Question

Solve the system by back substitution- 2x - 4y - 2 - 3w = 4- 5y - 302 - 10w = - 40- 42 - 4w = - 85w= 15The solution set is {0.00}. (Type an integer or a simplified fraction.)

Answer 1

1) To solve this system you have to start with the fourth equation because it has only one variable:

$5w=15$

-Divide both sides by 5 to determine the value of "w"

$\begin{gathered} \frac{5w}{5}=\frac{15}{5} \\ w=3 \end{gathered}$

2) Replace the value of "w" into the third equation and solve for z

$\begin{gathered} -4z-4w=-8 \\ -4z-4\cdot3=-8 \\ -4z-12=-8 \end{gathered}$

-Add 12 to both sides of the equation

$\begin{gathered} -4z-12+12=-8+12 \\ -4z=4 \end{gathered}$

-Divide both sides by -4 to determine the value of "z"

$\begin{gathered} -\frac{4z}{-4}=\frac{4}{-4} \\ z=-1 \end{gathered}$

3) Replace the values of "w" and "z" into the second equation and calculate the value of "y"

$\begin{gathered} -5y-30z-10w=-40 \\ -5y-(30(-1))-(10\cdot3)=-40 \\ -5y+30-30=-40 \\ -5y=-40 \end{gathered}$

-Divide both sides by -5 to determine the value of "y"

$\begin{gathered} -\frac{5y}{-5}=\frac{-40}{-5} \\ y=8 \end{gathered}$

4) Finally replace the values obtained for "w", "y", and "z" into the first equation to determine the value of x:

$\begin{gathered} -2x-4y-z-3w=4 \\ -2x-(4\cdot8)-(-1)-(3\cdot3)=4 \\ -2x-32+1-9=4 \\ -2x-40=4 \end{gathered}$

-Add 40 to both sides of the equation

$\begin{gathered} -2x-40+40=4+40 \\ -2x=44 \end{gathered}$

-Divide both sides by -2

$\begin{gathered} \frac{-2x}{-2}=\frac{44}{-2} \\ x=-22 \end{gathered}$

The solution for the system is:

$\mleft\lbrace w,x,y,z=3,-22,8,-1\}\mright?$