Question

Solve the following pairs of linear (simultaneous) equations using method of elimination by substitution:

Answer 1

Step-by-step explanation:

$y \div 15 + y \div 12 = 4 - 19 \div 8 \\ 12y + 15y = 13 \times 15 \times 12 \div 8 \\ 27y = 13 \times 45 \ \div 2 \\ 54y = 585 \\ y = 585 \div 54 \\ y = 585 \div 54$

Answer 2

$\frac{x}{6} + \frac{y}{15} = 4 \\ = > 5x + 2y = 120 \\ = > 5x = 120 - 2y \\ = > 20x = 480 - 8y \: \: \\ and \\ \frac{x}{3} - \frac{y}{12} = 4 \frac{3}{4} \\ = > \frac{4x - y}{12} = \frac{19}{4} \\ = > 4x - y = 57 \\ = > 20x - 5y = 285 \\ puttin g \: the \: value \: of \: 20x \: \: we \: get \: \: \\ = > 480 - 8y - 5y = 285 \\ = > 13y = 480 - 285\\ = > y = \frac{ 195 }{ 13} \\ = > y = 15 \\ \\ therefore \\ 5x = 120 - 2(15) \\ = > x = \frac{90}{5} \\ = > x = 16$

therefore x=16 and y=15