Question

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

${\displaystyle x=\frac{4-2y}{3}...\left(iii\right)}$

Putting this value in equation (i), we get

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

32 - 16y +15y = 27

-y = -5

y = 5 ... (iv)

Putting this value in equation (ii), we get

3x + 10 = 4

x = -2

Hence, x = -2, y = 5

By cross multiplication again, we get

8x + 5y -9 = 0

3x + 2y - 4 = 0

${\displaystyle \frac{x}{-20-(-18)}=\frac{y}{-27-(-32)}=\frac{1}{16-15}}$

${\displaystyle \frac{x}{-2}=\frac{y}{5}=\frac{1}{1}}$

x/-2 = 1 and y/5 = 1

x = -2 and y = 5