System of linear equations: x + 2y+6z= 4 -3x+2y-z= -4 4x+ 2z=16
Pick one variable to eliminate. Let us say we want to eliminate z. Find z from the last equation in terms of x put it in the first 2 equations and you will have two equations and 2 unknowns, namely, x and y. Now pick another variable to eliminate from these two equations and you would have solved for one variable. Then you can put back in the equations and find the other two.
So I can use the first two equations and eliminate z?
Yes. Find z from the third equation and substitute that for z in first 2 equations.
Am I supposed to get x by itself by eliminating y? or do i use substitution
Whichever method you find convenient. What are the 2 equations you are getting with x and Y?
After canceling out z using the first two equations...I got x+2y=4 -18x+12y=-24
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After substituting z I am getting: -x + 2y = 4 -11x + 2y = -44
You eliminated z by multiplying by 11?
4x + 2z = 16 2z = 16 - 4x z = 8 - 2x
Ok I see where that is coming from. Now where do i plug in the equation we found for z?
z goes into the first and second equations.
so with x+2y+6(-2x+4)=4 What do I do to the y still left
simplify the above first. keep x and y on the left and take the constant to the right.
so on the first one -13x+2y=-20?
no. try again.
z = 8 - 2x Make the correct substitution in first equation.
let me check my work
-11x+2y=-44?
yes. Now we have one equation with no z. Sub z in equation 2.
for the second one : -x+2y=4?
yes. Now you can easily eliminate y from these two equations by subtracting.
I got x=4.8
yes. put that in the last equation, -x+2y=4, and solve for y.
Then put x in z = 8 - 2x and solve for z.
z=-1.6?
yes all three values are correct.
If you want to know whether you got the right answers you can always put the x,y,z values you got into each of the equations and see if it satisfies all three.
thank you so much for being patient! i just checked back and plugged in all the values in the equations and checked them. They are all correct!This was much appreciated
you are very welcome.
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