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Penjelasan dengan langkah-langkah:

Dengan cara substitusi

1) 2× + y =3

× - 3 y =5 => x = 5+3y

substitusi x = 5+3y ke persamaan 2x+y=3

2(5+3y) +y=3

10+6y+y=3

7y=3–10=–7

y = –7/7 = –1

substitusi y = –1 ke persamaan x=5+3y

x=5+3(–1) = 5–3 = 2

jadi, X = 2 dan y= –1

2) × - 2y =7 => x=7+2y

3× - 2y =3

substitusi x=7+2y ke persamaan 3× - 2y =3

3(7+2y)–2y=3

21+6y–2y = 3

4y=3–21 = –18

y = –18/4 = –4, 5

substitusi y = –4, 5 ke persamaan x = 7+2y

x = 7+2(–4, 5)

x = 7 –9= –2

jadi, x = –2 dan y = –4, 5

Selesaikan SPLDV berikut dengan cara eliminasi

3) × + 2y =5

3x + y =5

eliminasi variabel x

× + 2y =5 |×3| 3x+6y=15

3x + y =5 |×1| 3x+ y =5 –

0+5y=10

y=10/5 = 2

eliminasi variabel y

× + 2y =5 |×1| x+2y=5

3x + y =5 |×2| 6x+2y =10 –

–5x+0= –5

x= –5/–5 = 1

jadi, x=1 dan y=2

4) -2× - 5y =3

3× + 8y = -6

eliminasi variabel x

-2× - 5y =3 |×3| –6×–15y=9

3× + 8y = -6 |×2| 6x+16y= –12 +

0 + y = –3

y = –3

eliminasi variabel y

-2× - 5y =3 |×8| –16×–40y=24

3× + 8y = -6 |×5| 15x+40y= –30 +

–x + 0 = –6

× = 6

jadi, x=6 dan y=–3

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