To solve: ((x-2y=2),(2x+3y=11),(y-4z=-7))

Calculation:
We have to solve
${\displaystyle x-2y=2\dots \left(1\right)}$
${\displaystyle 2x+3y=11\dots \left(2\right)}$
${\displaystyle y-4z=-7\dots \left(3\right)}$
We have to choose equation 1 and equation 2 and eliminate the same variable x.
Multiply both sides of first equation by -2 and we have to add equation 1 and 2, we get
${\displaystyle \left(1\right)\times -2\to -2x+4y=-4}$
${\displaystyle \left(2\right)\to 2x+3y=11}$
${\displaystyle 7y=7}$
${\displaystyle y=1}$
Now we have to choose a different pair of equations and eliminate the same variable.
Substitude ${\displaystyle y=1}$ in equation 1, we get
${\displaystyle x-2y=2}$
${\displaystyle x-2\left(1\right)=2}$
${\displaystyle x-2=2}$
${\displaystyle x=2+2}$
${\displaystyle x=4}$
Substitute ${\displaystyle y=1}$ in equation 3, we get
${\displaystyle y-4z=-7}$
${\displaystyle 1-4z=-7}$
${\displaystyle -4z=-7-1}$
${\displaystyle -4z=-8}$
Divide both sides by -4
${\displaystyle \frac{-4z}{-4}=\frac{-8}{-4}}$
${\displaystyle z=2}$
The solution set for the system is {(4, 1, 2)}
To check:
Substitute ${\displaystyle x=4,y=1}$ in equation 1
${\displaystyle x-2y=2}$
${\displaystyle 4-2\left(1\right)=2}$
${\displaystyle 4-2=2}$
${\displaystyle 2=2}$[True]
Substitute ${\displaystyle x=4,y=1}$ in equation 2
${\displaystyle 2x+3y=11}$
${\displaystyle 2\left(4\right)+3\left(1\right)=11}$
${\displaystyle 8+3=11}$
${\displaystyle 11=11}$ [True]
Substitute ${\displaystyle y=1,z=2}$ in equation 3
${\displaystyle y-4z=-7}$
${\displaystyle 1-4\left(2\right)=-7}$
${\displaystyle 1-8=-7}$
${\displaystyle -7=-7}$ [True]
The solution set ${\displaystyle =\left\{(4,1,2)\right\}}$
Conclusion:
The solution set for the system is {(4, 1, 2)}