Question

Solve the following equations:
$2x-3y+5z=11$
$5x+2y-7z=-12$
$-4x+3y+z=5$

• A. <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $x=10,y=1,z=9$
• B. <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $x=2,y=1,z=3$
• C. <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $x=10,y=6,z=3$
• D. <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $x=1,y=2,z=3$

Answer 1

The correct option is D <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $x=1,y=2,z=3$

We have
$2x-3y+5z=11$...........(1)
$5x+2y-7z=-12$............(2)
$-4x+3y+z=5$............(3)

Adding equations (1) and (3), we get

$-2x+6z=16$..............(4)

Multiplying equation (2) by 3 and (3) by 2 , we have

$15x+6y-21z=-36$...............(5)
and $-8x+6y+2z=10$..........(6)

Subtracting equation (6) from (5), we get

$23x-23z=-46$............(7)

Now, from equations (4) and (7), we have
$-2x+6z-16=0$..............(8)
$23x-23z+46=0$............(9)

Therefore, by cross multiplication, we have

$\frac{x}{276-368}=\frac{z}{-368+92}=\frac{1}{46-138}$

$\Rightarrow \frac{x}{-92}=\frac{z}{-276}=\frac{1}{-92}$

$\Rightarrow x=\frac{-92}{-92},z=\frac{-276}{-92}$

$\Rightarrow x=1,z=3$

Putting $x=1andz=3$ in equation (1), we get

$2-3y+15=11$

$\Rightarrow 3y=6$

$\Rightarrow y=2$

Thus, we have $x=1,y=2,andz=3$