Solve the system by back substitution.- 2x - 4y - Z - 3w = -5- 4y - 24z - 8w = - 32- 42 - 4w = -8W= 3The solution set is? (Type an integer or a simplified fraction.

Solve the system by back substitution 2x 4y Z 3w 5 4y 24z 8w 32 42 4w 8W 3The solution set is Type an integer or a simplified fraction class=

Respuesta :

[tex]\begin{gathered} -2x-4y-z-3w=-5 \\ -4y-24z-8w=-32 \\ -4z-4w=-8 \\ w=3 \end{gathered}[/tex]

Above are four equations we can use to solve for the values of x, y, z, and w.

Since it is already shown that w = 3, let's solve for "z" using the third equation.

[tex]\begin{gathered} -4z-4w=-8 \\ -4z-4(3)=-8 \\ -4z-12=-8 \\ \text{Add 12 on both sides.} \\ -4z-12+12=-8+12 \\ -4z=4 \\ \text{Divide both sides by -4.} \\ \frac{-4z}{-4}=\frac{4}{-4} \\ z=-1 \end{gathered}[/tex]

From the above solution, the value of z = -1.

To solve for "y", let's use the second equation and the values of w and z.

[tex]\begin{gathered} -4y-24z-8w=-32 \\ -4y-24(-1)-8(3)=-32 \\ -4y+24-24=-32 \\ -4y=-32 \\ \text{Divide both sides by -4.} \\ \frac{-4y}{-4}=\frac{-32}{-4} \\ y=8 \end{gathered}[/tex]

The value of y is 8 as shown above.

Lastly, to solve for "x", let's use the first equation and the values of w, z, and y.

[tex]\begin{gathered} -2x-4y-z-3w=-5 \\ -2x-4(8)-(-1)-3(3)=-5 \\ -2x-32+1-9=-5 \\ -2x-40=-5 \\ \text{Add 40 on both sides.} \\ -2x-40+40=-5+40 \\ -2x=35 \\ \text{Divide both sides by -2.} \\ \frac{-2x}{-2}=\frac{35}{-2} \\ x=-17.5 \end{gathered}[/tex]

The value of x is -17.5 or -35/2.

To summarize the results, we have:

x = -35/2

y = 8

z = -1

w = 3

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