Question

How do you solve `\frac { 56} { x - 2} = x - 3`?

Answer 1

`x_1=10` and `x_2=-5`

Explanation:

`56/(x-2)=x-3`

`56=(x-2)(x-3)`

`56=x^2-3x-2x+6`

`x^2-5x-50=0`

Discriminant is `Δ=b^2-4ac=(-5)^2-4*1*(-50)=25+200=225`

`x_(1,2)=(-b+-sqrtΔ)/(2a)=(5+-sqrt225)/2=(5+-15)/2`

Therefore, `x_1=10` and `x_2=-5`

Answer 2

`x=-5, x=10`

Explanation:

Solve like an equation, to get rid of the divide we mutiply. Therefore multiply (x-2) and (x-3)

`(x-2)(x-3)`

`x xx x=x^2`

`x xx -3=-3x`

`-2 xx x=-2x`

`-2 xx -3=6`

Collect like terms:

`-3x-2x=-5x`

Then put into an equation:

`x^2-5x+6`

Then plug into previous equation, replacing what was before, gets us:

`56=x^2-5x+6`

Then as we have a x^2 it would be very hard to solve so therefore we would have to factorise the simplified expression below if possible, in this instance it is so we do. If not complete the square or use the quadratic formula. We take 56 as we want a new equation to be formed:

`x^2-5x-50 (-56)`

Factorising: Needs 2 number that adds to -5, and multiplies to -50. One number should be negative, one should be positive as:
`-1 xx -1=1`

`5-10-5`
`5 xx -10=-50`

Therefore these numbers work, so the factorised form should be:

`(x+5)(x-10)`

But then you finally, you have to solve by setting each bracket equal to 0.

`x+5=0`

`x=-5(-5)`

`x-10=0`

`x=10 (+10)`

So therefore these are the two solutions.

`x=-5, x=+10`