Question

Selesaikan sistem persamaan linear dua variabel berikut dengan metode subtitusi.
1. x=-3y+11; 5x+3y=19
2.x-2+1=0; 2x+3y=12
3.6p-q=1; 4p-3q+4=0

Answer 1

Nomor 1.
x = -3y + 11
Substitusikan pada persamaan 2.
5x + 3y = 19
5(-3y+11) + 3y = 19
-15y + 55 + 3y = 19
-12y = 19 - 55
-12y = -36
y = 3
Sehingga:
x = -3(3) + 11
x = -9+11
x = 2

HP = {(2,3)}

Nomor 2.
x - 2y + 1 = 0
x = 2y - 1

Lalu,
2x + 3y = 12
2(2y - 1) + 3y = 12
4y - 2 + 3y = 12
7y = 14
y = 2

Substitusi untuk mendapatkan x,
x = 2y - 1
x = 2(2) - 1
x = 4 - 1
x = 3

HP={(3,2)}

Nomor 3.
6p - q = 1
q = 6p - 1
Substitusikan,
4p - 3q + 4 = 0
4p - 3(6p-1) + 4 = 0
4p - 18p + 3 + 4 = 0
-14p + 7 = 0
14p = 7
p = 1/2

Maka,
q = 6p - 1
q = 6(1/2) - 1
q = 3-1
q = 2

HP={(1/2, 2)}