# Thread: Solving an equation where the unknown is included in fractions

1. ## Solving an equation where the unknown is included in fractions

How do you do this equation if someone can please explain this step by step
2x+5 -x-4 =1
x+4 x-5

2. Originally Posted by skml02
How do you do this equation if someone can please explain this step by step
2x+5 -x-4 =1
x+4 x-5
Step 1

multiply both sides by th common denominator

Step b

Set the equation to zero

Step $c_3$

Let me know when you're done with steb b.

3. so I got 2x+5 over 1
and 4x-16 over x-5 = 4
but I don't get set the equation to zero

4. Originally Posted by skml02
so I got 2x+5 over 1
and 4x-16 over x-5 = 4
but I don't get set the equation to zero
The common denominator is (x+4)(x-5), so multiplying both sides by this gives:

$(x+4)(x-5)\left[\frac{2x+5}{x+4}-\frac{x-4}{x-5}\right]=1(x+4)(x-5)$

$(2x+5)(x-5)-(x-4)(x+4)=(x+4)(x-5)$

$2x^2-5x-25-x^2+16=x^2-x-20$

Oh, this is easier than I previously thought...The squared terms dissapear...

$-5x-9=-x-20$

$4x=11$

$x=\frac{11}{4}$