Question

Question 3
Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8x +5y = 9
3x +2y = 4

Answer 1

8x +5y = 9 ... (i)
3x +2y = 4 ... (ii)
From equation (ii), we get
$x=\frac{4-2y}{3}$ ... (iii)
Putting this value in equation (i), we get
$8\times \frac{4-2y}{3}+5y=9$
$\Rightarrow$ 32 - 16y +15y = 27
$\Rightarrow$ -y = -5
$\Rightarrow$ y = 5 ... (iv)
Substituting this value in equation (ii), we get
3x + 10 = 4
$\Rightarrow$ x = -2
Hence, x = -2, y = 5
By cross multiplication again, we get
8x + 5y -9 = 0
3x + 2y - 4 = 0
$\frac{x}{-20-(-18)}=\frac{y}{-27-(-32)}=\frac{1}{16-15}$
$\frac{x}{-2}=\frac{y}{5}=\frac{1}{1}$
$\frac{x}{-2}=1$ and $\frac{y}{5}=1$
$x=-2andy=5$