help me please :( :( :( solve the system by elimination x + 5y - 4z = -10 2x - y + 5z = -9 2x - 10y - 5z = 0 1. (5, -1, 0) 2. (-5, 1, 0) 3. (-5, -1. 0) 4. (-5, -1, -2)
set al the acquation =0 than subtract al the acqations zo that one or two variable = 0 and sovle left over the vartiable
can you help me do that?
10 + 1x + 5y - 4z = 0 9 + 2x - 1y + 5z = 0 0 + 2x - 10y - 5z =0
Do i put this in slope intercept form? idk where to put the z though
multiply equations so when you withdraw al the equations, you wil be left with one variable
im still confused.. do i have to graph these too to get those points?
are you sure it is 2x both times? not one 2x and one -2x? that will make the question a lot easier
Pair the 1st Equation with the 2nd Equation x + 5y - 4z = -10 2x - y + 5z = -9 Pair the 2nd Equation with the 3rd Equation 2x - y + 5z = -9 2x - 10y - 5z = 0 Eliminate the same variable in both, preferably z. Afterwards, you're left with a system of two equations of variables x and y...which you should be able to solve.
@hellothere2
is it (-5, 1, 0) @Hero
Did you set up the system in accordance with my suggestion?
Of course, the easiest way to solve this would have been to just plug the points.
(-5,1,0) can't be it because it fails for the first equation: x + 5y - 4z = -10 -5 + 5(1) -4(0) = -10 -5 + 5 - 0 = -10 0 = -10 False
If they give you the points, keep plugging in until you find the right point.
so -5, -1, 0? or -5, -1, 2?
I suppose you're going to keep guessing until I say which one is correct. If you actually try doing the problem, you'll be able to figure out which one is correct on your own.
i plugged it in and i do believe it is (-5, -1, 0)
@Hero
Very good
thanks for helping me set it up, i knew they were both negatives thats why i asked
If they don't give you any possible points, then you can try the other method I suggested
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