Question

How do you determine whether a linear system has one solution, many solutions, or no solution when given 3x-8y=0 and -2x+5y=-2?

Answer 1

I found `x=16` and `y=6`

Explanation:

A fast way to "see" the various possibilities of solutions of your system is to look at the numerical coefficients remembering that, basically, here you are dealing with two equations representing two straight lines (that can cross each other, be parallel or one on top of the other):

Case 1: the coefficients are all the same:
example:
`{(3x+2y=4),(3x+2y=4):}`
In this case you have the same equations representing the same lines.
Although a bit weird you can see that a line crosses itself in an `oo` number of points corresponding to `oo` solutions!!!

Case 1a: the coefficients of one equation are multiples of the ones of the other (i.e., they are multiplied by a number):
example:
`{(3x+2y=4),(6x+4y=8):}`
The coefficients of the second equations are equal to the ones of the first multiplied by `2`: the two lines described by the two equations are SUPERIMPOSED one over the other so, again, they cross each other in an `oo` number of points corresponding to `oo` solutions!!!

Case 2: the coefficients of `x` and `y` are either equal or multiples and only the independent coefficients are different:
example:
`{(3x+2y=4),(6x+4y=7):}`
In this case the two lines are PARALLEL !!! So, they cannot cross and so they will not have solutions in common and you will have a system with no solutions.

Case 3: The coefficients of `x` and `y` of the two equations have no connection or fixed relationship:
example:
`{(3x-8y=0),(-2x+5y=-2):}`
In this case the two lines will cross in one point of coordinates `x,y` that represents the solution of your system:

isolate `x` from the first:
`x=8/3y`
substitute into the second:
`-2(8/3y)+5y=-2` rearranging:
`-16y+15y=-6`
`y=6`
back into: `x=8/3y`
`x=8/3xx6=16`.