Question

Solve this system by elimination : -5x + 3z =8 x- 4y = 7 5y - 4z = -14 find the x , y and z using elimination

Answer 1

Answer

• x = –1

• y = –2

• z = 1

Explanation

Given the system:

$\begin{gathered} -5x+3z=8\text{, equation 1} \\ 5y-4z=-14,\text{ equation 2} \\ x-4y=7,\text{ equation 3} \end{gathered}$

We have to divide equation 1 over 5 and add it to equation 2:

$\begin{gathered} -(5x)+0y+3z=8 \\ x-4y+0z=7 \\ 0x+5y-4z=-14 \end{gathered}$$\begin{gathered} \frac{-(5x)+0y+3z=8}{5} \\ x-4y+0z=7 \\ 0x+5y-4z=-14 \end{gathered}$$\begin{gathered} -x+0y+\frac{3}{5}z=\frac{8}{5} \\ x-4y+0z=7 \\ 0x+5y-4z=-14 \end{gathered}$

Now, we have to add 1/5(equation 1) to equation 2:

$\begin{gathered} -x+0y+\frac{3}{5}z=\frac{8}{5} \\ x-4y+0z=7 \\ --------- \\ 0x-4y+\frac{3}{5}z=\frac{43}{5} \end{gathered}$

Next, we multiply the equation 2 obtained previously times 5:

$\begin{gathered} (0x-4y+\frac{3}{5}z=\frac{43}{5})\cdot5 \\ -20y+3z=43 \end{gathered}$

Then, we divide equation 2 over 4:

$\begin{gathered} \frac{-20y+3z=43}{4} \\ -5y+\frac{3}{4}z=\frac{43}{4} \end{gathered}$

We add it to equation 3:

$\begin{gathered} -5y+\frac{3}{4}z=\frac{43}{4} \\ 5y-4z=-14 \\ ------- \\ 0y-\frac{13}{4}z=-\frac{13}{4} \end{gathered}$

Then, we are left with:

$\begin{equation*} -\frac{13}{4}z=-\frac{13}{4} \end{equation*}$

Simplifying:

$-13z=-13$$z=\frac{-13}{-13}=1$$z=1$

Now that we have the value of z (z = 1), we can replace it in the modified equation 2 and solve for y:

$-20y+3z=43$$-20y+3(1)=43$$-20y+3=43$$-20y=43-3$$-20y=40$$y=\frac{40}{-20}$$y=-2$

Finally, calculating the value of x with any of the equations (as we already have the other two values):

$x-4(-2)=7$$x+8=7$$x=7-8$$x=-1$