Question 1203437: 3x+6y-z=7
2x-3y+5z=-12
8x+9y-2z=12
Found 3 solutions by josgarithmetic, math_tutor2020, greenestamps:
Answer by josgarithmetic(39630)   (Show Source):
You can put this solution on YOUR website! Not being patient to try, instead using wolframalpha.com for systems of linear equations,
The page also shows a button for "step-by-step solution" in case you are interested or need.
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
Solve the 1st equation for z.
3x+6y-z = 7
3x+6y = 7+z
3x+6y-7 = z
z = 3x+6y-7
Plug this into the 2nd equation.
2x-3y+5z = -12
2x-3y+5(3x+6y-7) = -12
2x-3y+15x+30y-35 = -12
17x+27y-35 = -12
17x+27y = -12+35
17x+27y = 23
Do the same for the 3rd equation.
8x+9y-2z=12
8x+9y-2(3x+6y-7)=12
8x+9y-6x-12y+14=12
2x-3y+14=12
2x-3y=12-14
2x-3y=-2
The task is now to solve this reduced system
17x+27y = 23
2x-3y = -2
We can solve for x in the equation 2x-3y = -2 to get x = 1.5y-1
Then plug this into the other equation to solve for y
17x+27y = 23
17(1.5y-1)+27y = 23
25.5y-17+27y = 23
52.5y-17 = 23
52.5y = 23+17
52.5y = 40
y = 40/52.5
y = 400/525
y = 16/21
Use this value of y to find x.
Then use the values of x and y to find z.
I'll let the student take over from here.
Answer by greenestamps(13215)  (Show Source):
You can put this solution on YOUR website!
The button on wolframalpha.com for seeing the step-by-step solution does not work unless you have acquired special privileges on their site. And the beginning of the step-by-step solution you can see shows a rigid solution method that might not be the easiest.
And with a system of 3 equations I would not use substitution as the other tutor does... although that is a valid method.
Given three linear equations in three variables, I would definitely use elimination instead of substitution.
The general solution method for a system of any number of linear equations is to eliminate one variable at a time.
Our equations are
 [1]
 [2]
 [3]
Use the "-z" in the first equation to eliminate the variable z in the other two equations, using elimination.
 [1], multiplied by 5
[2]
 [4] [the sum of those two equations]
and
 [1], multiplied by -2
[2]
 [5] [the sum of those two equations]
We have reduced the system of 3 equations in 3 variables to a system of 2 equations in 2 variables. Use elimination again to eliminate one of the variables; solve for the remaining variable, and back substitute to find the final answer.
[4]
 [5], multiplied by 9
 [the sum of those two equations]
Substitute x = 1/7 in [5] and solve for y
Substitute x = 1/7 and y = 16/21 in [1] and solve for z
ANSWER: (x,y,z) = (1/7,16/21,-2)
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