someone please help me solve this step by step
3x+y=15
4x+5y=9

- anonymous

someone please help me solve this step by step
3x+y=15
4x+5y=9

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- anonymous

By graphing? or algebraically?

- anonymous

by graphing i tried x=0 solve for y but its not right

- anonymous

You have to graph each line separately

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## More answers

- anonymous

the first equation i came up with (4,3)
and did the x=0 solve y for the second

- 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

Now do the same for the second equation. and see where the two lines cross.

- anonymous

So what do you get for y when you plug in 0 for x in the second equation?

- anonymous

1.8

- anonymous

Just leave it as a fraction for now. It's easier to see =)

- anonymous

9/5 is correct though. Now plug in 0 for y and solve for x.

- anonymous

2.3

- 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

9/4

- anonymous

Ok, so what are the two points you have found on this line?

- anonymous

idk im confused, i hate this stuff

- anonymous

(0,1.8) (2.3,0)

- anonymous

Yes, exactly right!

- anonymous

except you keep rounding 2.25 to 2.3 for some reason.

- anonymous

thought i was suppose to

- anonymous

So if you draw a line between those two points it'll give you the line that corresponds to the second equation.

- anonymous

That's fine

- 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

Now you're wanting to find where these two lines intersect.

- anonymous

according to my graph they dont

- 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

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

equation 1 i got (0,15) (5,0) drew the line
2nd i got (0,1.8) (2.25,0)

- radar

Look real close around the point (6,-3)

- anonymous

i only drew the lines to the points not thru to the end of the graph

- radar

You must extend the lines to the point where they intersect. That point is the solution.

- anonymous

ok so then it is considered consistent and dependent

- anonymous

cant figure out how to use my new graphing calc

- 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

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

This system is consistent (they aren't parallel) and independent ( they aren't the same line)

- radar

Ah so, so you are right tina=ann

- anonymous

yeah thats it, and you can just call me tina

- radar

ok

- radar

I believe tina uses her calculator and that is why she comes up with decimal answers rather than the exact fraction

- 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

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

your profile says you graduated from Strayer, that is where i am going

- radar

I guess you use the book and try and follow their examples, eventually you will learn it.

- radar

Mine is the TI-86

- radar

Oh really, in VA or DC

- anonymous

ok new problem solve by substitution
6x-2y=-56
5x+48=y

- anonymous

chesapeke Va

- anonymous

mine is TI-Nspire

- 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

i think so
i got (-40,-192) for the solution

- 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

yes you did i must of had a brain fart because i forgot to divide

- 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

and then you add 60+48 to get 108 right?

- anonymous

oh yeah im way off

- anonymous

is that 10 or -10

- radar

It is a -50 added to 48.
Please go over what I did and question me on any thing I did.

- anonymous

im just relearning this and i wasnt ever good at it to begin with

- radar

You question the solution of -10 for x?

- 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

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

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

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

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

Two angles are supplementary if their sum is what? Is it 90 or 180 degrees?

- anonymous

180 complementary is 90

- radar

Let x = one of the angles this then means that 180-x is the other angle. Does seem sensible?

- 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

I am looking at your work

- 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

Did you find your error in that next step

- anonymous

i added 3 instead of subtracting?

- radar

Well you should of added 3 to both sides and then you would of had
3y=183

- anonymous

i did and 3y=183
3(61)=183

- anonymous

so y=61

- radar

I just saw where you had 3y=180 some how you got the right answer lol

- radar

The other angle is supplementary so subtract the 61 from 180 and it is solved!

- 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

Did you say that you have graduated from Strayer or that you are presently going.

- anonymous

presently going

- 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

Good luck tina

- anonymous

thank you, im not good at this kind of math

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