Question

¿Me podrian resolver estos problemas por metodo de igualacion?

1) 5x+7y=-1
-3x+4y=-24

2) 15x+11y=32
7y-9x=8

Answer 1

1)
 5X + 7Y = -1  (1)

-3X + 4Y = -24 (2)

7Y = - 1 - 5X

Y = -1/7 - 5X/7

4Y = 3X - 24

Y = 3X/4 - 6

Y = Y: -1/7 - 5X/7 = 3X/4 - 6

-1/7 + 6 = 3X/4 + 5X/7

-1/7 + 6 = -1/7 + 42/7 = 41/7

3X/4 + 5X/7 = [7(3X) + 4(5X)]/28

[21X + 20X]/28: 41X/28

41X/28 = 41/7

7X = 28

X = 28/7

X = 4

Reemplazo el valor de X = 4

Y = 3X/4 - 6

Y = 3(4)/4 - 6

Y = 3 - 6

Y = -3

Rta: X = 6; Y = -3

2)

15X + 11Y = 32 (1)

7Y - 9X = 8 (2)

15X = 32 - 11Y

X = 32/15 - 11Y/15

9X = 7Y - 8

X = 7Y/9 - 8/9

X = X:

32/15 - 11Y/15 = 7Y/9 - 8/9

32/15 + 8/9 = 11Y/15 + 7Y/9

11Y/15 + 7Y/9 = [3(11Y) + 5(7Y)]/45 = [33Y + 35Y]/45 = 68Y/45

32/15 + 8/9 = [3(32) + 5(8)]/45 = [96 + 40]/45 = 136/45

136/45 = 68Y/45

136 = 68Y

Y = 136/68

Y = 2

Reemplazo el valor de Y = 2 en X = 7Y/9 - 8/9

X = 7(2)/9 - 8/9

X = 14/9 - 8/9

X = 6/9

X = 2/3

Rta: X = 2/3 y Y = 2