Question 998985
I assume your system is
{{{system((4/3)x+(3/2)y=-4/3,-(16/9)x-2y=16/9)}}}
There are fractions in those equations, and they make mistakes more probable, so I would get rid of them first.
To get rid of fractions in an equation, we can multiply both sides of the equal sign times a multiple of all the denominators to get an equivalent equation.
Shown baby step, by baby step, it looks like this:
{{{(4/3)x+(3/2)y=-4/3}}}-->{{{6*((4/3)x+(3/2)y)=6*(-4/3)}}}-->{{{6*(4/3)x+6*(3/2)y=-8)}}}-->{{{8x+9y=-8)}}} .
Working with the other equation in a similar way we get to a much nicer system.
{{{system((4/3)x+(3/2)y=-4/3,-(16/9)x-2y=16/9)}}}--->{{{system(8x+9y=-8,-16x-18y=16)}}} .
The equation {{{-16x-18y=16}}} can be simplified by
dividing both sides of the equal sign by 2:
{{{-16x-18y=16}}}-->{{{-16x/2-18y/2=16/2}}}-->{{{-8x-9y=8}}}
With that we get an equivalent system that is really user-friendly:
{{{system((4/3)x+(3/2)y=-4/3,-(16/9)x-2y=16/9)}}}--->{{{system(8x+9y=-8,-16x-18y=16)}}}--->{{{system(8x+9y=-8,-8x-9y=8)}}} .
Now you realized that the second equation is the first equation multiplied times {{{-1}}} .
They are equivalent equation that would graph as the same line.
There are infinite solutions: all the points on that line.
The system is dependent.