solve the pair of linear equation by method of elimination by substitution
Answers
Answer:
given,
eq.1 = x/6 + y/15 = 4
eq.2 = x/3 - y/12 = 19/4
Solution -:
From eq.1 :
by taking LCM = 30,
(5x + 2y)/30 = 4
5x + 2y = 120 ... (i)
5x = 120 - 2y
x = (120 - 2y) /5 .... (ii)
From eq.2 :
by taking LCM = 12,
(4x - y)/12 = 19/4
4x - y = 57 ..... (iii)
putting eq.ii in eq.iii,
4[(120-2y)/5] - y = 57
(480 - 8y)/5 - y = 57
480 - 8y - 5y = 57 × 5
480 - 13y = 285
480 - 285 = 13y
195 = 13y
or, y = 195/13 = 15
putting value of y in eq.i,
5x + 2y = 120
5x + 2×15 = 120
5x = 120 - 30
5x = 90
x = 90/5 = 18
Hence,
x = 18 and y = 15
Step-by-step explanation:
Answer:
given,
eq.1 = x/6 + y/15 = 4
eq.2 = x/3 - y/12 = 19/4
Solution -:
From eq.1 :
by taking LCM = 30,
(5x + 2y)/30 = 4
5x + 2y = 120 ... (i)
5x = 120 - 2y
x = (120 - 2y) /5 .... (ii)
From eq.2 :
by taking LCM = 12,
(4x - y)/12 = 19/4
4x - y = 57 ..... (iii)
putting eq.ii in eq.iii,
4[(120-2y)/5] - y = 57
(480 - 8y)/5 - y = 57
480 - 8y - 5y = 57 × 5
480 - 13y = 285
480 - 285 = 13y
195 = 13y
or, y = 195/13 = 15
putting value of y in eq.i,
5x + 2y = 120
5x + 2×15 = 120
5x = 120 - 30
5x = 90
x = 90/5 = 18
Hence,
x = 18 and y = 15