Question

Solve the following equations:
$2x-4y+9z=28$
$7x+3y-5z=3$
$9x+10y-11z=4$

• A. $x=2,y=3,z=4$
• B. $x=7,y=3,z=4$
• C. $x=2,y=3,z=5$
• D. $x=2,y=4,z=4$

Answer 1

The correct option is A $x=2,y=3,z=4$

We have
$2x-4y+9z=28$...........(1)
$7x+3y-5z=3$............(2)
$9x+10y-11z=4$............(3)

Multiplying equation (1) by 3, and (2) by 4, we have

$6x-12y+27z=84$...............(4)
and $28x+12y-20z=12$..........(5)

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

$34x+7z=96$............(6)

Again, multiplying (2) by 10, and (3) by 3, we have

$70x+30y-50z=30$...............(7)
and $27x+30y-33z=12$..........(8)

Subtracting equations (8) from (7), we get

$43x-17z=18$............(9)

Now, we have
$34x+7z=96$........(6)
$43x-17z=18$........(9)

Multiplying equation (6) by 17, and (9) by 7, we have

$578x+119y=1632$...............(10)
and $301x-119y=126$..........(11)

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

$879x=1758$

$\Rightarrow x=2$

Putting $x=2$ in equation (9), we get

$86-17z=18$

$\Rightarrow 17z=68$
$\Rightarrow z=4$

Putting $x=2andz=4$ in equation (1), we get

$4-4y+36=28$

$\Rightarrow 4y=12$
$\Rightarrow y=3$

Thus, we have $x=2,y=3,andz=4$