Question 998985
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(4/3)x   + (3/2)y = -4/3,
(-16/9)x -     2y = 16/9.

Multiply the first equation by &nbsp;{{{4/3}}}&nbsp; (both sides). You will get 

 {{{(16/9)x}}} + {{{2y}}} = {{{-16/9}}}
{{{(-16/9)x}}} - {{{2y}}} =  {{{16/9}}}

Now add both equations. You will get

0*x + 0*y = 0.

In this way you excluded x. But you excluded y in the same time, too.

It is because your original system is dependent: the equations are proportional, and the right sides are proportional with the same coefficient of proportionality.

As a result, your system has infinitely many solution.

You can assign any value to x and then to determine an appropriate value of y from, let say, the first equation of your original system. 
Then this pair of values (x,y) will be the solution of the second equation, too.

Is it clear to you?

If you have question, you can put it into the "Student comment" section. 
Do not forget to point the number of this problem (# 998985) in order I could identify it.

Good luck.
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