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Pavlova-9 [17]
3 years ago
9

Solve the system below

Mathematics
1 answer:
bija089 [108]3 years ago
4 0

Answer: x=-1, y=8 and z=-5.

Step-by-step explanation:

The given system of equations:

3x - 5y + 2z = -53   (i)

x - 7y - 4z = -37     (ii)

5x + 9y + 2z = 57   (iii)

Multiply 3 to equation (ii)

3x-21y-12z= -111   (iv)

Eliminate (iv) from (i) , we get

-5y+21y+2z+12z=-53+111

⇒ 16y+14z=58

⇒ 8y+7z=29     (v) [divide equation by 2]

Multiply 5 to equation (ii)

5x-35y-20z= -185   (vi)

Eliminate (vi) from (iii) , we get

9y+35y+2z+20z=57+185

⇒ 44y+22z=242

⇒ 2y+z=11    [divide equation by 22]

⇒ z= 11-2y (vii)

Put value of z in (v) , we get

8y+7(11-2y)=29

  ⇒ 8y+77-14y=29

 ⇒ 6y=48

⇒ y=8

Put value of y in (vii), we get z= 11-2(8) =11-16=-5

i.e. z=-5

Put y=8 and z=-5 in (ii) , we get

x - 7(8) - 4(-5) = -37

⇒x-56+20=-37

⇒ x-36=-37

⇒ x = -37+36

⇒ x = -1

Hence, x=-1, y=8 and z=-5.

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4 years ago
How would you go about finding the solution to this system of three equations in three variables? Be specific. For example, you
kupik [55]
I:2x – y + z = 7
II:x + 2y – 5z = -1
III:x – y = 6

you can first use III and substitute x or y to eliminate it in I and II (in this case x):
III: x=6+y
-> substitute x in I and II:
I': 2*(6+y)-y+z=7
12+2y-y+z=7
y+z=-5

II':(6+y)+2y-5z=-1
3y+6-5z=-1
3y-5z=-7

then you can subtract II' from 3*I' to eliminate y:
3*I'=3y+3z=-15

3*I'-II':
3y+3z-(3y-5z)=-15-(-7)
8z=-8
z=-1

insert z in II' to calculate y:
3y-5z=-7
3y+5=-7
3y=-12
y=-4

insert y into III to calculate x:
x-(-4)=6
x+4=6
x=2

so the solution is
x=2
y=-4
z=-1
3 0
4 years ago
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