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By graphing? or algebraically?
by graphing i tried x=0 solve for y but its not right
You have to graph each line separately
the first equation i came up with (4,3) and did the x=0 solve y for the second
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.
Now do the same for the second equation. and see where the two lines cross.
So what do you get for y when you plug in 0 for x in the second equation?
Just leave it as a fraction for now. It's easier to see =)
9/5 is correct though. Now plug in 0 for y and solve for x.
Not quite. As I said, leave it as a fraction. \[y =0 \implies 4x + 5(0) = 9 \implies 4x = 9 \implies x = 9/4 = 2.25\]
Ok, so what are the two points you have found on this line?
idk im confused, i hate this stuff
Yes, exactly right!
except you keep rounding 2.25 to 2.3 for some reason.
thought i was suppose to
So if you draw a line between those two points it'll give you the line that corresponds to the second equation.
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.
Now you're wanting to find where these two lines intersect.
according to my graph they dont
They must. They don't have the same slope, so they are not parallel. Any two distinct non-parallel lines must intersect somewhere.
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.
equation 1 i got (0,15) (5,0) drew the line 2nd i got (0,1.8) (2.25,0)
Look real close around the point (6,-3)
i only drew the lines to the points not thru to the end of the graph
You must extend the lines to the point where they intersect. That point is the solution.
ok so then it is considered consistent and dependent
cant figure out how to use my new graphing calc
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??
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.
This system is consistent (they aren't parallel) and independent ( they aren't the same line)
Ah so, so you are right tina=ann
yeah thats it, and you can just call me tina
I believe tina uses her calculator and that is why she comes up with decimal answers rather than the exact fraction
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
yes i use calc, cant figure out how to use my new graphing calc, just spent 150 cant figure out how to use it
your profile says you graduated from Strayer, that is where i am going
I guess you use the book and try and follow their examples, eventually you will learn it.
Mine is the TI-86
Oh really, in VA or DC
ok new problem solve by substitution 6x-2y=-56 5x+48=y
mine is TI-Nspire
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?
i think so i got (-40,-192) for the solution
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?
yes you did i must of had a brain fart because i forgot to divide
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
and then you add 60+48 to get 108 right?
oh yeah im way off
is that 10 or -10
It is a -50 added to 48. Please go over what I did and question me on any thing I did.
im just relearning this and i wasnt ever good at it to begin with
You question the solution of -10 for x?
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
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?
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
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
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.
Two angles are supplementary if their sum is what? Is it 90 or 180 degrees?
180 complementary is 90
Let x = one of the angles this then means that 180-x is the other angle. Does seem sensible?
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?
I am looking at your work
You did good until you got to 3y-3=180 good so far. Now look at that step after that and find your error
Did you find your error in that next step
i added 3 instead of subtracting?
Well you should of added 3 to both sides and then you would of had 3y=183
i did and 3y=183 3(61)=183
I just saw where you had 3y=180 some how you got the right answer lol
The other angle is supplementary so subtract the 61 from 180 and it is solved!
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
Did you say that you have graduated from Strayer or that you are presently going.
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
Good luck tina
thank you, im not good at this kind of math