Question

Solve the system by elimination
-2x+2y+3z=0
-2x-y+z=-3
2x+3y+3z=5 show all work

Answer 1

`\large\boxed{x=1,\ y=1,\ z=0}`

Explanation:

`\left\{\begin{array}{ccc}-2x+2y+3z=0&(1)\\-2x-y+z=-3&(2)\\2x+3y+3z=5&(3)\end{array}\right\\\\\text{add both sides of equations (1) and (3):}\\\\\underline{+\left\{\begin{array}{ccc}-2x+2y+3z=0&(1)\\2x+3y+3z=5&(3)\end{array}\right}\\.\qquad\qquad5y+6z=5\qquad(*)\\\\\text{add both sides of equations (2) and (3):}\\\\\underline{+\left\{\begin{array}{ccc}-2x-y+z=-3&(2)\\2x+3y+3z=5&(3)\end{array}\right}\\.\qquad\qquad2y+4z=2\qquad(**)`

`\text{We have new system of equations with two variables}\ y\ \text{and}\ z:\\\\\left\{\begin{array}{ccc}5y+6z=5&\text{multiply both sides by 2}\\2y+4z=2&\text{multiply both sides by (-5)}\end{array}\right\\\\\underline{+\left\{\begin{array}{ccc}10y+12z=10\\-10y-20z=-10\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad\qquad-8z=0\qquad\text{divide both sides by (-8)}\\.\qquad\qquad \boxed{z=0}\\\\\text{Put it to the first equation:}`

`5y+6(0)=5\\5y+0=5\\5y=5\qquad\text{divide both sides by 5}\\\boxed{y=1}\\\\\text{Put the values of}\ y\ \text{and}\ z\ \text{to (1)}:\\\\-2x+2(1)+3(0)=0\\-2x+2+0=0\qquad\text{subtract 2 from both sides}\\-2x=-2\qquad\text{divide both sides by (-2)}\\\boxed{x=1}`