Posted by **jas20** on Saturday, March 24, 2007 at 8:15pm.

Can someone correct these for me.PLZ

Problem#1

Directions solve equation

ãx+4 = 3

My answer: x = 5

Probelm #2

Directions solve equation

ã(4x+1) + 3 = 0

My answer x = 2

Problem #3

Directions solve equation

ã(2y+7)+4=y

My answer y= 1 and y = 9

the symbol a is a radical it came out wrong.

the first one is correct

Use brackets to show which parts are below the square root sign

in the second:

√(4x+1)+3=0

√(4x+1)=-3

square both sides etc.

You should have been told that if you obtain algebraic answers after "squaring" , all solutions have to be verfied.

If you check your answer of x=2, it does not satisfy the original equation.

So the second equation has no real solution.

The same is true for your third equation.

Even though y=1 and y=9 are algebraic solutions, when you check the answers in the original equation, only y=9 works, the other answer is not valid.

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