# Solve equations asap please?

1

3/(1-3x) + 5/(2-x )+ (9-2x)/(3x^2-7x+2) =0

#2

2/(2-y) + 3/(y+2) = 2y/ (y^2-4)

#3

1/3= 3/(x+2) - 2/(x-3)
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asked Dec 28, 2012

1 ) 3/(1-3x) + 5/(2-x )+ (9-2x)/(3x^2-7x+2) =0

The LCM of 1-3x, 2-x and 3x^2-7x+2 is 3x^2-7x+2

(3(2-x) + 5(1-3x) + (9-2x))3x^2-7x+2 = 0

3(2-x) + 5(1-3x) + (9-2x) = 0

6 - 3x + 5 - 15x + 9 - 2x=0

-20x + 20 = 0

x = 1

answered Dec 28, 2012

2) 2/(2-y) + 3/(y+2) = 2y/ (y^2-4)

Subtract 2y/ (y^2-4) from each side

-2/(y-2) + 3/(y+2) - 2y/ (y^2-4) = 0

The LCM of y - 2 , y+2 and  y^2 - 4 is y^2 - 4

Multiply each side by y^2 - 4

(-2(y+2) + 3(y - 2) -2y ) y^2-4= 0

- 2(y+2) + 3(y - 2) -2y = 0

-2y - 4 +3y - 6 - 2y = 0

-y - 10 = 0

y = -10

answered Dec 28, 2012

3) 1/3= 3/(x+2) - 2/(x-3)

3/(x+2) - 2/(x-3) = 1/3

Subtract 1/3 from each side

3/(x+2) - 2/(x-3) - 1/3 = 0

The LCM of x+2, x-3 and 3 is 3(x^2 - x - 6)

Multiply each side by 3(x^2 - x - 6)

3(x-3)3 - 2 ( x+2)3 - 1 (x^2 - x - 6) = 0

9x - 27 - 6x - 12 - x^2 + x + 6 = 0

-x^2 + 4x + 33 = 0

x^2 - 4x - 33 = 0

x =( 4 ± sqrt (166-132))2

x = 2 ±i sqrt i(29)

x = 2  + i sqrt i(29) and 2 - i sqrt i(29)

answered Dec 28, 2012

) 3/(1-3x) + 5/(2-x )+ (9-2x)/(3x^2-7x+2) =0

The LCM of 1-3x, 2-x and 3x^2-7x+2 is 3x^2-7x+2

(3(2-x) + 5(1-3x) + (9-2x))/3x^2-7x+2 = 0

3(2-x) + 5(1-3x) + (9-2x) = 0

6 - 3x + 5 - 15x + 9 - 2x=0

-20x + 20 = 0

x = 1

answered Dec 29, 2012
+1 vote

3) 1/3= 3/(x+2) - 2/(x-3)

3/(x+2) - 2/(x-3) = 1/3

Subtract 1/3 from each side

3/(x+2) - 2/(x-3) - 1/3 = 0

The LCM of x+2, x-3 and 3 is 3(x^2 - x - 6)

Multiply each side by 3(x^2 - x - 6)

3(x-3)3 - 2 ( x+2)3 - 1 (x^2 - x - 6) = 0

9x - 27 - 6x - 12 - x^2 + x + 6 = 0

-x^2 + 4x + 33 = 0

x^2 - 4x - 33 = 0

x =( 4 ± sqrt (16-132))/2

x = 2 ±i sqrt (29)

x = 2  + i sqrt(29) and 2 - i sqrt (29)

answered Dec 29, 2012