Answer:
(-2, 4, 2)
Where x = -2, y = 4, and z = 2.
Step-by-step explanation:
We are given the system of three equations:
[tex]\displaystyle \left\{ \begin{array}{l} 4x -y -2z = -8 \\ -2x + 4z = -4 \\ x + 2y = 6 \end{array}[/tex]
And we want to find the value of each variable.
Note that both the second and third equations have an x.
Therefore, we can isolate the variables for the second and third equation and then substitute them into the first equation to make the first equation all one variable.
Solve the second equation for z:
[tex]\displaystyle \begin{aligned} -2x+4z&=-4 \\ x - 2 &= 2z \\ z&= \frac{x-2}{2}\end{aligned}[/tex]
Likewise, solve the third equation for y:
[tex]\displaystyle \begin{aligned} x+2y &= 6\\ 2y &= 6-x \\ y &= \frac{6-x}{2} \end{aligned}[/tex]
Substitute the above equations into the first:
[tex]\displaystyle 4x - \left(\frac{6-x}{2}\right) - 2\left(\frac{x-2}{2}\right)=-8[/tex]
And solve for x:
[tex]\displaystyle \begin{aligned} 4x+\left(\frac{x-6}{2}\right)+(2-x) &= -8 \\ \\ 8x +(x-6) +(4-2x) &= -16 \\ \\ 7x-2 &= -16 \\ \\ 7x &= -14 \\ \\ x &= -2\end{aligned}[/tex]
Hence, x = -2.
Find z and y using their respective equations:
Second equation:
[tex]\displaystyle \begin{aligned} z&=\frac{x-2}{2} \\ &= \frac{(-2)-2}{2} \\ &= \frac{-4}{2} \\ &= -2\end{aligned}[/tex]
Third equation:
[tex]\displaystyle \begin{aligned} y &= \frac{6-x}{2}\\ &= \frac{6-(-2)}{2}\\ &= \frac{8}{2}\\ &=4\end{aligned}[/tex]
In conclusion, the solution is (-2, 4, -2)
Answer:
x = -2
y =4
z=-2
Step-by-step explanation:
4x−y−2z=−8
−2x+4z=−4
x+2y=6
Solve the second equation for x
x = 6 -2y
Substitute into the first two equations
4x−y−2z=−8
4(6-2y) -y -2 = 8
24 -8y-y -2z = 8
-9y -2z = -32
−2(6-2y)+4z=−4
-12 +4y +4z = -4
4y+4z = 8
Divide by 4
y+z = 2
z =2-y
Substitute this into -9y -2z = -32
-9y -2(2-y) = -32
-9y -4 +2y = -32
-7y -4 = -32
-7y =-28
y =4
Now find z
z = 2-y
z = 2-4
z = -2
Now find x
x = 6 -2y
x = 6 -2(4)
x =6-8
x = -2
