Solve the following pair of linear equations by the substitution and cross-multiplication methods:

Solution:

By substitution method,

8x + 5y = 9 ...(i)

3x + 2y = 4 ...(ii)

From (ii), we get

$x=\frac{4-2 y}{3}$

Substituting x from (iii) in (i), we get

$8\left(\frac{4-2 y}{3}\right)+5 y=9$

= 32 – 16y + 15y = 27

= 5 = y

Substituting y = 5 in (ii) we get

3x + 2(v) = 4

= 3x = – 6

= x = – 2

Hence, x = – 2, y = 5

By cross multiplication method

8x + 5y = 9 ...(i)

3x + 2y = 4 ...(ii)

$\frac{x}{5 \times(-4)-2(-9)}=\frac{y}{3 \times(-9)-8(-4)}=\frac{1}{8 \times 2-3 \times 5}$

$\Rightarrow x=\frac{-\mathbf{2 0}+\mathbf{1 8}}{\mathbf{1}}=-2$

$\Rightarrow y=\frac{-\mathbf{2 7}+\mathbf{3 2}}{\mathbf{1}}=5$

Hence, x = –2, y = 5