Question 1199023
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The rule on this website is that you need to post one problem at a time (or one problem per post). 
I'll do problem 1 to get you started.


The system we have is
{{{system(x-2y=5,2x+3y=-32)}}}
If we were to add the equations straight down, then:
x+2x = 3x
-2y+3y = 1y = y
None of the variables cancel. 
Same goes if we were to subtract the equations.


What we need is to have the coefficients of either x or y match up so they cancel.


Let's say we doubled everything in the 1st equation
We'd go from x-2y = 5 to 2x-4y = 10


Now we have this updated equivalent system
{{{system(2x-4y=10,2x+3y=-32)}}}
We can subtract the equations straight down
2x-2x = 0x = 0, the x terms go away
-4x-3y = -7y
10-(-32) = 10+32 = 42


We end up with the equation -7y = 42 which solves to y = -6.


Then use that y value to find x.
Pick either equation with x and y.
Let's say we chose the 1st equation.
x - 2y = 5
x - 2(-6) = 5
x + 12 = 5
x = 5-12
x = -7
Or you can pick on the 2nd equation.
2x+3y = -32
2x+3(-6) = -32
2x-18 = -32
2x = -32+18
2x = -14
x = -14/2
x = -7
Either way you should get the same x value. 
This helps confirm you have the correct solution.


A visual way to confirm the answer is to graph each equation onto the same xy grid.
I recommend either Desmos or GeoGebra as two graphing options.
{{{graph(400,400,-10,10,-10,10,-100,(5-x)/(-2),(-32-2x)/(3))}}}
x - 2y = 5 in green and 2x + 3y = -32 in blue
The two lines intersect at (-7, -6)


Or you could plug each coordinate of (-7,-6) back into the original equations. 
You should get true results after simplifying everything.


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Answer: (x,y) = (-7, -6)
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