# Thread: equation with fractions problem

1. ## equation with fractions problem

1
- x - 5 =
2

1
- x + 3 *(I can't write out the equation horizontally without the layout screwing up!)
4

I have no idea what to do with equations and fractions set out like this. Do i need to cancel them out of the equation before i make the first step to solving the unknown?

thanks
J

2. You need to make it clear what you mean.

For example is it: $\dfrac{11}{x-5} = \dfrac{24}{x+3}$

3. the layout of the equation is the fraction (1/2) x - 5 = (1/4) x + 3

*brackets being the fraction

4. Originally Posted by johnnerz
the layout of the equation is the fraction (1/2) x - 5 = (1/4) x + 3

*brackets being the fraction
multiply both sides by 4 to clear the fractions ... then solve for x.

5. That's a better layout.

I would go about this problem by multiplying each term by the lowest common denominator of 2 and 4 to clear the fraction.

6. So is the first step (1/2) x -5 *4 = (1/4) x + 3 * 4 ?

7. Originally Posted by johnnerz
So is the first step (1/2) x -5 *4 = (1/4) x + 3 * 4 ?
$4\left(\dfrac{1}{2}x - 5\right) = 4\left(\dfrac{1}{4}x+3\right)$

8. Originally Posted by johnnerz
So is the first step (1/2) x -5 *4 = (1/4) x + 3 * 4 ?
$\frac{1}{2} x - 5 = \frac{1}{4} x + 3$

$4(\frac{1}{2} x - 5) = 4(\frac{1}{4} x + 3)$

Can you continue? Literally multiply each term by 4.

9. x= 8 ?

10. Originally Posted by johnnerz
1
- x - 5 =
2

1
- x + 3 *(I can't write out the equation horizontally without the layout screwing up!)
4

I have no idea what to do with equations and fractions set out like this. Do i need to cancel them out of the equation before i make the first step to solving the unknown?

thanks
J
Fractions are not as bad as the publicity they get...

$\displaystyle\left(\frac{1}{2}\right)x-5=\left(\frac{1}{4}\right)x+3\Rightarrow\left(\fra c{1}{4}\right)x+\left(\frac{1}{4}\right)x-5=\left(\frac{1}{4}\right)x+3$

Therefore

$\displaystyle\left(\frac{1}{4}\right)x-5=3\Rightarrow\left(\frac{1}{4}\right)x=5+3$

A quarter of what number is 8 ?

11. Originally Posted by johnnerz
x= 8 ?
$4\left(\dfrac{1}{2}x - 5\right) = 4\left(\dfrac{1}{4}x + 3\right)$

$2x - 20 = x + 12$

subtract x from both sides; add 20 to both sides

$x = 32$

you shouldn't have to ask if a solution is correct ... always check solutions by substituting them into the original equation to see if it makes it a true statement.