36x-15y+50z= -10
2x+25y=40
54x-5y+30z= -160
2006-10-09 04:33:59 · 5 answers · asked by JP 1 in Education & Reference ➔ Homework Help
Usually, people use matrices to solve this type of problem. That is WAY to involved to explain here, in my opinion. Do a google on matrices -- lots of online help and guidance.
2006-10-09 04:41:42 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Solve the first and last equation for z. Make those two equations equal to each other. Solve the middle equation for y and substitute in the other equation. Solve for X. Put that number into the middle equation and solve for Y. Put Y and X into either the first equation or the second equation and solve for Z.
2006-10-09 05:22:23 · answer #2 · answered by msbedouin 4 · 0⤊ 0⤋
Do you know how do solve systems with two equasions? If so, just take two of the equasions and eliminate a variable so that there are only two variables in it. Then take a different pair and eliminate the same variable. With these two, eliminate a variable so that you only have one left. It takes a long time, but it works.
2006-10-09 04:55:13 · answer #3 · answered by inventor245 2 · 0⤊ 0⤋
u can use metrics
or can do conventionally
36x-15y+50z = -10 ---------------------[1]
2x+25y = 40 --------------------[2]
54x-57y+30z = -160 -------------------[3]
[1]*3
108x-45y+150z= -30 ----------------------[4]
[3]*5
270x-325y+150= -800 ---------------------[5]
[5]-[4]
162x-280y = -770 ---------------------[6]
[2]*81
162x+2025y = 3240---------------------[7]
[6]-[7]
-2305y = -4010
y = -4010/-2305
y=1.74
now u can continue with the value of y in [2]
2006-10-09 05:10:27 · answer #4 · answered by pmmrahman 1 · 0⤊ 0⤋
The easiest way to do it is with matrices, but this would be very difficult to explain here. Try googling matrices for instructions on their us.
2006-10-09 05:08:10 · answer #5 · answered by Buzlite 2 · 0⤊ 0⤋