How to solve for x/z and y/z here?

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...

• 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. Jan 30 '16 at 16:08
• Rename $\frac xz=u$ and $\frac yz=v$ and rewrite the initial equations. Jan 30 '16 at 16:08
• 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. Jan 30 '16 at 16:44
• 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? Jan 30 '16 at 16:56
• 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. 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?

• it seems we get a different thing, did I do something wrong in my original post? I edited it in. Jan 30 '16 at 16:42
• 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. Jan 30 '16 at 16:48
• 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. 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:

$$-b_1a_2v-c_1a_2=-b_2a_1v-c_2a_1.$$

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

$$v(a_1b_2-a_2b_1)v=a_2c_1-a_1c_1.$$

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:

$$a_1u=-b_1\frac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}-c_1.$$

Let us make that RHS into a single fraction:

$$a_1u=\frac{\overline{-b_1a_2c_1}+b_1a_1c_2-c_1a_1b_2+\overline{c_1a_2b_1}}{a_1b_2-b_1a_2}=\frac{a_1(b_1c_2-c_1b_2)}{a_1b_2-b_1a_2}.$$

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 :).

• 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. Jan 30 '16 at 18:29