SOLUTION: page 718 of chapter 8 section 3 question #16 #18 #28 #38 I am double checking my work and noticed after doing these 4 yesterday, they are confusing the heck out of me today. Can

Question 90937: page 718 of chapter 8 section 3 question #16 #18 #28 #38
I am double checking my work and noticed after doing these 4 yesterday, they are confusing the heck out of me today. Can you explain how to solve these so I can refresh my mind and be assured that I done them properly before I turn in my assignments today? (I am giving four different styles)
#16 solve by substitution 5x-2y=-5 then y-5x=3 now question #20 solve by substitution 8x-4y=16 then y=2x-4 now question #28 solve by substitution 4x-12y=5 then -x+3y=-1 and question #38 solve by either addition or substitution. If a unique solution does not exist, stste wether it is dependent or inconsistent. 10x+2y=7 and y=-5x+3 THANKS! My answers that I had for #16 was x=-1/5 and y=2 then I did it over today and got x =0.6 and y =4 so I am just lost now.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
#16 solve by substitution
1st:5x-2y=-5
2nd: y-5x=3
--------------
Rewrite as:
1st: 5x-2y=-5
3rd: y = 5x+3
------------
Substitute 3rd into 1st to get:
4th: 5x-2(5x+3) = -5
5x-10x-6 = -5
-5x=1
x= -1/5
-------------
Substitute into 3rd to solve for y:
y = 5(-1/5)+3
y = -1+3
y = 2
==================
Check your answer in 1st:
1st: 5x-2y=-5
5(-1/5)-2(2)= -5
-1-4=-5
-5=-5
It checks. You could do this on each of your other problems.
============
Cheers,
Stan H.

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