Question 1208973
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The original system is
{{{system(-x+2y=-4,2x+y=3)}}}
Let's say we wanted to eliminate the x variable. To do so, we can double both sides of the 1st equation to get this equivalent system
{{{system(-2x+4y=-8,2x+y=3)}}}
At this point the x coefficients are equal in magnitude but opposite sign.
Adding straight down has the x terms go to 0. 
The y terms add to 5y. The right hand sides add to -5.


So 5y = -5 leads to y = -1
Then plug this into any equation mentioned so far. Isolate x.
-x+2y = -4
-x+2(-1) = -4
-x-2 = -4
-x = -4+2
-x = -2
x = 2


We have x = 2 pair up with y = -1
The ordered pair solution is <font color=red size=4>(2, -1)</font> 
This is where the two lines intersect.


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Another approach.


Let's isolate y in the 2nd equation
2x+y = 3 turns into y = -2x+3
That is then plugged into the other equation of the original system so we can solve for x.
-x+2y=-4
-x+2(-2x+3)=-4
-x-4x+6=-4
-5x+6=-4
-5x=-4-6
-5x=-10
x = -10/(-5)
x = 2


Then,
y = -2x+3
y = -2*2+3
y = -4+3
y = -1
We arrive at (x,y) = <font color=red size=4>(2, -1)</font> once again.


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Answer: <font color=red size=4>(2, -1)</font>
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