anonymous
  • anonymous
someone please help me solve this step by step 3x+y=15 4x+5y=9
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
By graphing? or algebraically?
anonymous
  • anonymous
by graphing i tried x=0 solve for y but its not right
anonymous
  • anonymous
You have to graph each line separately

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anonymous
  • anonymous
the first equation i came up with (4,3) and did the x=0 solve y for the second
anonymous
  • anonymous
For the first equation: \[x =0 \implies 3(0) + y = 15 \implies y = 15\] \[y =0 \implies 3x + 0 = 15 \implies 3x = 15 \implies x = 5\] This gives the pionts (0,15) and (5,0) So now you can graph the line that corrisponds to the equation 3x+y=15 by drawing the line between those two points.
anonymous
  • anonymous
Now do the same for the second equation. and see where the two lines cross.
anonymous
  • anonymous
So what do you get for y when you plug in 0 for x in the second equation?
anonymous
  • anonymous
1.8
anonymous
  • anonymous
Just leave it as a fraction for now. It's easier to see =)
anonymous
  • anonymous
9/5 is correct though. Now plug in 0 for y and solve for x.
anonymous
  • anonymous
2.3
anonymous
  • anonymous
Not quite. As I said, leave it as a fraction. \[y =0 \implies 4x + 5(0) = 9 \implies 4x = 9 \implies x = 9/4 = 2.25\]
anonymous
  • anonymous
9/4
anonymous
  • anonymous
Ok, so what are the two points you have found on this line?
anonymous
  • anonymous
idk im confused, i hate this stuff
anonymous
  • anonymous
(0,1.8) (2.3,0)
anonymous
  • anonymous
Yes, exactly right!
anonymous
  • anonymous
except you keep rounding 2.25 to 2.3 for some reason.
anonymous
  • anonymous
thought i was suppose to
anonymous
  • anonymous
So if you draw a line between those two points it'll give you the line that corresponds to the second equation.
anonymous
  • anonymous
That's fine
anonymous
  • anonymous
I think 2 and 1/4 is easier to graph than 2 and 3/10'ths but it depends on your graph paper I suppose.
anonymous
  • anonymous
Now you're wanting to find where these two lines intersect.
anonymous
  • anonymous
according to my graph they dont
anonymous
  • anonymous
They must. They don't have the same slope, so they are not parallel. Any two distinct non-parallel lines must intersect somewhere.
anonymous
  • anonymous
You will need to continue the line beyond the points that you graphed. It will go straight through those 2 points, but continue on infinitely.
anonymous
  • anonymous
equation 1 i got (0,15) (5,0) drew the line 2nd i got (0,1.8) (2.25,0)
radar
  • radar
Look real close around the point (6,-3)
anonymous
  • anonymous
i only drew the lines to the points not thru to the end of the graph
radar
  • radar
You must extend the lines to the point where they intersect. That point is the solution.
anonymous
  • anonymous
ok so then it is considered consistent and dependent
anonymous
  • anonymous
cant figure out how to use my new graphing calc
radar
  • radar
The each equation by itself has many solutions all points on the line are solutions. But when you consider the two equations together, the point they intersect is the only point that is a solution to both equations. I am not familiar with term consistent, maybe you can tell me what that term means??
anonymous
  • anonymous
Consistent means that the equations are either the same line (overlapping) or non-parallel. If they are parallel and do not overlap, they are not consistent because there will be no solution. If they are the same line (each one overlaps the other) then they are dependent.
anonymous
  • anonymous
This system is consistent (they aren't parallel) and independent ( they aren't the same line)
radar
  • radar
Ah so, so you are right tina=ann
anonymous
  • anonymous
yeah thats it, and you can just call me tina
radar
  • radar
ok
radar
  • radar
I believe tina uses her calculator and that is why she comes up with decimal answers rather than the exact fraction
radar
  • radar
tina, I have a graphing calculator, bought it new, still in blister pack at a garage sale for $10.00. I haven't learnt how to use it! lol
anonymous
  • anonymous
yes i use calc, cant figure out how to use my new graphing calc, just spent 150 cant figure out how to use it
anonymous
  • anonymous
your profile says you graduated from Strayer, that is where i am going
radar
  • radar
I guess you use the book and try and follow their examples, eventually you will learn it.
radar
  • radar
Mine is the TI-86
radar
  • radar
Oh really, in VA or DC
anonymous
  • anonymous
ok new problem solve by substitution 6x-2y=-56 5x+48=y
anonymous
  • anonymous
chesapeke Va
anonymous
  • anonymous
mine is TI-Nspire
radar
  • radar
I went to the Manassas,VA campus, graduated at the Kennedy Center in DC..Ok lets look at the substitution problem. I think we should write the last equation like; y=5x+48 do you agree?
anonymous
  • anonymous
i think so i got (-40,-192) for the solution
radar
  • radar
Humor me as I wade through this, you are ahead of me! lol I want to subsitute for y in the first equation so it will look like this: 6x -2(5x+48)=-56 6x-10x-96=-56 -4x=40 x=-10 Did I do that right?
anonymous
  • anonymous
yes you did i must of had a brain fart because i forgot to divide
radar
  • radar
I don't think I violated any laws of algebra! Now going back to the second equation where we had y=5x+48 substitute are newly found value of x in that we get: y=5(-10)+48=-2 y=-2 The solution according to the above : x=-10, y=-2
anonymous
  • anonymous
and then you add 60+48 to get 108 right?
anonymous
  • anonymous
oh yeah im way off
anonymous
  • anonymous
is that 10 or -10
radar
  • radar
It is a -50 added to 48. Please go over what I did and question me on any thing I did.
anonymous
  • anonymous
im just relearning this and i wasnt ever good at it to begin with
radar
  • radar
You question the solution of -10 for x?
radar
  • radar
I will put it here so you can ask away. I want to subsitute for y in the first equation so it will look like this: 6x -2(5x+48)=-56 6x-10x-96=-56 -4x=40 x=-10
radar
  • radar
It didn't paste so well lol should be a new line after the -56 Note that the (5x+48) was substituted for y inthe first equation. Do you see that?
radar
  • radar
However, in the first equation the coefficient of y was a -2 So you have to multiply (5x+48) by -2 It then becomes -10x-96
anonymous
  • anonymous
ok i got it i new a new one for you two angles are supplementary one angle is 3 degrees less than twice the other. find the measures of the angles
radar
  • radar
So now the first equation with the substitution of y looks like this: 6x-10x-96=-56 Combining similar terms it becomes: -4x-96=-56 and so on.
radar
  • radar
Two angles are supplementary if their sum is what? Is it 90 or 180 degrees?
anonymous
  • anonymous
180 complementary is 90
radar
  • radar
Let x = one of the angles this then means that 180-x is the other angle. Does seem sensible?
anonymous
  • anonymous
i was taught x+y=180 x=2y-3 (2y-3)+y=180 3y-3=180 3y=180 y=61 x=2(61)-3 x=119 is this right?
radar
  • radar
I am looking at your work
radar
  • radar
You did good until you got to 3y-3=180 good so far. Now look at that step after that and find your error
radar
  • radar
Did you find your error in that next step
anonymous
  • anonymous
i added 3 instead of subtracting?
radar
  • radar
Well you should of added 3 to both sides and then you would of had 3y=183
anonymous
  • anonymous
i did and 3y=183 3(61)=183
anonymous
  • anonymous
so y=61
radar
  • radar
I just saw where you had 3y=180 some how you got the right answer lol
radar
  • radar
The other angle is supplementary so subtract the 61 from 180 and it is solved!
anonymous
  • anonymous
i wrote it down right on my paper i got lots of math problems. i just need someone to help me solve them or keep it on track so i get the right answers
radar
  • radar
Did you say that you have graduated from Strayer or that you are presently going.
anonymous
  • anonymous
presently going
radar
  • radar
I graduated around 1992. I no longer live in VA or work in DC, I am retired and live in the Ozarks of MO. Way out in the woods! lol
radar
  • radar
Good luck tina
anonymous
  • anonymous
thank you, im not good at this kind of math

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