Question

How do you solve the pair of equations `1/2x+1/3y=2` and `1/(3x) + 1/(2y) = 13/6` by reducing them to a pair of linear equations?

Answer 1

See explanation

Explanation:

`color(brown)("Reduction of an equation "->" rewriting in a simpler form")`

Consider `1/(3x)+1/(2y)=13/6`

`1/(2y)=13/6-1/(3x)`

`1/(2y)=(13x-2)/(6x)`

Turn the whole thing upside down

`2y=(6x)/(13x-2)`

`y=(6x)/(2(13x-2)) = (3x)/(13x-2)`....................Equation(1)

Not sure the above is in a 'simpler form'!
,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Consider:`" "x/2+y/3=2`

`y/3=2-x/2`

`y=-3/2x+6` ...................................Equation(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Equating both through y

`-3/2x+6=y=(3x)/(13x-2)`

`-3/2x+6=(3x)/(13x-2)`

Multiply both sides by 2

`-3x+12=(6x)/(13x-2)`

Multiply both sides by `(13x-2)`

`(-3x+12)(13x-2)=6x`

`-39x^2+162x-24=6x`

`-39x^2+156x-24=0`

Divide throughout by 3

`-13x^2+52-8=0`

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`=>x=(-52+-sqrt( 2704-4(-13)(-8)))/(2(-13))`

`x=2+-sqrt(2288)/(-26)`

Factors of 2288 are `4^2xx143`

`x=2+-4sqrt(143)/(-26)`

I am running out of time so I will let you take it from hear

Answer 2

`x/2+y/3=2->3x+2y=12`
`1/(3x)+1/(2y)=13/6->3x+2y=13 xy` then `13 xy = 12`

Now

`13/3y(3x+2y=12)-> 13 xy +26/3y^2=52y`
`->26/3y^2-52y+12 = 0`

Solving for `y` gives

`y = 3/13 (13 pm sqrt[143])`

substituting in `13 xy = 12` gives

`x = 4/(13pmsqrt[143])`

pairing we have

#((x = 4/(13 + sqrt[143]),y = 3/13 (13 + sqrt[143])), (x = 4/(13 - sqrt[143]),y =3/13 (13 - sqrt[143]) ))#