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

468

step-by-step explanation: