## mskyeg 2 years ago What is the solution of the system of equations?

1. mskyeg

2. mskyeg

Please show steps so I learn :)

3. Aylin

What class are you in?

4. mskyeg

Im in algebra 2

5. Aylin

Ok. So you have$2x+2y+3z=-6$$3x+5y+4z=3$$2x+3y+4z=-10$There are two ways to solve this, elimination and substitution. I'll go over substitution first and if you're interested I can also go over elimination. So with substitution, the idea is to solve one of the equations for one of the variables. Then we'll know that that variable is the same as the the other side of the equal sign, so in a second equation we can substitute in that other side of the equal sign in place of that variable. I'll show an example:$2x+2y+3z=-6$$2x=-2y-3z-6$$x=-y-\frac{3}{2}z-3$So now I know that x is the same as -y-3z/2-3, so in the second equation instead of x I can put that in.$3(-y-\frac{3}{2}z-3)+5y+4z=3$Notice that this equation is only in terms of y and z. I'm going to solve this for y.$-3y-\frac{9}{2}z-9+5y+4z=3$$2y-\frac{1}{2}z=12$$2y=\frac{1}{2}z+12$$y=\frac{1}{4}z+3$Now we're almost done. In the third equation, I'm going to replace x with -y-3z/2-3 again:$2(-y-\frac{3}{2}z-3)+3y+4z=-10$Now I can replace all of the ys with z/4+3:$2(-(\frac{1}{4}z+3)-\frac{3}{2}z-3)+3(\frac{1}{4}z+3)+4z=-10$Now I can solve this for z:$2(-\frac{1}{4}z-3-\frac{6}{4}z-3)+3(\frac{1}{4}z+3)+4z=-10$$2(-\frac{7}{4}z-6)+3(\frac{1}{4}z+3)+4z=-10$$-\frac{14}{4}z-12+\frac{3}{4}z+9+\frac{16}{4}z=-10$$\frac{5}{4}z-3=-10$$\frac{5}{4}z=-7$$z=-\frac{28}{5}$And so after all of that we know what z is, we know what y is in terms of z, and we know what x is in terms of y and z. So you can use z to find y, and then you can use z and y to find x.

6. mskyeg

Thats great but I don't know what any of the variable equal in the equation im trying to solve

7. mskyeg

thats the problem. Ive done these types of problems and i know the steps, but i cant figure out this specific problem :/