Question 1179096: Solve the following system of equations for the unknown variables.
3x + 3y = 3
-3x - 2z = -11
y + z = -1
so far i have
y=1-x
x=(2z-11)/-3
and z=-1-y
i cannot figure how to get these equations to solve for a single variable. I am used to problems that have all variables in at least two of the equations. each equation having only two variables has thrown me for a loop...
please help!
Thank you!
Found 4 solutions by MathLover1, mananth, greenestamps, MathTherapy:
Answer by MathLover1(20849)   (Show Source):
Answer by mananth(16946)  (Show Source):
You can put this solution on YOUR website!
3x + 3y = 3
divide by 3
x+y=1
x=1-y
-3x - 2z = -11
substitute x=1-y in above
-3(1-y)-2z =-11
-3+3y-2z=-11
3y-2z =-8
3y-2z =-8-------------1
y+z=-1
multiply by 2 and add to the equation above
2y+2z=-2--------------2
5y=-10
y=-2
plug y=-2 in y+z=-1
z=1
x+y=1 given
x+ -2 = 1
x=3
x=3,y=-2, z=1
Answer by greenestamps(13196)  (Show Source):
You can put this solution on YOUR website!
It is very common for a beginning student to be "thrown for a loop" when each equation does not contain all the variables....
You have received two very different responses from tutors who both use some form of substitution to solve the system of equations.
I would never do that unless one of the given equations expressly gave one variable in terms of the other, like "z=x+2y-3". In my younger days, when I tried solving a system of equations that way, I would often end up with a useless equation like "x+2y=x+2y". That's easy to do if you use substitution and are not careful what direction you go with your work.
The three given equations are all in the form Ax+By+Cz=D, although some of the coefficients are 0. For me at least, it is ALWAYS easiest to solve the system using elimination when the equations are all in that form.
To address your problem of not being able to figure out how to solve a system when some of the coefficients are zero, remember that the first step -- regardless of whether some of the coefficients are zero -- is to eliminate one of the variables.
So now that I have given my speech, let's look at your specific problem.
3x + 3y = 3
-3x - 2z = -11
y + z = -1
Certainly dividing the first equation by 3 will make it easier to see where to go with the problem. So now we have
x + y = 1
-3x - 2z = -11
y + z = -1
What I see here is "y" by itself in the 1st and 3rd equations, and no "y" in the second. So my choice is to eliminate y between the 1st and 3rd equations to give me a second equation in x and z.
Multiply the first equation by -1 and add it to the 3rd equation:
-x -y = -1
y + z = -1
---------------
-x + z = -2
Now my two equations in x and z are
-3x-2z=-11
-x+z=-2
Note again that our objective was finding a way to eliminate one variable to "reduce" the system of three equations in three variables to a system of two equations in two variables. While there are many different ways we could have done that, the "y" with coefficient 1 in two of the equations dictated my choice of how to proceed.
The path from there to the solution should be more familiar. Eliminate z by multiplying the second equation by 2 and adding to the first:
-3x -2z = -11
-2x +2z = -4
--------------
-5x = -15
So x=3.
Now go back to the original first equation with x=3 to find y=-2; then go back to the original third equation with y=-2 to find z=1.
ANSWER: (x,y,z) = (3,-2,1)
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For tutor @MathTherapy....
When are you going to get over your ego trip?
There is no one best way to solve this problem, or any similar problem. The solutions shown by the other tutors are perfectly valid. Stop calling them rubbish because you found a different way to solve the problem.
You found a particular way to solve this system of three equations which works fine but doesn't follow a well-defined process. My response presented a solution by a well-defined process, and it included discussions of WHY the process works.
For a student just learning how to solve problems like this, a response describing HOW to solve this kind of problem is far more useful than a solution that presents a particular solution to this specific problem.
Grow up!!
Answer by MathTherapy(10549)  (Show Source):
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