Q solve hte following by eliminating method..

1) 99x+101y=499

101x+99y=501

2) 23x-29y=98

29x-23y=110

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Q solve hte following by eliminating method
1) 99x+101y=499
101x+99y=501
2) 23x-29y=98
29x-23y=110

Ajanta Trivedi · Accepted Answer

1 the given linear equations are: $99 x + 101 y = 499 (1) 101 x + 99 y = 501 (2)$ multiplying eq(1) by 101 and eq(2) by 99 and then subtracting : $99 * 101 x + 101^{2} y = 499 * 101 99 * 101 x + 99^{2} y = 99 * 501 (101^{2} - 99^{2}) y = 499 * 101 - 99 * 501 (101 - 99) (101 + 99) y = (500 - 1) (100 + 1) - (100 - 1) (500 + 1)$ $2 * 200 y = 50000 - 100 + 500 - 1 - [50000 - 500 + 100 - 1] 400 y = - 100 - 1 + 500 + 500 - 100 + 1 400 y = 1000 - 200 400 y = 800 y = \frac{800}{400} = 2$ put y =2 in eq(1): $99 x + 101 * 2 = 499 99 x + 202 = 499 99 x = 499 - 202 99 x = 297 x = \frac{297}{99} = 3$ thus the solutions of the eq(1) and eq(1) are x = 3 and y = 2 2 the given linear equations in two variables are $23 x - 29 y = 98 (1) 29 x - 23 y = 110 (2)$ multiplying eq(1) by 29 and eq(2) by 23 and then subtracting: $29 * 23 x - 29^{2} y = 29 * 98 23 * 29 x - 23^{2} y = 23 * 110 {(23}^{2} - 29^{2}) y = 29 * 98 - 23 * 110 (23 - 29) (23 + 29) y = 2842 - 2530 - 6 * 52 y = 312 y = - \frac{312}{6 * 52} = - 1$ put y = -1 in eq(1): $23 x - 29 * (- 1) = 98 23 x + 29 = 98 23 x = 98 - 29 = 69 x = \frac{69}{23} = 3$ thus x = 3 and y = -1 is the solution of the given equations hope this helps you cheers!!

Q solve hte following by eliminating method.. 1) 99x+101y=499 101x+99y=501 2) 23x-29y=98 29x-23y=110

Q solve hte following by eliminating method..

1) 99x+101y=499

101x+99y=501

2) 23x-29y=98

29x-23y=110