I thought that solving a 2x2 linear system of equations using either the substitution method or the addition method (adding the two equations to eliminate a variable, then using the substitution method to get the value of the other variable) would lead to the same result even if the equations contained parameters (which I treat like any regular numbers). The two equations I was trying to solve were:
1) $ax-by = a^2+b^2$
2) $2bx-ay = 2b^2+3ab-a^2$
($x$ and $y$ are the variables, $a$ and $b$ are the parameters)
While trying to solve this system using just the substitution method, I must've made a mistake along the way, but it reminded me that I should (?) be able to technically get the same answer, even though using the addition method requires less work in this particular case. So I set up two simpler equations to test this:
1) $xa+yb = 3b$
2) $xb+ya = 20a$
When I solve this supposedly simpler system using the addition method (multiply (1) by $a$ and (2) by $-b$ to eliminate $y$, add, isolate $x$ and simplify etc.), I get:
$x = -{17ab\over a^2-b^2}$ and
$y = 20 + {17b^2\over a^2-b^2}$
(details on how I got these answers below)
This seemed correct to me because it means that for any $a$ and $b$ there is a solution, except when $a^2 = b^2$ (part of this exercise was learning to identify parameter values for which the system does not have a solution).
When I try to solve this using just the substitution method however, I get completely different answers. What am I doing wrong? Below is my detailed work:
Substitution method:
Isolate $x$ in (2):
(2) $xb + ya = 20a$
$\Rightarrow xb = 20a - ya$
$\Rightarrow x = {a(20-y)\over b}$
Now substitute $x$ into (1) and isolate $y$ (UPDATED):
(1) $xa + yb = 3b$
$\Rightarrow ({a(20-y)\over b})a + yb = 3b$
$\Rightarrow {a^2(20-y)\over b} + yb = 3b$ (multiply by $b$)
$\Rightarrow {a^2(20-y)} + yb^2 = 3b^2$ (multiply out parentheses)
$\Rightarrow 20a^2-ya^2+yb^2 = 3b^2$ (subtract $20a^2$)
$\Rightarrow -ya^2+yb^2 = 3b^2 - 20a^2$ (factor out $y$)
$\Rightarrow y(-a^2+b^2) = 3b^2 - 20a^2$ (divide by $b^2-a^2$)
$\Rightarrow y = {3b^2 - 20a^2\over b^2-a^2}$
Since this answer for $y$ in terms of just $a$ and $b$ is still different from the answer above (when I used the addition method), I got confused...
This is what I did in detail for the addition method:
Eliminate $y$ by multiplying both equations with parameters to make $y$ have opposite coefficients:
(1) $xa+yb=3b \rightarrow *a$
$\Rightarrow xa^2+yab = 3ab$
(2) $xb+ya=20a \rightarrow *-b$
$\Rightarrow -xb^2-yab=-20ab$
Now add the two equations to eliminate $y$, then isolate $x$:
$xa^2-xb^2 = -17ab$ (factor out x from LS)
$\Rightarrow x(a^2-b^2) = -17ab$
$\Rightarrow x = -{17ab\over a^2-b^2}$
Now plug $x$ in terms of $a$ and $b$ into one of the two equations to get $y$:
$(-{17ab\over a^2-b^2})b + ya = 20a$
$\Rightarrow -{17ab^2\over a^2-b^2} + ya = 20a$
$\Rightarrow ya = 20a + {17ab^2\over a^2-b^2}$
$\Rightarrow y = 20 + {17b^2\over a^2-b^2}$
Any help with this would be highly appreciated, thanks in advance!
Your expression $$ y = 3 - {a^2(20-y)\over b^2} $$ still has $y$ showing up on the RHS. So you can not compare it directly to your earlier expression for $y$. You have to finish solving for $y$ first, then compare.