1. solve 40-x / 3 = 4+x
2. simplify fully 4x^2 -6x / 4x^2 - 9
2007-02-25 04:11:45 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
1. 40-x/3= 4+x
To have common denominators multiply 4 + x by 3
40-x= 12 + 3x
Then
40= 12 + 4x
28= 4x
7=x
2. Separate the numerator and denominator into two parts
4x^2 - 6x becomes x (4x -6) and
4x^2 -9 becomes (2x-3) (2x+3)
2007-02-25 04:19:14 · answer #1 · answered by najelpszy 2 · 1⤊ 0⤋
1)Multiply both sides by 3:
40-x=3(4+x)
40-x=12+3x
Subtract 3x from both sides:
40-4x=12
Subtract 40 from both sides:
-4x=-28
Divide -4 from both sides:
x=7
Check:
(40-7)/3=4+7
33/3=11
11=11
2)Factor them out:
2x(2x-3)/(2x+3)(2x-3)
Cancel out the 2x-3:
2x/(2x+3)
2007-02-25 12:28:57 · answer #2 · answered by Anonymous · 0⤊ 0⤋
4+x+x/3 = 40
x+x/3=36
4x/3=36
x=27
2007-02-25 12:14:46 · answer #3 · answered by mradigan747 2 · 0⤊ 0⤋
40 - x/3=4 +x
Multiply by 3 both sides
3(40-x)=3(4+x)
120-3x=12+3x
120-120-3x=12-120+3x\-3x=
-3X=-108+3x
-3x-3x=-108+3x-3x
-6x=-108
-6x/-6=-108/-6
x=18
4x^2-6x/4x^2-9
2x(2x-3)/(2x-3)(2x+3)
2x/2x+3
2007-02-25 12:21:13 · answer #4 · answered by Dave aka Spider Monkey 7 · 0⤊ 0⤋
1. x = 7
2. 2x/2x+3
2007-02-25 12:15:58 · answer #5 · answered by choco taco 3 · 1⤊ 0⤋