I got stuck solving these two equations: $$a_1(x/z) + b_1 (y/z) + c_1 = 0$$ and, $$a_2(x/z) + b_2 (y/z) + c_2 = 0$$ for $$x/z$$ and $$y/z$$.

The desired result would be: $${x \over z} = {b_1c_2 - b_2c_1\over a_1b_2 - a_2b_1}$$,

$${y \over z} = {c_1a_2 - c_2a_1\over a_1b_2 - a_2b_1}$$

How do I get there? I keep getting dead ends.

Edit: I basically came this far: $$u = v{-b_1 \over a_1} - {c_1 \over a_1}$$ and, $$u = v{-b_2 \over a_2} - {c_2 \over a_2}$$

And ofcourse $$u = x/z$$, $$v = y/z$$

Now i got here, $$v{-b_1 \over a_1} - {c_1 \over a_1} = v{-b_2 \over a_2} - {c_2 \over a_2}$$

Then I took all the v's to one side and factored v outside: $$v({-b_2 \over a_2} + {b_1 \over a_1}) = {c_1 \over a_1} - {c_2 \over a_2}$$

now divide both sides by that b and a thingy on the left, but now it gets messy...

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    $\begingroup$ Can you show which dead end you run into? It should be simple enough to substitute your proposed solutions into the equations and simplify to show that they work. $\endgroup$ Jan 30 '16 at 16:08
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    $\begingroup$ Rename $\frac xz=u$ and $\frac yz=v$ and rewrite the initial equations. $\endgroup$ Jan 30 '16 at 16:08
  • $\begingroup$ Hint: take the last thing, check that it is correct (the second term on the left should be $\frac{b_2}{a_2}$, and the plus on the right should be a minus), then multiply by $a_1a_2$ on both sides then solve for $v$. You just obtained $\frac yz$ as in the solution you give :). Edit You just fixed the sign while I typed. $\endgroup$
    – MickG
    Jan 30 '16 at 16:44
  • $\begingroup$ I suppose I am stuck on how to divide c1/a1 - c2/a2 by that thing on the left now. Any clever way to do that? $\endgroup$ Jan 30 '16 at 16:56
  • $\begingroup$ See my answer below. I started from a little earlier than that equation, but to make the division, I suggest you multiply both sides by $a_1a_2$, and then divide. The multiplication eliminates all denominators. This is why it is advisable. $\endgroup$
    – MickG
    Jan 30 '16 at 17:04

HINT: with $$\frac xz=u$$ and $$\frac yz=v$$ we get $$a_1u+b_1v+c_1=0$$ $$a_2u+b_2v+c_2=0$$ if $$b_1\ne 0$$ we have $$v=-\frac{c_1}{b_1}-\frac{a_1}{b_1}u$$ and we get an equation for $$u$$ $$a_2u+b_2\left(-\frac{c_1}{b_1}-\frac{a_1}{b_1}u\right)+c_2=0$$ can you proceed?

  • $\begingroup$ it seems we get a different thing, did I do something wrong in my original post? I edited it in. $\endgroup$ Jan 30 '16 at 16:42
  • $\begingroup$ I urge you to decide if you mean to use slash fractions $a/b$ or normal fractions $\frac ab$, because mixing the two can be quite confusing for readers. @TheProgramMAN123 I think you just followed a different path than he did. You solved both for $u$ and then equalled the two expressions, he solved one for $v$ and then substituted the expression in the other one. $\endgroup$
    – MickG
    Jan 30 '16 at 16:48
  • $\begingroup$ If you solve the last equation in this post for $u$, you get the solution in the question, with signs changed both in numerator and in the denominator. $\endgroup$
    – MickG
    Jan 30 '16 at 16:51

Let me just continue your attempt. You got to:

$$\left\{\begin{array}{c} u=-\frac{b_1}{a_1}v-\frac{c_1}{a_1} \\ u=-\frac{b_2}{a_2}v-\frac{c_2}{a_2} \\ \end{array}\right.$$

If we equal these two expressions for $u$ and multiply both sides by $a_1a_2$, we get:


Bringing the RHS's $v$ term to the left and the LHS's "constant" term to the right, this becomes:


Yes, I also factored out $v$ from the LHS and reordered the factors in the various $a,b,c$ products. Is this not precisely the solution you gave, that is:

$$\frac yz=v=\frac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}?$$

Wonderful! To complete, we substitute this into the equation for $u$, one of the two, I mean. To simplify the calculations, we multiply by $a_1$ the first equation of the system at the start of the answer, and then substitute, getting:


Let us make that RHS into a single fraction:


The overlines simply indicate those two terms cancel. Oh, but if we divide both sides by $a_1$, we get the desired result, don't we? Very good. We are done.

Bottom line: when you have coefficients with fractions, the trick is often to sum all fractions and see if you can multiply/divide to simplify numerators or denominators. Don't be afraid of complicated coefficients: just have patience and do the algebra to the end, and you will get to the desired result :).

  • $\begingroup$ thanks a bunch! I guess the lesson for me here is to remove those fractions as soon as you can. They make things confusing and messy. i guess multiplying both sides by a1a2 is a clever way to do that. the solution really is painfully simple then. $\endgroup$ Jan 30 '16 at 18:29

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