Solving xy-3x+4y=12

[apple2357]

I am trying to get my head around how to express the solution to

xy-3x+4y=12

I can see how to factorise this to get

xy-3x+4y-12=0

(x+4)(y-3)=0

so x=-4 or y= 3?

But when x=-4, y can be anything ( by substituting back into the first equation), So do i express the solutions as (-4, y)

and (x, 3) if it was pairs of solutions i was interested in?

So are there infinite solutions to this problem?

 

[apple2357]

Also, something else i don't understand.

I could approach the above problem in a different way and rearrange for x ( say) to get

x(y-3) = 12-4y
x = 4(3-y)/(y-3) so x=-4 but y cannot equal 3 ?

How does this not confirm the above?

Am i doing two different things?

 

[lev888]

S(x,y)=xy-3x+4y-12 is a surface in 3d space.
When you solve S=0 you find where the surface intersects the z=0 plane. And the solution is 2 lines: -4,y,0 and x,3,0 (infinite number of points).

 

[mmm4444bot]

Consider: x + y = 12

One equation; two variables; infinite solutions possible.

 

[Steven G]

I am trying to get my head around how to express the solution to

xy-3x+4y=12

I can see how to factorise this to get

xy-3x+4y-12=0

(x+4)(y-3)=0

so x=-4 or y= 3?

But when x=-4, y can be anything ( by substituting back into the first equation), So do i express the solutions as (-4, y)

and (x, 3) if it was pairs of solutions i was interested in?

So are there infinite solutions to this problem?

Yes, everything that you said is true. Good job. Sometimes answers are strange looking. This is the property or zero. Any number times zero = 0. Consider (x-5)(x-6)=0 We can immediately say that x=6 is a solution without even knowing what 6-5 equals! In your example you have (x+4)(y-3)=0. Clearly y=3 is a solution while x-3 we don't care what it equals. This time (unlike my example above) x-3 is NOT a function of y and as a result of this, x can be any value.Please think about all this and ask any question you might have.

 

[Harmless Drudge]

I am trying to get my head around how to express the solution to

xy-3x+4y=12

I can see how to factorise this to get

xy-3x+4y-12=0

(x+4)(y-3)=0

so x=-4 or y= 3?

But when x=-4, y can be anything ( by substituting back into the first equation), So do i express the solutions as (-4, y)

and (x, 3) if it was pairs of solutions i was interested in?

So are there infinite solutions to this problem?

The intersection of an unbounded surface with an unbounded plane will often have an infinite number of points as solutions, comprising and expressed as curves. In this case, those curves are formed as lines. You've described the intersections by parameters. You might be expected to use a specific syntax or notation, but you've cracked the nut entirely already.