Question 39330: Solve for x,y in the following system of three
x + 10y + z = 52
5x + y + 4z =15
x + 2y - 3z = 12
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! SEE THE FOLLOWING EXAMPLE AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK
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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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