Question

solve the following system of equations algebraically for all values of x,y,z x+3y+5z=456x-3y+2z=-10-2x+3y+8z=72

Answer 1

The equations are given as shown below:

`\begin{gathered} x+3y+5z=45\text{ ------------(1)} \\ 6x-3y+2z=-10\text{ ------------(2)} \\ -2x+3y+8z=72\text{ -------------(3)} \end{gathered}`

From Equation (1):

`x=45-3y-5z\text{ ----------(4)}`

Substitute for x into equations (2) and (3):

`\begin{gathered} \text{Equation 2:} \\ 6(45-3y-5z)-3y+2z=-10 \\ 270-18y-30z-3y+2z=-10 \\ -21y-28z=-10-270 \\ -21y-28z=-280 \\ \text{Dividing all through by -1} \\ 21y+28z=280\text{ -------------(5)} \end{gathered}`
`\begin{gathered} \text{Equation 3:} \\ -2(45-3y-5z)+3y+8z=72 \\ -90+6y+10z+3y+8z=72 \\ 9y+18z=72+90 \\ 9y+18z=162 \\ \text{Dividing all through by 9:} \\ y+2z=18\text{ -------------(6)} \end{gathered}`

Solving Equations (5) and (6) simultaneously:

From equation (6):

`y=18-2z\text{ ------------(7)}`

Substitute for y into equation (5):

`\begin{gathered} 21(18-2z)+28z=280 \\ 378-42z+28z=280 \\ -14z=280-378 \\ -14z=-98 \\ z=(-98)/(-14) \\ z=7 \end{gathered}`

Substitute for z into equation (7):

`\begin{gathered} y=18-2(7) \\ y=18-14 \\ y=4 \end{gathered}`

Substitute for y and z into equation 4:

`\begin{gathered} x=45-3(4)-5(7) \\ x=45-12-35 \\ x=-2 \end{gathered}`

The solutions are:

`\begin{gathered} x=-2 \\ y=4 \\ z=7 \end{gathered}`