five doesn't work only 3
Can you please check-I didn't put all the steps for the second one-it was too long

2 answers

you forgot about the 3 in front of 3√(y+2) when you squared

should be

9(y+2) = 9
y+2 = 1
y = -1

check:
LS = 3√(-1+2) = 3√1 = 3
RS = 3

so y = -1

for the second watch your simplifation, my final quadratic was
y^2 - 8y + 15 = 0
(y-3)(y-5) = 0
y = 3 or y = 5

You should have had y = -3

you are right in the conclusion, only x = 3 verifies.
Thanks I think I somewhat get it-Thanks again for the explanations
Similar Questions
  1. Find f X g , g X f , and f X ff(x) = 3sqrt x-1 g(x) = x^3 + 1 For f(x) the 3 is little above the sqrt sign. My answers were:
    1. answers icon 0 answers
  2. Find f X g , g X f , and f X ff(x) = 3sqrt x-1 g(x) = x^3 + 1 For f(x) the 3 is little above the sqrt sign. My answers were:
    1. answers icon 0 answers
  3. Find f X g , g X f , and f X ff(x) = 3sqrt x-1 g(x) = x^3 + 1 For f(x) the 3 is little above the sqrt sign. My answers were:
    1. answers icon 0 answers
  4. Find f X g , g X f , and f X ff(x) = 3sqrt x-1 g(x) = x^3 + 1 For f(x) the 3 is little above the sqrt sign. My answers were:
    1. answers icon 0 answers
more similar questions