Question

Solve the system by elimination
-2x+2y+3z=0
-2x-y+Z=-3
2x+3y+3z=5

Answer 1

x = 1

y = 1

z = 0

Explanation:

Given system of equations:

`\begin{cases}-2x+2y+3z=0\\-2x-y+z=-3\\2x+3y+3z=5\end{cases}`

To solve the given system of equations by the method of elimination, begin by adding the second and third equations to eliminate the x-terms:

`\begin{array}{rccrcrcl}&-2x&-&y&+&z&=&-3\\\vphantom{\frac12}+&(2x&+&3y&+&3z&=&\;\;\;5)\\\cline{2-8}\vphantom{\frac12}&&&2y&+&4z&=&\;\;\:2\\\cline{2-8}\end{array}`

Now, subtract the second equation from the first equation to eliminate the x-terms:

`\begin{array}{rccrcrcl}&-2x&+&2y&+&3z&=&\;\;\:0\\\vphantom{\frac12}-&(-2x&-&y&+&z&=&-3)\\\cline{2-8}\vphantom{\frac12}&&&3y&+&2z&=&\;\;\:3\\\cline{2-8}\end{array}`

This gives us two equations with terms in y and z.

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

Subtract twice the second equation from the first to eliminate the terms in z:

`\begin{array}{crcccl}&2y&+&4z&=&\;\;\;2\\\vphantom{\frac12}-&2(3y&+&2z&=&\;\;\;3)\\\cline{2-6}\vphantom{\frac12}&-4y&&&=&-4\\\cline{2-6}\end{array}`

Solve for y by dividing both sides of the equation by -4:

`\begin{aligned}(-4y)/(-4)&=(-4)/(-4)\\\\y&=1\end{aligned}`

Therefore, y = 1.

Now, substitute y = 1 into one of the equations with terms in y and z, then solve for z:

`\begin{aligned}2y+4z&=2\\\\y=1\implies 2(1)+4z&=2\\2+4z&=2\\4z&=0\\z&=0\end{aligned}`

Therefore, z = 0.

Finally, substitute the found values of y and z into one of the original equations and solve for x:

`\begin{aligned}2x+3y+3z&=5\\\\y=1,z=0 \implies 2x+3(1)+3(0)&=5\\2x+3+0&=5\\2x+3&=5\\2x&=2\\x&=1\end{aligned}`

Therefore, x = 1.