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Mathematics 15 Online
OpenStudy (anonymous):

Tell whether the ordered pair is a solution of the linear equation. (-3, 1) x=y=-2 x+5y=2 I will gladly appreciate if you help, because I don't know how to do this at all. So can you explain the step for me please. And thank you.

OpenStudy (anonymous):

please help

OpenStudy (anonymous):

help plz

jimthompson5910 (jim_thompson5910):

x=y=-2 has too many equal signs

jimthompson5910 (jim_thompson5910):

is it x+y=-2 or is it x = y + -2, or is it x = y+2

OpenStudy (anonymous):

x+y=-2 my bad

jimthompson5910 (jim_thompson5910):

ok first solve for y in x+y=-2 x+y=-2 y = -2 - x y = -x - 2

OpenStudy (anonymous):

Just need to know how to do this so I can figure out the other problems as well.

jimthompson5910 (jim_thompson5910):

now plug this into the second equation x+5y=2 x+5(-x - 2)=2 ... replace y with -x - 2 (this is possible since y = -x - 2) x - 5x - 10 = 2 I'll let you finish

OpenStudy (anonymous):

and Oh yeah I was suppose to put in a graph as well.

jimthompson5910 (jim_thompson5910):

so you will need to solve both for y

OpenStudy (anonymous):

y = -2 - x y = -x - 2 How come you flipped it like that?

jimthompson5910 (jim_thompson5910):

x+5y=2 5y=2 - x 5y = -x + 2 y = -x/5 + 2/5 y = -1/5x + 2/5

jimthompson5910 (jim_thompson5910):

because the general form is y = mx+b and you can compare it easier that way

OpenStudy (anonymous):

oh yeah...hahhaha I remember....that XD equation.

jimthompson5910 (jim_thompson5910):

you use that to help you graph

OpenStudy (anonymous):

need help

jimthompson5910 (jim_thompson5910):

ok where are you stuck

OpenStudy (anonymous):

Fraction part

jimthompson5910 (jim_thompson5910):

you mean this y = -1/5x + 2/5 right?

OpenStudy (anonymous):

Sorry I'm confused how to do this problem to be honest

jimthompson5910 (jim_thompson5910):

you graph the two equations where they cross is where your solution is at

OpenStudy (anonymous):

yseh how did you get the fraction

jimthompson5910 (jim_thompson5910):

well you can use 0.4 in place of 2/5 if you want better yet, do you have a graphing calculator?

OpenStudy (anonymous):

nope. But its ok. I know how to do the calculation. Just ned to know the step hwo to do this problem and that's it.

OpenStudy (anonymous):

Give me two minutes to understand what you wrote ok? Because I am tired and did soooo much homework today. XD Stressed pretty much. XD

jimthompson5910 (jim_thompson5910):

y = -1/5x + 2/5 y = (-1/5)(0) + 2/5 y = 0 + 2/5 y = 2/5 So when x = 0, y is y = 2/5, which means you have one point (0, 2/5)

jimthompson5910 (jim_thompson5910):

alright

OpenStudy (anonymous):

For the x- 5x-10=2 is it-4x-20=2?

jimthompson5910 (jim_thompson5910):

x- 5x-10=2 turns into -4x - 10 = 2 not sure where you got the 20 from

OpenStudy (anonymous):

-4x - 10= 2 becomes -4x=12then its x=3?

jimthompson5910 (jim_thompson5910):

-4x = 12 x = -3

jimthompson5910 (jim_thompson5910):

not x = 3

OpenStudy (anonymous):

Was a typo. Was about to fix it.

jimthompson5910 (jim_thompson5910):

i gotcha

OpenStudy (anonymous):

Why do you have to plug the first equation to the second equation

jimthompson5910 (jim_thompson5910):

because that's one method of getting rid of one variable so you can solve for the other

jimthompson5910 (jim_thompson5910):

notice you went from 2 variables (x and y) to just one variable (x) and you solved for it you would then use that solution for x to find the solution for y

OpenStudy (anonymous):

So the fraction part would equal 0.2 and -1.44?

OpenStudy (anonymous):

wait it would be 2.14?

OpenStudy (anonymous):

I don't really know how to graph it on the graph. XD

jimthompson5910 (jim_thompson5910):

we found x to be x = -3 remember how we solved for y? well we found that y = -x - 2 so... y = -x - 2 y = -(-3) - 2 ... plug in x = -3 (replace x with -3) y = 3 - 2 y = 1

jimthompson5910 (jim_thompson5910):

this means that the solution is (-3, 1)

jimthompson5910 (jim_thompson5910):

visually, the two graphs will cross at the point (-3, 1)

OpenStudy (anonymous):

Then what is the point for the second equation?

OpenStudy (anonymous):

Is the second equation for another x and y?

jimthompson5910 (jim_thompson5910):

we used it to find x

jimthompson5910 (jim_thompson5910):

no there's only one solution in this case

OpenStudy (anonymous):

For them to cross?

jimthompson5910 (jim_thompson5910):

yes, you need two lines to have them cross

jimthompson5910 (jim_thompson5910):

at a single point

OpenStudy (anonymous):

That's what I mean

jimthompson5910 (jim_thompson5910):

btw, once you find x, you can use either equation that has y in it to plug in x = -3 and solve for y

OpenStudy (anonymous):

What do you do after y= -1/5 + 2/5?

jimthompson5910 (jim_thompson5910):

well if you wanted to graph, you would solve each for y

jimthompson5910 (jim_thompson5910):

then you would plug in any x value you want to get a corresponding y value

jimthompson5910 (jim_thompson5910):

this will give you a point that lies on the line

jimthompson5910 (jim_thompson5910):

do it twice and you have two points in which you can use to graph

OpenStudy (anonymous):

Just show me how to do it. plez........................draw it XDD

OpenStudy (anonymous):

I really need to get this done XD

jimthompson5910 (jim_thompson5910):

ok say you wanted to graph y = 2x + 7 you would plug in x = 0 to get y = 2(0) + 7 = 7 so one point is (0, 7) when x = 1, y = 2(1) + 7 = 9 another point is (1,9)

jimthompson5910 (jim_thompson5910):

so to graph y = 2x + 7, you plot the two points (0,7) and (1,9) then you draw a straight line through them

OpenStudy (anonymous):

So one line for the first equation and another one line for the second quation right? If its only one equation you use it for 2 lines?

OpenStudy (anonymous):

TO see if it crosses and get a single point answer

jimthompson5910 (jim_thompson5910):

you got it

OpenStudy (anonymous):

But where did you get the 0 though? o.O

jimthompson5910 (jim_thompson5910):

I just randomly picked an x value (a small x value to make things easy)

jimthompson5910 (jim_thompson5910):

x could be 2, 7, 11, whatever you want it to be

jimthompson5910 (jim_thompson5910):

just make sure it is within the window of view

OpenStudy (anonymous):

So its kinda like a free for all in away DXD

jimthompson5910 (jim_thompson5910):

pretty much

jimthompson5910 (jim_thompson5910):

x could be a million, but that's way off the charts, so we won't go there

OpenStudy (anonymous):

That kinda makes more sense to me now. XD I just need a little more explaination XD I may have to ask you give me another 2 minutes

jimthompson5910 (jim_thompson5910):

alrigt

jimthompson5910 (jim_thompson5910):

alright*

OpenStudy (anonymous):

Let me ask you a question though. How do you get a line from a fraction again? XD

OpenStudy (anonymous):

That one confuses me the most

jimthompson5910 (jim_thompson5910):

say x = 0 when x = 0, then y is... y = -1/5x + 2/5 y = (-1/5)(0) + 2/5 y = 0 + 2/5 y = 2/5

jimthompson5910 (jim_thompson5910):

so when x = 0, y is y = 2/5

jimthompson5910 (jim_thompson5910):

2/5 is the same as 0.4 so when x = 0, y is y = 0.4

jimthompson5910 (jim_thompson5910):

so you have the point (0, 2/5) or (0, 0.4)

OpenStudy (anonymous):

That's what I did so the x can be anything right?

OpenStudy (anonymous):

Or any number

jimthompson5910 (jim_thompson5910):

yes, 0 is the easiest to work with though

OpenStudy (anonymous):

mm ok

OpenStudy (anonymous):

Do you have answer ? and can I give you the answer and tell if its correct?

jimthompson5910 (jim_thompson5910):

yes I do, and I posted the answer previously

jimthompson5910 (jim_thompson5910):

but tell me what you got

OpenStudy (anonymous):

WHat was answer though?

jimthompson5910 (jim_thompson5910):

what did you get? don't worry about being correct or not

OpenStudy (anonymous):

so -3,1 and the other one is (0.4,0)?

jimthompson5910 (jim_thompson5910):

(-3,1) is the only solution

jimthompson5910 (jim_thompson5910):

(0,0.4) is a point that's used to help graph y = (-1/5)x + 2/5

jimthompson5910 (jim_thompson5910):

but it's not the solution

OpenStudy (anonymous):

So its not asking to make the graph then?

jimthompson5910 (jim_thompson5910):

remember, these two lines only cross at one point

jimthompson5910 (jim_thompson5910):

no they just want the solution how you get the solution is up to you

OpenStudy (anonymous):

I know...XD but you put 0, 0.4 on the graph thas what Im asking

jimthompson5910 (jim_thompson5910):

you could graph, use substitution, elimination, etc etc

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