Question

Solve the following linear equations.3x-5y+34=0 and 8x-7y+21=0​

Answer 1

Step-by-step explanation:

3x-5y+34 =0

3x - 5y = -34 (eqn no. 1)

8x-7y+21 = 0

8x - 7y = - 21 (eqn. no. 2)

(equal the value of x or y in both the eqn,

I'm going to equal the value of x)

multiplying eqn no. 1 by 8

= ( 3x - 5y = -34 ) × 8

= 24x - 40y = - 272 (eqn no. 3)

multiplying eqn no. 2 by 3

= ( 8x - 7y = - 21 ) × 3

= 24x -21y = - 63 (eqn no. 4)

here, now the value of x is same in both the eqn.

subtracting eqn no. 4 from eqn no. 3

(24x - 40y) - (24x - 21y) = -272 - (-63)

24x - 40y - 24x + 21y = -272 + 63

-19y = - 209

y = (-209)/(-19)

y = 11

putting value of y in eqn. no. 1

3x - 5y = -34

3x - 5(11) = -34

3x - 55 = -34

3x = -34 + 55

3x = 21

x = 21/3

x = 7

hence the value of x = 7 & y = 11

Answer 2

Step-by-step explanation:

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