Question

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Answer 1

Answer:

3x+2y=16………(1)

y+z=13………….(2)

5x-z=2………….(3)

x+y+z=?………..(4)

From eq (1)

3x=16–2y

x=16–2y/3

From eq (2)

z=13-y

Put the values of x & z in eq (3) we get

5x-z=2

5(16–2y/3)-(13-y)=2

80–10y/3–13+y=2

80–10y-39+3y/3=2 …..by taking L.C.M

41–7y/3=2

41–7y=6

-7y=6–41

-7y=-35 …..- &- cancel each other

7y=35

y=35/7

y=5

put in eq (1) we get

3x+2y=16

3x+2(5)=16

3x=16–10=6

x=6/3

x=2

by puting the value of x in eq (3) we get

5x-z=2

5(2)-z=2

10-z=2

-z=2–10

-z=-8

z=8

put the values of x ,y &z in eq(4)

=x+y+z

=2+5+8

=15 Ans

Just change the numbers

Step-by-step explanation:

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Answer 2

Answer:

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