Question

Solve the system using any algebraic method!!!!!! HELP I BEG U

2x+y-z=-1
x+2y+2z=10
2x+z=3

Answer 1

Use matrices

Answer 2

no i have to use elimination can u show me step by step how do do it please
?

Answer 3

I thought it said to use any algebraic method...but now you have to use elimination method only?

Answer 4

yeah our teacher told us to use elimination......can u please draw out the stepss?

Answer 5

I'll do it haha

Answer 6

We have:
\[2x+y-z=-1\phantom{..}\rightarrow\phantom{..}y=z-1-2x\]
We also have:
\[\eqalign{
&x+2y+2z=10 \\
&x+2(z-1-2x)=10 \\
&x+2z-2-4x=10 \\
&-3x+2z=12 \\
&2z=12+3x \\
&z=\frac{1}{2}(12+3x) \\
}\]
We cal also redefine y as a result of isolating for \(z\):
\[\eqalign{
y&=z-1-2x \\
&=(1/2)(12+3x)-1-2x \\
&=6+(3/2)x-1-2x\\
&=5-(1/2)x
}\]
So now we have z and y in terms of x, let's slove for x:
\[\eqalign{
&2x+z=3 \\
&2x+(1/2)(12+3x)=3 \\
&2x+6+(3/2)x=3 \\
&(7/2)x=-3 \\
&x=-\frac{6}{7} \\
}\]
So now that we have x, and since y and z were in terms of x, we can find z and y:
\[z=\frac{1}{2}(12+3x)=\frac{1}{2}\left(12+3*-\frac{6}{7}\right)=\frac{1}{2}\left(12-\frac{18}{7}\right)=\frac{1}{2}\times\frac{66}{7}=\frac{66}{14}\]
\[y=5-\frac{1}{2}x=5-\frac{1}{2}\times-\frac{6}{7}=5+\frac{6}{14}=\frac{76}{14}\]

Answer 7

So therefore, the solution to the system \(S\):
\[S=\left\{\eqalign{
&2x+y-z=-1 \\
&x+2y+2z=10\\
&2z+x=3\\
}\right\}\]
Is the single point:
\[SOLUTION_{\{S\}}=\left(x,y,z\right)=\left(-\frac{6}{7},\frac{76}{14},\frac{66}{14}\right)\]

Answer 8

WHEW! Cool?