I have a system of linear equations (two equations, two variables) and an unknown coefficient a. I need to find which values of a make the system of equations have a single solution. These are the equations:
ax + 2y = 5
8x + ay = 10
My general approach for this is to isolate both x and y separately after eliminating one and adding the equations together and to then look if a is in any of the denominators which could cause a null division. I'm having trouble finding a way to eliminate the x or y variable, so when I add the equations together and isolate x or y, the other variable always ends up on the other side and I get wrong answers (checked through graphing the equations). Is there any way I can isolate x or y and then isolate them to find out where a ends up? Thanks in advance!
For example: After adding the equations together (can't eliminate any variables), I get $x(8+a) + y(2+a) = 15$. Then I can isolate x or y to get for example $$x = \frac{15 - y(2+a)}{(8+a)}$$I believe it's wrong to conclude from that step that a cannot have the value -8.
Solution (using @Emilio Novate's method):
After doing the multiplications I isolated x to get: $$x=\frac{5(a-4)}{a^2-16}$$ While 4 works for a, -4 doesn't! Now the same for y. I had to multiply the first equation by -8 and the second by a. After eliminating x, I got $$y = \frac{10a-40}{(a^2-16)}$$ This confirms that the value of -4 for a does not work :)
A value of 4 for a will make both the x and y equations evaluate to 0/0. What is the significance of this?
Hint for your method:
1) Multiply the first equation by $a$
2) multiply the second equation by $-2$
3) sum the two equations and find: $x(a^2-16)=5(a-4)$
4) find for what values of $a$ there is a solution.