Question

BY SUBSTITUTION METHOD 7x - 15y=2 and x+2y=3

Answer 1

7x-15y=2 eq(1)
x+2y=3 eq(2)
from eq(1)
7x=2+15y
x=2+15y/7
putting x=2+15y/7 in eq(2)
x+2y=3
(2+15y/7)+2y=3
(2+15y+14y)/7=3
2+29y/7=3
2+29y=3×7
2+29y=21
29y=21-2
29y=19
y=19/29

y=19/29 putting in eq(2)
x+2y=3
x+2(19/29)=3
x+38/29=3
(29x+38)/29=3
29x+38=3×29
29x+38=87
29x=87-38
29x=49
x=49/29

Answer 2

Answer:

$7x - 15y = 2 \\ \: \implies7x = 2 + 15y \\ \implies \: x = \frac{2 + 15y}{7} \: \: \: \: \: \rightarrow(i) \\ \\ x + 2y = 3 \\ \implies\frac{2 + 15y}{7} + 2y = 3 \: (putting \: x = \frac{2 + 15y}{7} ) \\ \implies \frac{2 + 15y + 14y}{7} = 3 \\ \implies15y + 14y = 3 \times 7 - 2 \\ \implies29y = 21 - 2 \\ \implies \: y = \frac{19}{29} \\ \\ \\ \: \: substituting \: the \: value \: of \: y \: in \: equation \: (i) \\ 7x - 15 \times \frac{19}{29} = 2 \\ \implies7x - \frac{285}{29} = 2 \\ \implies7x = 2 + \frac{285}{29} \\ \implies7x = \frac{58 + 285}{29} \\ \\ \: \implies7x = \frac{343}{29 } \\ \\ \implies \: x = \frac{343}{203} \\ \\ \implies \: x = \frac{49}{29} \\ \\ \therefore \: x = \frac{49}{29} \\ \\ y = \frac{19}{29}$