dengan metode substitusi, tentukan siatem persamaan dengan caranya: 1) 5a +2b=5 dan 6a +b=1 2) 6s-t-2=0 dan 3s-2t +5=0 tolong bantu pake Carany terima kasih

1) 5a + 2b = 5
6a + b = 1 --------> b = 1-6a
maka :
5a + 2b = 5
5a + 2(1-6a) = 5
5a + 2 - 12a = 5
-7a = 3
a = - 3/7
5a + 2b = 5
5(-3/7) + 2b = 5
2b = 5 + 15/7
b = 50/7 : 2
b = 25/7

2) 6s-t-2 = 0 --->t=6s-2
3s - 2t + 5 = 0
maka :
3s - 2t + 5 = 0
3s - 2(6s -2) + 5 = 0
3s - 12s + 4 + 5 = 0
-9s = -9
s = 1
3s -2t +5 = 0
3(1) -2t +5 = 0
-2t = -5-3
t = -8/-2
t = -4

1. 5a + 2b = 5
6a+ b = 1 ➡ b = -6a+1

subtitusikan 2b
5a +2b = 5
5a + 2(-6a+1) = 5
5a + -12a +2 = 5
-7a = 5-2
-7a = -3
a = -3/-7
a = 3/7

6a + b = 1
(3/7×6) +b = 1
18 +b = 7
b = 7-18
b = -11

hp = {3/7,-11}

2. 6s -t = 2 ➡ 6s -2 = t
3s -2t = -5

subtitusikan t
3s -2t = -5
3s -2(6s -2) = -5
3s -12s+4 = -5
-9s = -5-4
-9s = -9
s = -9/-9
s = 1

6s -t = 2
(6×1)-t = 2
-t = 2-6
-t = -4
t = -4/-1
t = 4

hp = {1,4}