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

To solve:
$\left(\begin{array}{c}x-2y=2\\ 2x+3y=11\\ y-4z=-7\end{array}\right)$

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