tolong kerjakan dengan metode substitusi dan metode eliminasi

a)
×1 | 2x + 3y = 5
×3 | 12x - 3y = 9
14x = 14
x = 14/14 = 1

4x - y = 3
4(1) - y = 3
y = 3-4 = -1

b)
×5 | 20x - 5y = 15
×1 | ⅔x + 5y = -1
19,5x = 16
x = 16/19,5 = 0,82

4x - y = 3
4(0,82) - y = 3
y = 3 - 3,28 = -0,28

c)
×5 | 10x + 50y = 70
×2 | 10x - 18y = 2
68y = 68
y = 68/68 = 1

2x + 10y = 14
2x + 10(1) = 14
2x = 14-10
x = 4/2 = 2