Question

How do you solve the system `x/2+(2y)/3=32` and `x/4-(5y)/9=40`?

Answer 1

`x = 100 and y = -27`

Explanation:

At first glance this horrifying because of the fractions!

Luckily with equations you can always get rid of any fractions by multiplying each term by the LCM of the denominators.

`xx6 rarr " "x/2+(2y)/3=32" "rarr 3x+4y = 192`

`xx 36rarr" "x/4-(5y)/9=40" "rarr 9x -20y = 1440`

The equations look much better, now we can solve them:

`color(white)(..............)3x+4y = 192......................................A`
`color(white)(..............)9x-20y= 1440...................................B`

`A xx-3:" "color(blue)(-9x)-12y = -576......................C`
`color(white)(........................)color(blue)(9x)-20y= 1440.........................B`

`C+B:" "-32y =864`
`color(white)(................................)color(red)(y=-27)`

Substitute `-27` for `y` in `A`

`" "3x+4(-27) = 192......................................A`
`" "3x-108 = 192`
`" "3x = 192+108`
`" "3x = 300`
`" "color(blue)(x = 100)`

Check the solutions in equation B

`" Is " 9(100)-20(-27)= 1440 ?`
`" "900+540= 1440`
`" Indeed! "1440 = 1440`

We can even check in the original equation for A;

`x/2+(2y)/3=32`

`100/2 +(2xx-27)/3`

`=50-18`

`=32" "larr` the answer is correct.