Question

How can one halfx − 5 = one thirdx + 6 be set up as a system of equations?

Answer 1

The system of equations is

`x-2y=10`

`x-3y=-18`

Explanation:

we have

`(1)/(2)x-5=(1)/(3)x+6`

Let set both sides of the equation separately equal to y

so

Left side

`y=(1)/(2)x-5`

Multiply both sides by 2 to remove the fraction

`2y=x-10`

Rewrite as standard equation

`x-2y=10` -----> equation A

Right side

`y=(1)/(3)x+6`

Multiply both sides by 3 to remove the fraction

`3y=x+18`

Rewrite as standard equation

`x-3y=-18` -----> equation B

therefore

The system of equations is

`x-2y=10`

`x-3y=-18`