finding the solution of this system: a1x+b1y=c1 a2x+b2y=c2?

Answer 1

(a1)x + (b1)y = c1
(a2)x + (b2)y = c2

Let's solve this by elimination. Multiply the first equation by (a2) and the second equation by -a1, to get

(a1)(a2)x + (b1)(a2)y = (c1)(a2)
-a1(a2)x - a1(b2)y = (-a1)(c2)

Adding them together should eliminate the x variable.

(b1)(a2)y - a1(b2)y = (c1)(a2) - (a1)(c2)

Factoring out y, we get

y [(b1)(a2) - (a1)(b2)] = (c1)(a2) - (a1)(c2). Therefore,

y = [(c1)(a2) - (a1)(c2)] / [(b1)(a2) - (a1)(b2)]

Now, let's eliminate the y variable, in the system of equations:

(a1)x + (b1)y = c1
(a2)x + (b2)y = c2

Multiply the first equation by (b2) and the second equation by (-b1), to get

(a1)(b2)x + (b1)(b2)y = (c1)(b2)
(a2)(-b1)x + (-b1)(b2)y = (c2)(-b1)

Adding them together should eliminate the y, leaving us with

(a1)(b2)x - (a2)(b1)x = (c1)(b2) - (c2)(b1)

Factor x out,

x ( (a1)(b2) - (a2)(b1) ) = (c1)(b2) - (c2)(b1)

x = [(c1)(b2) - (c2)(b1)] / [(a1)(b2) - (a2)(b1)]

To conclude, our solutions again are:

x = [(c1)(b2) - (c2)(b1)] / [(a1)(b2) - (a2)(b1)]
y = [(c1)(a2) - (a1)(c2)] / [(b1)(a2) - (a1)(b2)]


Note: This is under the assumption that none of the denominators end up being 0, which will occur when
(a1)(b2) = (a2)(b1), OR
(b1)(a2) = (a1)(b2).

Answer 2

a1x+b1y=c1 <-- Eq 1
a2x+b2y=c2 <-- Eq 2
Multiply Eq 1 by a2 and Eq 2 by a1 getting:
a1a2x +a2b1y = c1a2 <-- Eq 3
a1a2x +a1b2y = a1c2 <-- Eq 4
Subtract Eq 4 from Eq 3 getting:
a2b1y-a1b2y = a2c1- a1c2
y = (a2c1- a1c2)/(a2b1-a1b2)

Plug this value of y into Eq 1 and solve for x

Answer 3

Assuming a1,a2, b1, and b2 are unkonwn variable.
Multiply one or both by different number so that the y term of each are equal. Now subtract one from the other and the y term will dissappear and leave you with a simple equation you can solve for x.

Answer 4

a1x+b1y=c1 => y=(c1-a1x)/b1
a2x+b2y=c2 => a2x+ b2*(c1-a1x)/b1=c2

b1a2x+b2c1-a1b2x=c2b1

x=(c2b1-c1b2)/(b1a2-a1b2)
y=(c1-a1*(c2b1-c1b2)/(b1a2-b2a1))/b1