Solve: ax+by=c and bx+ay=1+c [S.Chand Class 9 ICSE Fundamentals of Mathematics Chapter 4 Exercise 4.2 Simultaneous Linear Equations of Two Variables Solution]
Substitution Method
ax+by=c —-(i)
bx+ay=1+c —-(ii)
From Equation (i) we get,
ax= c-by
⟹ x = $\frac{c – by}{a}$ —(iii)
Substituting the Value of x in Equation (ii) We get,
$b\left(\frac{c – by}{a}\right) + ay = 1 + c$
$\Rightarrow \frac{bc – b^2y}{a} + ay = 1 + c$
$\Rightarrow \frac{bc – b^2y + a^2y}{a} = 1 + c$
$\Rightarrow bc + a^2y – b^2y = a + ac$
$\Rightarrow a^2y – b^2y = a + ac – bc$
$\Rightarrow {\mathrm{y}}\left(a^2 – b^2\right) = {\mathrm{a}} + {\mathrm{c}}({\mathrm{a}} – {\mathrm{b}})$
$\Rightarrow {\mathrm{y}} = \frac{{\mathrm{a}} + {\mathrm{c}}({\mathrm{a}} – {\mathrm{b}})}{\left(a^2 – b^2\right)}$
From Equation (i) we get,
${\mathrm{ax}} + {\mathrm{b}} \times { \frac{{\mathrm{a}} + {\mathrm{c}}({\mathrm{a}} – {\mathrm{b}})}{{\mathrm{a}}^2 – {\mathrm{b}}^2}} = {\mathrm{c}}$
$\Rightarrow {\mathrm{ax}} + \frac{{\mathrm{ab}} + {\mathrm{bc}}({\mathrm{a}} – {\mathrm{b}})}{{\mathrm{a}}^2 – {\mathrm{b}}^2} = {\mathrm{c}}$
$\Rightarrow {\mathrm{ax}} + \frac{{\mathrm{ab}} + {\mathrm{bca}} – {\mathrm{b}}^2{\mathrm{c}}}{{\mathrm{a}}^2 – {\mathrm{b}}^2} = {\mathrm{c}}$
$\Rightarrow {\mathrm{ax}} = {\mathrm{c}} – \frac{\mathrm{ab} + \mathrm{bca} – {\mathrm{b}}^2{\mathrm{c}}}{{\mathrm{a}}^2 – {\mathrm{b}}^2}$
$\Rightarrow ax = \frac{c({\mathrm{a}}^2 – {\mathrm{b}}^2) – \mathrm{ab} – \mathrm{bca} + {\mathrm{b}}^2{\mathrm{c}}}{{\mathrm{a}}^2 – {\mathrm{b}}^2}$
$\Rightarrow ax = \frac{ca^2 – cb^2 – ab – bca + b^2c}{{\mathrm{a}}^2 – {\mathrm{b}}^2}$
$\Rightarrow ax = \frac{a(ca – b – bc)}{{\mathrm{a}}^2 – {\mathrm{b}}^2}$
$\Rightarrow x = \frac{{ c(a – b) – b} }{{\mathrm{a}}^2 – {\mathrm{b}}^2}$
∴ Required Solutions are x = $\frac{{ c(a – b) – b} }{{\mathrm{a}}^2 – {\mathrm{b}}^2}$ and y = $\frac{{\mathrm{a}} + {\mathrm{c}}({\mathrm{a}} – {\mathrm{b}})}{\left(a^2 – b^2\right)}$
Elemination Method
ax+by=c —-(i)
bx+ay=1+c —-(ii)
Multiplying Equation (i) by a and Equation (ii) by b we get,
a2x+aby=ac —(iii)
b2x +aby= b+bc —-(iv)
Subtracting Equations (iii) and (iv) we get,
(a2x+aby)-(b2x +aby)=ac-(b+bc)
⟹ a2x+aby -b2x -aby =ac-b-bc
⟹ a2x-b2x =ac-bc-b
⟹ x(a2-b2) =c(a-b)-b
⟹x=$\frac{{ c(a – b) – b} }{{\mathrm{a}}^2 – {\mathrm{b}}^2}$
Now Substituting the value of x in equation (i) we get,
$a \times \frac{c(a – b) – b}{a^2 – b^2} + by = c$
$\Rightarrow \frac{ac(a – b) – ab}{a^2 – b^2} + by = c$
$\Rightarrow \frac{a^2c – abc – ab}{a^2 – b^2} + by = c$
$\Rightarrow by = c – \frac{a^2c – abc – ab}{a^2 – b^2}$
$\Rightarrow by = \frac{c\left(a^2 – b^2\right) – \left(a^2c – abc – ab\right)}{a^2 – b^2}$
$\Rightarrow by = \frac{ca^2 – cb^2 – a^2c + abc + ab}{a^2 – b^2}$
$\Rightarrow by = \frac{abc + ab – cb^2}{a^2 – b^2}$
$\Rightarrow y = \frac{ac + a – bc}{a^2 – b^2}$
$\Rightarrow y = \frac{c(a – b) + a}{a^2 – b^2}$
∴ Required Solutions are x = $\frac{{ c(a – b) – b} }{{\mathrm{a}}^2 – {\mathrm{b}}^2}$ and y = $\frac{{\mathrm{a}} + {\mathrm{c}}({\mathrm{a}} – {\mathrm{b}})}{\left(a^2 – b^2\right)}$