# Math Help - Faster way to work out multiples

1. ## Faster way to work out multiples

Hi all, my fractions are not great to be honest but here is my question.

Using addition, solve for a and b:

$4a-6b=15$
$6a-4b=10$

So I find a common multiple of b, I choose to use 24.

$16a-24b=60$
$36a-24b=60$

$16a-24b=60$
$-36a+24b=-60$

I'm left with: $-20a=0 == a=0$

Is this correct so far? If so then I plug the answer back into the equation and get:

$4-6b=15$

I only know from using a website that $b=-\frac{5}{2}$

My question is I just don't know how to go about finding the correct fraction, is there a simple method?

2. ## Re: Faster way to work out multiples

After you've found your answers plug them back in the two equations and see if LHS = RHS

3. ## Re: Faster way to work out multiples

I don't know what LHS = RHS means. Could you elaborate?

4. ## Re: Faster way to work out multiples

LHS means left hand side, RHS means right hand side

5. ## Re: Faster way to work out multiples

But my question was with regards to finding the quick way to the correct fraction

6. ## Re: Faster way to work out multiples

Hi Uperkurk.

Your answer also seems to be correct. You are left with $a=0$ So by plugging that into either one of your equations (I'm choosing the first one for my example), you get $4\cdot 0-6b\right)=15$ It seems that by solving this equation you get $b=\frac{-5}{2}$ which is harmonious with the answer.

For a more in-depth view on how to solve the equation

$\\4\cdot 0-6b=15\\\\0-6b=15\\\\-6b=15\\\\b=\frac{15}{-6}\\\\b=-\frac{5}{2}$

Does this clear it up for you?

7. ## Re: Faster way to work out multiples

Originally Posted by Paze
Hi Uperkurk.

Your answer also seems to be correct. You are left with $a=0$ So by plugging that into either one of your equations (I'm choosing the first one for my example), you get $4\cdot 0-6b\right)=15$ It seems that by solving this equation you get $b=\frac{-5}{2}$ which is harmonious with the answer.
$\\4\cdot 0-6b=15\\\\0-6b=15\\\\-6b=15\\\\b=\frac{15}{-6}\\\\b=-\frac{5}{2}$