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I really forgot how to solve a equation like:

$$-4A-3B=-2$$ $$3A-4B=0$$

I tried hard and I got $A=24/25$ and $B=24/75$. Can someone please verify this result? Also would be nice if you can show me some tricks to use in such cases. The only trick I know is to multiply one term by a number to make one part of sum of two terms zero, but I couldnt do it for this case.

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3 Answers 3

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You can multiply both equations by constants before adding. Here if you multiply the first equation by $3$ and the second by $4$, you get

$$\left\{\begin{align*}-12A-9B&=-6\\12A-16B&=0\;,\end{align*}\right.$$

which you can add to get $-25B=-6$. Dividing both sides by $-25$ now gives you $B=\frac6{25}$, which you can substitute into either original equation to solve for $A$. I’ll use the second:

$$3A-4\cdot\frac6{25}=0\;,$$ so $$3A=\frac{24}{25}\;,$$ and $A=\frac8{25}$.

To check a potential answer, just substitute it into the original equations and see whether it ‘works’. Here we have $$-4A-3B=-4\cdot\frac{8}{25}-3\cdot\frac6{25}=\frac{-32-18}{25}=-2\;,$$ as we should, and you can easily check that the second equation is also satisfied by these numbers.

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I am afraid you made a slip somewhere. Cramers' formulae (although it is a slight overkill for this problem, but you asked for tricks) give $$A=\frac{\left|\begin{array}{cc} -2 & -3\\ 0 & -4 \end{array}\right|}{\left|\begin{array}{cc} -4 & -3\\ 3 & -4 \end{array}\right|}=\frac{8}{25}$$ $$B=\frac{\left|\begin{array}{cc} -4 & -2\\ 3 & 0 \end{array}\right|}{\left|\begin{array}{cc} -4 & -3\\ 3 & -4 \end{array}\right|}=\frac{6}{25}$$

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$$-4A-3B=-2$$$$3A-4B=0$$Multiply first eq. by $\,3\,$ and 2nd one by $\,4\,$: $$-12A-9B=-6$$$$12A-16B=0$$Add both eq's above: $$-25B=-6\Longrightarrow B=\frac{6}{25}$$ Substitute now this in any of the eq's above, say the second one: $$3A-4\frac{6}{25}=0\Longrightarrow A=\frac{24}{75}=\frac{8}{25}$$

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