# Equivalent fractions question [Solved!]

**Michael** 24 Nov 2015, 10:03

### My question

Is your answer to exercise 4 on Equivalent Fractions correct? If you multiply by minus on the bottom must you not also multiply by minus on top? Substituting real numbers for the algebraic terms does not work out!!

### Relevant page

5. Equivalent Fractions

### What I've done so far

Tried the examples on the above page.

X

Is your answer to exercise 4 on Equivalent Fractions correct? If you multiply by minus on the bottom must you not also multiply by minus on top? Substituting real numbers for the algebraic terms does not work out!!

Relevant page
<a href="/factoring-fractions/5-equivalent-fractions.php">5. Equivalent Fractions</a>
What I've done so far
Tried the examples on the above page.

## Re: Equivalent fractions question

**Murray** 25 Nov 2015, 04:18

Hi Michael

Good question. I have modified the solution on that page and I hope it makes sense now why I multiply the bottom of the fraction by minus, but not the top.

As for substituting, let's try `x = 2` and `y = 5`

In the original question:

`\text{LHS} = \frac{2^2 - 5^2}{5 - 2} = \frac{4 - 25}{3} = -\frac{21}{3} = -7`

RHS (the final answer) `= -2 - 5 = -7`

Hope that makes sense

Regards

Murray Bourne

X

Hi Michael
Good question. I have modified the solution on that page and I hope it makes sense now why I multiply the bottom of the fraction by minus, but not the top.
As for substituting, let's try `x = 2` and `y = 5`
In the original question:
`\text{LHS} = \frac{2^2 - 5^2}{5 - 2} = \frac{4 - 25}{3} = -\frac{21}{3} = -7`
RHS (the final answer) `= -2 - 5 = -7`
Hope that makes sense
Regards
Murray Bourne

## Re: Equivalent fractions question

**Michael** 25 Nov 2015, 14:38

Thanks for your help, as always

X

Thanks for your help, as always

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