How can one halfx - 5 = one thirdx + 6 be set up as a system of equations? 2y + x = - 10 3y + x = 18 2y + 2x = - 10 3y + 3x = 18 2y - x = - 10 3y - x = 18 2y - 2x = - 10 3y - 3x = 18

2y-x = - 10

3y-x = 18

Step-by-step explanation:

The correct option is. 2y-x = - 10, 3y-x = 18

one half x - 5 = 1/2 (x-5)

one third x + 6 = 1/3 * (x+6)

1/2 x-5 = 1/3x+6 = y

y = 1/2x-5

y = 1/3x+6

Now,

y=x/2 - 5 equation 1

y = x/3 + 6 equation 2

By taking L. C. M of the first equation we get:

y=x/2 - 5

y = x-10/2

Now multiply both terms by 2.

2y=x-10

2y-x = - 10

Now lets solve second equation:

Take L. C. M of the second equation:

y = x/3 + 6

y=x+18/3

Multiply both sides by 3

3y = x+18

3y-x = 18

Therefore the system of equations we get is:

2y-x = - 10

3y-x = 18 ...