Question 1197248
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I'll do problem 1 to get you started.


The given system is this
x+4y = 16
3x+5y = 20


Triple everything in the first equation to go from 
x+4y = 16
to
3x+12y = 48


This is done so that the x terms have the same coefficient.


Here's what the system of equations looks like now
3x+12y = 48
3x+5y = 20


Now subtract the equations straight down
3x-3x turns into 0x or 0, so the x terms go away
12y-5y turns into 7y
48-20 turns into 28


We get 0x+7y = 28 or 7y = 28
Divide both sides by 7 to get y = 4


Now use this y value to find x.
We can use it with any equation involving x and y mentioned earlier.
I'll use the first equation
x+4y = 16
x+4*4 = 16
x+16 = 16
x = 16-16
x = 0


Answer: The solution is <font color=red>(x,y) = (0,4)</font>


Check:
Plug x = 0 and y = 4 into the first equation
x+4y = 16
0+4*4 = 16
16 = 16
That works out. Now do the same for the second equation
3x+5y = 20
3*0+5*4 = 20
20 = 20
That is confirmed as well. 
Both equations are true for x = 0 and y = 4, which fully confirms the solution.
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