$5 x-4 y+10 z=-3$
$10 \mathrm{x}+2 \mathrm{y}-15 \mathrm{z}=21$
$15 \mathrm{x}-2 \mathrm{y}+10 \mathrm{z}=0$
A) $\left(\frac{3}{5},-\frac{3}{2},-\frac{6}{5}\right)$
B) $\left(\frac{3}{2},-3, \frac{6}{5}\right)$
C) $\left(\frac{3}{2}, \frac{3}{5},-\frac{6}{5}\right)$
D) $\left(-\frac{3}{5}, \frac{3}{4}, \frac{6}{5}\right)$

Question

Quizplus · Accepted Answer

Explanation:Multiply the first equation by 3, the second equation by -1, and the third equation by 2. Then add the three equations together to get 20x = 18, so x = 9/10. Substitute this value of x into the first equation to get 45/10 - 4y + 10z = -3, so y = 3/5. Substitute the values of x and y into the third equation to get 135/10 - 2(3/5) + 10z = 0, so z = 6/5. Therefore, the solution is (9/10, 3/5, 6/5).

5x−4y+10z=−35 x-4 y+10 z=-35x−4y+10z=−3 10x+2y−15z=2110 \mathrm{x}+2 \mathrm{y}-15 \mathrm{z}=2110x+2y−15z=21 15x−2y+10z=015 \mathrm{x}-2 \mathrm{y}+10 \mathrm{z}=015x−2y+10z=0

$5 x-4 y+10 z=-3$ $10 \mathrm{x}+2 \mathrm{y}-15 \mathrm{z}=21$ $15 \mathrm{x}-2 \mathrm{y}+10 \mathrm{z}=0$