Question

Please say what will be the answer of this question. This is a 3-term linear equation.

Answer 1

Step 1

Given;

$\begin{gathered} -x+2y+2z=4----(1)_{} \\ -2x-y-3z=-3\text{ -----(2)} \\ x-y-z=-5-----(3) \end{gathered}$

Required; To find the value of x,y,z.

Step 2

Find the value of x from equation 1 and substitute it into equations (2) and 3

$\begin{gathered} \text{from 1} \\ 2y+2z-4=x \\ x=\text{ 2y+2z-4} \end{gathered}$$\begin{gathered} \text{Substituting in 2 and 3 gives} \\ -2(2y+2z-4)-y-3z=\text{ -3} \\ -4y-4z+8-y-3z=-3 \\ -5y-7z=-3-8 \\ -5y-7z=-11----(4) \\ \text{For 3} \\ 2y+2z-4\text{ -y-z=-5} \\ y+z=-5+4 \\ y+z=\text{ -1-----(5)} \end{gathered}$

Step 3

Find the value of y

Hence

$\begin{gathered} 2y=-18 \\ \frac{2y}{2}=-\frac{18}{2} \\ y=-9 \end{gathered}$

Step 4

Find the value of z

$\begin{gathered} \text{From equation 5} \\ y+z=-1 \\ -9+z=-1 \\ z=\text{ -1+9} \\ z\text{ = 8} \end{gathered}$

Step 5

Find the value of x

$\begin{gathered} \text{From equation 3} \\ x-y-z=-5 \\ x-(-9)-8=-5 \\ x+9-8=-5 \\ x=\text{ -5+8-9} \\ x=\text{ -6} \end{gathered}$

Hence

x= -6

y=-9

z=8