Faster way to work out multiples

Oct 2011
170
3
Hi all, my fractions are not great to be honest but here is my question.

Using addition, solve for a and b:

\(\displaystyle 4a-6b=15\)
\(\displaystyle 6a-4b=10\)

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

\(\displaystyle 16a-24b=60\)
\(\displaystyle 36a-24b=60\)

Now I go ahead and do the addition:

\(\displaystyle 16a-24b=60\)
\(\displaystyle -36a+24b=-60\)

I'm left with: \(\displaystyle -20a=0 == a=0\)

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

\(\displaystyle 4-6b=15\)

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

My question is I just don't know how to go about finding the correct fraction, is there a simple method?
 
Last edited:
Jun 2011
68
4
Canada
After you've found your answers plug them back in the two equations and see if LHS = RHS
 
Oct 2011
170
3
I don't know what LHS = RHS means. Could you elaborate?
 
Jun 2011
68
4
Canada
LHS means left hand side, RHS means right hand side
 
Oct 2011
170
3
But my question was with regards to finding the quick way to the correct fraction
 
Nov 2012
379
19
Iceland
Hi Uperkurk.

Your method is correct.

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

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

\(\displaystyle \\4\cdot 0-6b=15\\\\0-6b=15\\\\-6b=15\\\\b=\frac{15}{-6}\\\\b=-\frac{5}{2}\)

Does this clear it up for you?
 
Last edited:
Oct 2011
170
3
Hi Uperkurk.

Your method is correct.

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

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

\(\displaystyle \\4\cdot 0-6b=15\\\\0-6b=15\\\\-6b=15\\\\b=\frac{15}{-6}\\\\b=-\frac{5}{2}\)

Does this clear it up for you?
Perfect, thank you.