Diketahui sistem persamaan: ,2/a+2/b=1−3/c 4/a+4/b+1=−3/c 2/b−6/c=8/a−3 Nilai ab+bc+ca=… A. 3 B. 2 C. 1 D. 0 E. −1

Hallo Niko, kakak bantu jawab ya :) Jawaban untuk soal ini adalah D Kita gunakan metode eliminasi substitusi 2/a+2/b=1−3/c 4/a+4/b+1=−3/c 2/b−6/c=8/a−3 Misal: 1/a = x 1/b = y 1/c = z 2x + 2y = 1 - 3z 2x + 2y + 3z = 1 ... (1) 4x + 4y + 1 = -3z 4x + 4y + 3z = -1 ... (2) 2y - 6z = 8x - 3 -8x + 2y - 6z = -3 ... (3) Eliminasi pers. (1) dan (2) 2x + 2y + 3z = 1 4x + 4y + 3z = -1 --------------------- - -2x - 2y = 2 ... (4) Eliminasi pers. (2) dan (3) 4x + 4y + 3z = -1 |×2| 8x + 8y + 6z = -2 -8x + 2y - 6z = -3 |×1| -8x + 2y - 6z = -3 8x + 8y + 6z = -2 -8x + 2y - 6z = -3 ----------------------- + 10y = -5 y = -1/2 Substitusi ke pers. (4) -2x - 2y = 2 -2x - 2(-1/2) = 2 -2x + 1 = 2 -2x = 1 x = -1/2 Substitusi ke pers. (1) 2x + 2y + 3z = 1 2(-1/2) + 2(-1/2) + 3z = 1 -1 - 1 + 3z = 1 -2 + 3z = 1 3z = 3 z = 1 Substitusi ke pemisalan 1/a = x 1/a = -1/2 a = -2 1/b = y 1/b = -1/2 b = -2 1/c = z 1/c = 1 c = 1 Nilai ab + bc + ca = = (-2)(-2) + (-2)(1) + (1)(-2) = 4 - 2 - 2 = 0 Maka nilai ab + bc + ca = 0 Semoga membantu ya :)