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For the system of linear equations

x+y+z=5000x + y + z = 5000

6x+7y+8z=358006x + 7y + 8z = 35800

6x+7y−8z=70006x + 7y - 8z = 7000

the values of x, y and z are:

Solution

✅ Correct Option: 3

Given system:

x+y+z=5000x + y + z = 5000 ... (1)

6x+7y+8z=358006x + 7y + 8z = 35800 ... (2)

6x+7y−8z=70006x + 7y - 8z = 7000 ... (3)


Equations (2) and (3) both contain 6x+7y6x + 7y, with +8z+8z and −8z-8z respectively.

Adding equations (2) and (3):

(6x+7y+8z)+(6x+7y−8z)=35800+7000(6x + 7y + 8z) + (6x + 7y - 8z) = 35800 + 7000

12x+14y=4280012x + 14y = 42800

6x+7y=214006x + 7y = 21400 ... (4)


Subtracting equation (3) from (2):

(6x+7y+8z)−(6x+7y−8z)=35800−7000(6x + 7y + 8z) - (6x + 7y - 8z) = 35800 - 7000

16z=2880016z = 28800

z=1800z = 1800


Substituting z=1800z = 1800 into equation (1):

x+y+1800=5000x + y + 1800 = 5000

x+y=3200x + y = 3200 ... (5)


From equation (5): x=3200−yx = 3200 - y

Substituting into equation (4):

6(3200−y)+7y=214006(3200 - y) + 7y = 21400

19200−6y+7y=2140019200 - 6y + 7y = 21400

19200+y=2140019200 + y = 21400

y=2200y = 2200


From equation (5):

x=3200−yx = 3200 - y

x=3200−2200x = 3200 - 2200

x=1000x = 1000


Therefore: x=1000x = 1000, y=2200y = 2200, z=1800z = 1800

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