Question

Solve the following pair of linear equations..
(1) ax+by = c
bx+ay = 1+c
(2) (a-b) x + (a+b) y= a²-2ab-b²
(a+b) (x+y) = a²+ b²

Answer 1

Answer:

$y=\frac{a+ac-bc}{-b^2+a^2},\:x=\frac{-b-bc+ac}{-b^2+a^2}$

Step-by-step explanation:

$[tex]\begin{bmatrix}ax+by=c\\ bx+ay=1+c\end{bmatrix}\\\mathrm{Isolate}\:x\:\mathrm{for}\:ax+by=c:\quad x=\frac{c-by}{a};\quad \:a\ne \:0\\\mathrm{Subsititute\:}x=\frac{c-by}{a}\\\\\\\begin{bmatrix}b\frac{c-by}{a}+ay=1+c\end{bmatrix}\\\mathrm{Isolate}\:y\:\mathrm{for}\:b\frac{c-by}{a}+ay=1+c:\quad y=\frac{a+ac-bc}{-b^2+a^2};\quad \:a\ne \:0,\:a\ne \:b,\:a\ne \:-b\\\mathrm{For\:}x=\frac{c-by}{a}\\\mathrm{Subsititute\:}y=\frac{a+ac-bc}{-b^2+a^2}\\\\x=\frac{c-b\frac{a+ac-bc}{-b^2+a^2}}{a}$

Answer 2

Answer:

Step-by-step explanation:

Given system of equations:

ax+by=c ---(1)

bx+ay=1+c ----(2)

multiply equation (1) by a, and equation (2) by b , we get

a²x+aby=ac---(3)

b²x+aby=b+bc---(4)

Subtract (4) from (3) , we get

(a²-b²)x= ac-b-bc

Now,

multiply equation (1) by b, and equation (2) by a , we get

abx+b²y=bc---(5)

abx+a²y=a+ac---(6)

Subtract (5) from (6), we get

(a²-b²)y = a+ac-bc