Math, asked by floorvixen5700fy, 1 year ago

solve the pair of linear equation by method of elimination by substitution ​

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Answered by Aalima786
0

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

Answered by Anonymous
8

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

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