Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.

Hello,

I am new to Purplemath, and am experiencing some difficulty with the above subject. Here is the problem:

(Square root y + Square root y+7=7), I raised the entire equation to the power of 2, to elimnate the square root.
This is what I have so far:

y+y+7=49
Subtracted 7 from both sides
Gave me 2y=42
Divided 2 from both sides
Gave me y=21

But when plugged 21 in for y, didn't equal 7.
joeasy0823

Posts: 1
Joined: Sat Apr 07, 2012 5:56 pm

joeasy0823 wrote:(Square root y + Square root y+7=7), I raised the entire equation to the power of 2, to elimnate the square root.
This is what I have so far: y+y+7=49

if U sqr both sides U still have roots:
(sqrt[y] + sqrt[y+7])^2 = 7^2
y + 2sqrt[y[sqrt[y+7] + y+7 = 49
do the sqring rite & it should come out rite:
2sqrt[7]sqrt[y+7]=49-7-2y
2sqrt[7]sqrt[y+7]=42-2y
sqrt[7]sqrt[y+7]=21-y
7[y+7]=(21-y)^2
keep going til U get the ans!

little_dragon

Posts: 152
Joined: Mon Dec 08, 2008 5:18 pm

What you say you started with is sqrt(y)+sqrt(y)+7=7.

Steps can be,
$2sqrt(y)=7-7$
$2sqrt(y)=0$
$sqrt(y)=0$
$y=0$.
jg.allinsymbols

Posts: 41
Joined: Sat Dec 29, 2012 2:42 am