Maths equation help please!

I have the question:

Show that the equation 5/x+2 = 4-3x/x-1

can be arranged to give 3x^2+7x-13=0




When I try to work it out I get 4x^2+4x-1=0

Could someone please tell me where I'm going wrong and how to work it out properly?

Thank you

multiply both sides by the denominator of the other side to start with:
5(x-1) = (4-3x)(x+2)
now multiply out the brackets:
5x - 5 = 4(x+2) - 3x(x+2)
5x - 5 = 4x + 8 - 3x^2 - 6x
now put them all on one side of the equals sign (here it would be best to put them all on the left because if we can we want to try and make any x^2 terms positive):
5x - 5 - 4x - 8 + 3x^2 + 6x = 0
now simplify by grouping like terms:
3x^2 + 7x - 13 = 0

so yeah I'm pretty sure you're right, maybe the answer is wrong :/

Q. Show that the equation 5/x+2 = 4-3x/x-1

can be arranged to give 3x^2+7x-13=0
Sol: 5/(x+2) = {(4-3x)/(x-1)
by cross multiplication
5(x-1) = (x+2)(4-3x)
5x -5 = 4x -3x^2 +8 -6x
3x^2 +7x -13 = 0...................Ans

multiply it all through by x+2 to get
5 = (4-3x)(x+2)/x-1
then multiply through by x-1
5(x-1) = (4-3x)(x+2)

multiply out the brackets
5x-5 = 4x+8-3x^2-6x

move all to the right hand side
5x-5-4x-8+3x^2+6x =0
Simplify:
3x^2+7x-13 =0

5/x + 3 = 4 - 3x/x

5 + 3x = 4x - 3x

5 + 3x = x

5 = -2x

x = -2.5

I am guessing there needs to be some parenthesis in the original equation.

5(x-1)=(x+2)(4-3x)
5x-5=4x-3x@square+8-6x
5x+1=-2x-3xSQUARE
7x+1=-3x@SQUARE........
well what does '^'sign mean?