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ilikequadraticsokay
Solve the system by elimination; 4a - 2b + 3c = -10 5a + 2b - 2c = -28 -3a - 2b + c = 22
pic one variable of one equation and solve for it. LIke pick c of bottom most equation and solve for it then plug it into middle equation
how do you eliminate the a and b though?
i end up with 4a - 2b + 3c = -10 5a + 2b - 2c = -28 -3a - 2b + c = 22 4a - 2b + 3c = -10 5a + 2b - 2c = -28 c = 22 + 3a +2b
take the first two equations and add them...this will cancel out the b's
wait i was doing substitution sorry @ilikequadraticsokay
the link i gave has a great example
4a - 2b + 3c = - 10 5a + 2b - 2c = - 28 ----------------- 9a + c = - 38 now take the next two equations and eliminate b. 5a + 2b - 2c = - 28 -3a - 2b + c = 22 ------------------ 2a - c = - 6 now take the 2 equations you came up with...the ones you cancelled out the b's in. 9a + c = - 38 2a - c = - 6 ----------- 11a = - 44 a = - 4 now sub -4 in for a in either of the above equations 9a + c = - 38 9(-4) + c = - 38 - 36 + c = - 38 c = - 38 + 36 c = - 2 now sub known values into any of the original equations 4a - 2b + 3c = -10 4(-4) - 2b + 3(-2) = - 10 -16 - 2b - 6 = - 10 -2b = - 10 + 22 -2b = 12 b = -6 lets check... 5a + 2b - 2c = - 28 5(-4) + 2(-6) - 2(-2) = - 28 -20 - 12 + 4 = - 28 - 32 + 4 = - 28 - 28 = - 28 (correct) a = -4, b = -6, c = -2 Is there anything I have just done that you don't understand ? I will be happy to explain it to you.
You explained this perfectly, thank you!