Subtitusika (2) ke (1)
x + 3y = 10
2y – 5 + 3y = 10
5y – 5 = 10
5y = 10 + 5
5y = 15
y = 3
x = 2y – 5
= 2. 3 – 5
= 6 – 5
= 1
Nilai 5x – 2y yaitu
5.1 – 2.3 = 5 – 6
= –1
Jadi nilai dari 5x – 2y adalah –1
x – 3y – 5 = 0
x – 3y = 5 .... (2)
Dari (1) dan (2)
3x + 2y = 4 Ix1I 3x + 2y = 4
x – 3y = 5 Ix3I 3x – 9y = 15 _
11y = –11
y = –1
3x + 2y = 4
3.x + 2 (–1) = 4
3x – 2 = 4
3x = 4 + 2
3x = 6
x = 2
Jadi penyelesaiannya adalah (2, –1)