HW HELPPPPPPP! MATH!

the answer and work for :
solve: |2x-5|=1 and sq.rt(9-x) +3=x and 4/x+1-3/x+2=1

| 2x - 5 | = 1 WILL be as two solutions :

2x - 5 = 1 -----> 2x = 5 + 1 -----> 2x = 6 -----> x = 3
2x - 5 = -1 -----> 2x = 5 - 1 -----> 2x = 4 -----> x = 2

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√( 9 - x ) + 3 = x
√( 9 - x ) = x - 3 ----> square both sides

9 - x = x^2 - 6x + 3
0 = x^2 - 6x + x + 3 - 9
0 = x^2 - 5x - 6
0 = (x - 6)(x + 1) -----> x = -1 & 6

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[ 4/(x + 1) ] - [ 3/(x + 2) ] = 1 ----> LCD

[ 4(x + 2)/(x + 1)(x + 2) ] - [ 3(x + 1)/(x + 2)(x + 1) ] = 1
[ (4x + 8)/(x + 1)(x + 2) ] - [ (3x + 3)/(x + 2)(x + 1) ] = 1

[ (4x + 8) - (3x + 3) ] / (x + 2)(x + 1) = 1
[ (4x + 8 - 3x - 3) ] / (x + 2)(x + 1) = 1
[ (x + 5) ] / (x + 2)(x + 1) = 1

(x + 5) = (x + 2)(x + 1)
x + 5 = x^2 + x + 2x + 2
x + 5 = x^2 + 3x + 2
0 = x^2 + 3x - x - 5 + 2
0 = x^2 + 2x - 3
0 = (x + 3)(x - 1) ----> x = 1 & - 3

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