Question

−2x+3y−z=−23x+y=4−2y+2z=4

Enter your answer, in the form (x,y,z), in the boxes in simplest terms.

Answer 1

`(x,y,z)=\left(-(1)/(11),(21)/(11),(21)/(11)\right)`

Explanation:

Given:

`-2x+3y-z=-23x+y=4-2y+2z=4`

Therefore:

`\begin{cases}-2x+3y-z=4\\-23x+y=4\\4-2y+2z=4\end{cases}`

Rewrite Equation 2 to make x the subject:

`\implies -23x+y=4`

`\implies -23x+y-y=4-y`

`\implies -23x=4-y`

`\implies (-23x)/(-23)=(4)/(-23)-(y)/(-23)`

`\implies x=-(4)/(23)+(y)/(23)`

Rewrite Equation 3 to make z the subject:

`\implies 4-2y+2z=4`

`\implies 4-2y+2z-4=4-4`

`\implies -2y+2z=0`

`\implies -2y+2z+2y=0+2y`

`\implies 2z=2y`

`\implies (2z)/(2)=(2y)/(2)`

`\implies z=y`

Substitute the found expressions for y and z into Equation 1 and solve for y:

`\implies -2x+3y-z=4`

`\implies -2\left(-(4)/(23)+(y)/(23)\right)+3y-y=4`

`\implies (8)/(23)-(2)/(23)y+2y=4`

`\implies (8)/(23)+(44)/(23)y=4`

`\implies (8)/(23)+(44)/(23)y- (8)/(23)=4- (8)/(23)`

`\implies (44)/(23)y=(84)/(23)`

`\implies (44)/(23)y \cdot (23)/(44)=(84)/(23)\cdot (23)/(44)`

`\implies y=(84)/(44)`

`\implies y=(21)/(11)`

Therefore, as y = z:

`\implies z=(21)/(11)`

Substitute the found value of y into the found expression for x and solve for x:

`\implies x=-(4)/(23)+(y)/(23)`

`\implies x=-(4)/(23)+((21)/(11))/(23)`

`\implies x=-(4)/(23)+(21)/(253)`

`\implies x=-(1)/(11)`

Therefore, the final solution is:

`(x,y,z)=\left(-(1)/(11),(21)/(11),(21)/(11)\right)`