Question

Question 9 (v)
Solve the following pair of equations 43x + 67y = - 24 and 67x + 43y = 24.

Answer 1

Given pair of linear equations is
43x + 67y = - 24.....(i)
and 67x + 43y = 24......(ii)
On multiplying Eq. (i) by 43 and Eq. (ii) by 67 and then subtracting both of them, we get
$(67{)}^{2}x+43\times 67y=24\times 67(43{)}^{2}x+43\times 67y=-24\times 43\underset{–}{}\left\{\right(67{)}^2-\left(43{)}^2\right\}x=24(67+43)\Rightarrow (67+43)(67-43)x=24\times 110[\therefore (a^2-b^2)=(a-b\left)\right(a+b\left)\right]\Rightarrow 110\times 24x=24\times 110\Rightarrow x=1$
Now, put the value of $x$ in Eq. (i), we get
43 $\times$ 1 + 67 y = - 24
$\Rightarrow$ 67y = - 24 – 43
$\Rightarrow$ 67y = - 67
$\Rightarrow$ y = - 1
Hence, the required values of $x$ and $y$ are 1 and -1, respectively.