Question

What is the solution to the system?

1) 3x+y+z=7
2) x+3y-z=13
3) y=2x-1

A) (3, 2, −2)


B)(7, 13, −1)


C) (−2, 3, 2)


D)(2, 3, −2)

Answer 1

Answer:D

Explanation:

1) 3x+y+z=7 -----------1

2) x+3y-z=13 ----------2

3) y=2x-1 ----------------3

Put y= 2x-1 in eqn 1 and 2

3x +2x-1+ z =7

5x +z =8-------------4

x+3(2x-1)-z =13

x + 6x -3-z = 13

7x-z = 16------------5

Solving eqn 4 and 5 by elimination method, Add both equations

12x = 24

x = 2

y= 2x-1

y = 4-1 = 3

Put y = 3 and x = 2 in eqn 2

2+3×3-z=13

2+9-z =13

-z = 13-11=2

z =-2

The values are (2,3,-2)

Answer 2

D) (2, 3, -2)

Explanation:

For a question like this, it is often easiest just to check the answers to see which works. The easiest equation to use for that purpose is the one that has only two variables: y = 2x-1. Only selections B and D are successful in that equation.

Selection B doesn't work in equation (2), leaving answer D as the best choice.