Question

Solve the following equations:
$2x+3y+4z=16$
$3x+2y-5z=8$
$5x-6y+3z=6$

• A. <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $x=3,y=2,z=0$
• B. $x=3,y=1,z=1$
• C. $x=3,y=2,z=1$
• D. $x=4,y=2,z=1$

Answer 1

The correct option is C $x=3,y=2,z=1$

We have
$2x+3y+4z=16$...........(1)
$3x+2y-5z=8$..........(2)
$5x-6y+3z=6$...........(3)

Multiplying equation (1) by '-3', and (2) by 2, we have

$-6x-9y-12z=-48$...............(4)
and $6x+4y-10z=16$..........(5)

Adding equations (4) and (5), we get

$-5y-22z=-32$............(6)

Again, multiplying (1) by 5 and (3) by '-2', we have

$10x+15y+20z=80$...............(7)
and $-10x+12y-6z=-12$..........(8)

Adding equations (7) and (8), we get

$27y+14z=68$............(9)

Now, we have
$-5y-22z=-32$........(6)
$27y+14z=68$........(9)

Multiplying equation (6) by 27, and (9) by 5, we have

$-135y-594z=-864$...............(10)
and $135y+70z=340$..........(11)

Adding equations (10) and (11), we get

$-524z=-524$

$\Rightarrow z=1$

Putting $z=1$ in equation (9), we get

$27y+14=68$

$\Rightarrow 27y=54$
$\Rightarrow y=2$

Putting $z=1andy=2$ in equation (1), we get

$2x+6+4=16$

$\Rightarrow 2x=6$
$\Rightarrow x=3$

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