Question

4X+2y+z=14, x+5y-z = 10 , x+y+8z= 20.
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Answer 1

GIVEN :

The equations are 4x+2y+z=14, x+5y-z = 10 , x+y+8z= 20

TO FIND :

The values of x, y and z

SOLUTION :

Given equations are

4x+2y+z=14 ------ (1)

x+5y-z = 10 ------- (2)

x+y+8z= 20 -------- (3)

Solving the given equations by elimination method :

Adding the equations (1) and (2) we get,

4x+2y+z=14

x+5y-z = 10

_________

5x+7y=24 ----- (4)

Multiply the equation (2) into 8

8x+40y-8z=80 ---------- (5)

Now adding the equations (5) and (3) we get

8x+40y-8z=80

x+y+8z= 20

___________

9x+41y=100 ------- (6)

Multiply the equation (4) into 9 we get

45x+63y=216 ----- (7)

Multiply the equation (6) into 5 we get

45x+205y=500 ----- (8)

Subtracting the equations (7) and (8)

45x+63y=216

45x+205y=500 (-)

_____________

    -142y=-284

142y=284

$y=\frac{284}{142}$

∴ y=2

Substitute the value of y=2 in equation (6)

5x+7(2)=24

5x=24-14

5x=10

$x=\frac{10}{5}$

∴ x=2

Substitute the values of y=2 and x=2 in equation (1)

4(2)+2(2)+z=14

8+4+z=14

12+z=14

z=14-12

∴ z=2

∴ the values are x=2, y=2 and z=2

Answer 2

Answer:

4x+2y+z=14, x+5y-z=10, x+y+8z=20