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.
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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
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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
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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
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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
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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
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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?
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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
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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
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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?
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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
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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
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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)
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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
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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
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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
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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)
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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?
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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)?
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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
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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