I have a huge test tomorrow, but I still don't know how to solve 3 system of equations with 3 variables, somehow I always end up getting fractions. Can you please help me? Thank you.
4x+6y+5z=66
x+y+z=14
y=z
and
2T+1S+1B=21
3T+2B=31
1S=2B=19
4x+6y+5z=66
x+y+z=14
y=z
and
2T+1S+1B=21
3T+2B=31
1S=2B=19
-
I'll use different numbers just to teach you, so you don't copy it. once you learn it try it on your problem.
(1)4x + 5y - 3z = 10
(2) 2x - 4y + z = 6
(3) x + y - z = 8
pick a variable to cancel out with another equation. for example, if you choose equations 1 and 2, and you want to eliminate x, you need to turn one into +4 and one into -4. you could do this by multiplying equation 2 by -2 (of course, you need to do that to every term in the equation)
so you end up with (1) 4x + 5y - 3z = 10
(2) -4x + 8y -2z = -12
add the equations. the x terms will cancel out, which is your goal. you will end up with 13y -5z = -2
now do the same thing, eliminating the same variable (x) with another set of equations from the three.
if you were to use equations 2 and 3, you would need to change the x terms into +2 and -2. to do this, multiply equation 3 by -2.
so you end up with (2) 2x - 4y + z = 6
(3) -2x -2y +2z = -16
again, the x terms cancel out. so after adding, you end up with 2y +3z = -10
now that you have two equations with only two variables (after getting rid of x) you can eliminate another variable within these two equations.
the equations were 13y - 5z = -2 and 2y + 3z = -10
let's eliminate the z terms, since they're easier. multiply the first one by 3 and the second by 5, giving you z terms of 15 and -15.
39y - 15z = -6
10y + 15z = -50
again, the z terms cancel each other out so after adding you get 49y = -56
... these numbers are kinda nasty lol, since I made the equations up on the spot. but y would be equal to -56/49. plug this into one of the equations with two variables, and you will be able to find the other variable in the equation. plug both into one of the equations with three variables, and find the third. then list the numbers like so (x,y,z) and that's your answer.
(1)4x + 5y - 3z = 10
(2) 2x - 4y + z = 6
(3) x + y - z = 8
pick a variable to cancel out with another equation. for example, if you choose equations 1 and 2, and you want to eliminate x, you need to turn one into +4 and one into -4. you could do this by multiplying equation 2 by -2 (of course, you need to do that to every term in the equation)
so you end up with (1) 4x + 5y - 3z = 10
(2) -4x + 8y -2z = -12
add the equations. the x terms will cancel out, which is your goal. you will end up with 13y -5z = -2
now do the same thing, eliminating the same variable (x) with another set of equations from the three.
if you were to use equations 2 and 3, you would need to change the x terms into +2 and -2. to do this, multiply equation 3 by -2.
so you end up with (2) 2x - 4y + z = 6
(3) -2x -2y +2z = -16
again, the x terms cancel out. so after adding, you end up with 2y +3z = -10
now that you have two equations with only two variables (after getting rid of x) you can eliminate another variable within these two equations.
the equations were 13y - 5z = -2 and 2y + 3z = -10
let's eliminate the z terms, since they're easier. multiply the first one by 3 and the second by 5, giving you z terms of 15 and -15.
39y - 15z = -6
10y + 15z = -50
again, the z terms cancel each other out so after adding you get 49y = -56
... these numbers are kinda nasty lol, since I made the equations up on the spot. but y would be equal to -56/49. plug this into one of the equations with two variables, and you will be able to find the other variable in the equation. plug both into one of the equations with three variables, and find the third. then list the numbers like so (x,y,z) and that's your answer.
12
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