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...
 A: 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?
A: 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 :).
