Question

Solve the following equations:
$4x-3y+2z=40$
$5x+9y-7z=47$
$9x+8y-3z=97$

• A. $x=10,y=1,z=3$
• B. $x=10,y=2,z=3$
• C. $x=11,y=2,z=3$
• D. None of the above

Answer 1

The correct option is B $x=10,y=2,z=3$

We have
$4x-3y+2z=40$...........(1)
$5x+9y-7z=47$............(2)
$9x+8y-3z=97$............(3)

Multiplying equation (1) by 7, and (2) by 2, we have

$28x-21y+14z=280$...............(4)
and $10x+18y-14z=94$..........(5)

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

$38x-3y=374$............(6)

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

$12x-9y+6z=120$...............(7)
and $18x+16y-6z=194$..........(8)

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

$30x+7y=314$............(9)

Now, we have
$38x-3y=374$........(6)
$30x+7y=314$........(9)

Multiplying equation (6) by 7, and (9) by 3, we have

$266x-21y=2618$...............(10)
and $90x+21y=942$..........(11)

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

$356x=3560$

$\Rightarrow x=10$

Putting $x=10$ in equation (9), we get

$300+7y=314$

$\Rightarrow 7y=14$
$\Rightarrow y=2$

Putting $x=10andy=2$ in equation (1), we get

$40-6+2z=40$

$\Rightarrow 2z=6$
$\Rightarrow z=3$

Thus, we have $x=10,y=2,andz=3$