Question

Solve the following equations:
$12x+9y-7z=2$
$8x-26y+9z=1$
$23x+21y-15z=4$

• A. <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $x=4,y=1,z=3$
• 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;}--> $x=2,y=3,z=7$
• D. <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $x=9,y=1,z=3$

Answer 1

The correct option is C <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $x=2,y=3,z=7$

We have
$12x+9y-7z=2$..........(1)
$8x-26y+9z=1$............(2)
$23x+21y-15z=4$...........(3)

Multiplying equation (2) by 2,we have

$16x-52y+18z=2$...............(4)
also, $12x+9y-7z=2$..........(1)

Subtracting equation (1) from (4), we get

$4x-61y+25z=0$............(5)

Again, multiplying (1) by 2, we have

$24x+18y-14z=4$...............(6)
also, $23x+21y-15z=4$...........(3)
Subtracting equation (3) from (6), we get
$x-3y+z=0$............(7)

Now, we have
$4x-61y+25z=0$............(5)
$x-3y+z=0$............(7)

Therefore, by cross multiplication, we have

$\frac{x}{-61+5}=\frac{y}{25-4}=\frac{z}{-12+61}$

$\Rightarrow \frac{x}{14}=\frac{y}{21}=\frac{z}{49}$

$\Rightarrow \frac{x}{2}=\frac{y}{3}=\frac{z}{7}$

Let each fraction equal to k. we get
$x=2k,y=3k,z=7k$

Now, equation (1) $\Rightarrow k(24+27-49)=2$

$\Rightarrow 2k=2$
$\Rightarrow k=1$

Thus, we have $x=2,y=3,andz=7$