Question 14103: 2x+3y+4z=2
5x-2y+3z=0
x-5y-2z=-4
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! SEE THE FOLLOWING SOLUTION GIVEN BY ME TO A SIMILAR PROBLEM AND WORK THE SOLUTION YOUR SELF.
Answer 7295 by venugopalramana(11) on 2005-10-10 05:14:13 (Show Source):
Hi Tutors:
I have working on this problem for some time but I am still having a hard time working this one:
I am suppose to solve this system using elimination method
10x+6y+z=7 (1)
5x-9y-2z=3 (2)
15x-12y+2z=-5 (3)
20x-21y=-2 (4)
Please show me how to work this problem from this point.
Thank you
Good you have proceeded correctly and infact on the way to solving the problem by your self..you only need a little guidance on the path you should follow to solve the problem..o.k. ..let us see you have added equations 2 and 3 to get equation 3 ,which has accomplished elimination of one unknown z . The basic procedure is , if we start with 3 equations in 3 unknowns ,we try to eliminate one unknown taking one pair of equations at a time to get 2 new equations in 2 unknowns only.Then we take those 2 new equations to eliminate one another unknown to get one more new equation , but this time with one unknown only.This we can easily solve to find the unknown.Now , we travel backwards along the same path as we travelled to find the 2 other unknowns one after another by substituting the known values every time.Let us illustrate the procedure now with this example.Now that you have already got one new equation 4 from 2 and 3 to eliminate z., let us take equations 1 and 2 to eliminate the same unknown z.For this we observe the coefficients of z in the two equations which are 1 and -2 respectively.So we multiply equation 1 with 2 and add it to equation 2.
Eqn.1 * 2 gives us ...20x+12y+2z=14 .....(5)
Eqn.2 is .............5x-9y-2z = 3........(6)
Eqn.5 + Eqn.6 gives us .....25x+3y = 17....(7)
but from Eqn.4 we have .....20x-21y=-2......(4)..proceeding on the same basis ,we eliminate y from these 2 equations.
Eqn.7 * 7 gives us .........175x+21y=119....(8)
Eqn.8 + Eqn.4 gives us .....195x=117 ..or x= 117/195 = 39/65 = 3/5.....now substitute this value of x in eqn.4 to get y
y=(20*(3/5)+2)/21=14/21=2/3…….now substitute these values of x and y in eqn.1 to get z.
z=(7-10*(3/5)-6*(2/3))=-3………….. As a check ,you can substitute these values of x,y,and z in the 3 given equations to
verify that your answer is correct.
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