x-3y=-9 6x+y=3

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x-3y=-9 6x+y=3

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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i dont understand this @freckles
How does a 99 rated mathlete not know how to do system of equations O.o
Well when you haven't done math in a while u forget things.....

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|dw:1442268783821:dw|
ok i know the part where u add them and stuff but i just need help in the beginning. How do u find what to multiply them with?
when they are set out like that, just imagine adding them and knocking out an x or a y so, let's do it again
|dw:1442269218613:dw|
wait why did u use 1 and 2
if you meant, the (1) and (2) in brackets to identify the equations, and just a habit is that what you were referring to?
Because my teacher taught me to multiply them by a number and then add then and cancel the x or y u used etc
loads of ways to do these eg |dw:1442269699628:dw|
can i ask , do you work on the OS website??
whats the easiest way? can u do in equation form cause the drawing is losing me
do i work on the website?
yes
what does that mean?
are you part of the OS team?
Oh yes...
easiest way is substitution, IMHO. as in, it's pure method. so, from the first equation, which is \(x - 3y = -9\), you can say \(x = 3y - 9\) then in the second, which is \(6x + y = 3\), you can say \(6(3y - 9) +y = 3\) so \(18y - 54 + y = 3\) or \(19y = 57\), \(y = 3\)
so you code this site?!
OOOHOHO you have to change the x-3y=*9 in x=3y-9 and lets stay relevant :D u can message me the other questions
OK!!
But what if its x+3y=-9 does 3y become positive cause u move it?
if x + 3y = -9 then x = -3y - 9
because x+3y-3y = -9-3y
so i got 6(3y-9)+y=3 18y-54+y=3 +y +y 19y-54=3 +54 +54 19y=57 y=3
so now u plug it into the x too get x right? x-3(3)=-9 x+9=-9 -9 -9 x=-18 so we got (-18,3)
??
if you know that y = 3, plug that into one of the original equations to find x
eg, if x - 3y = -9 what is x if y = 3
or if 6x + y = 3 what is x if y = 3??
you plug it into the x-3y=-9 right?
into either, you want to know when they meet
it says -18,3 is wrong
wait don't i need another point?
i thought we had agreed 0, 3 ?!?!
as in x = 0
and y = 3
OOHOHOHO i did it wrong ok so i need another set of points to connect it.
indeed you might! gtg
So how do i get it?
really sorry @jagr2713 i gotta go sleep i am in europe, i guess you are somewhere else my condolensces!!!!
Its ok
are you meant to solve by either substitution or elimination?
Graphing them
i got the answer 0,3 just need another point
well... where the graph of two equations meet, is where the answer is :) hm if 0,3 is the answer.... isn't that what you needed? :)
Yea but thats only one point need 2 for a line
hmmm one sec
ok :D
here use this i did another question x-3y=-3 3x+y=1 i got (0,1)
..is that another one?
Yea i just change questions i got 0,1 but need another point
hmmm thought you were supposed to just graph them?
Yea it say solve the following system of equation by graphing them
well... I'd just graph them then :) lemme do a quick table
ok thanks
wait do we just do the x|y thing and put 0 for x and ory?
|dw:1442273690049:dw| like that?
yeap to make a line, all you need is 2 points so just pick two random "x" to get two "y"s
OHH so |dw:1442273743049:dw|
other is 0,3 2,0
hmm yes \(\begin{array}{llll} x-3y=-3\implies &\cfrac{x+3}{3}={\color{brown}{ y}} \\ \quad \\ 3x+y=1\implies &{\color{brown}{ y}}=1-3x \end{array} \\ \quad \\ \qquad \qquad \begin{array}{ccllll} x&y \\\hline\\ &\frac{x+3}{3}\\ \\\hline\\ &\frac{x+3}{3}\\ 0&1\\ -3&0 \end{array}\qquad \begin{array}{ccllll} x&y \\\hline\\ &1-3x \\\hline\\ 0&1\\ 4&-11 \end{array}\)
where did u get they -3 from?
x-3y=-3 <--- 3x+y=1
is this correct (0,3) (-9,0) and (0,3) (2,0) ?
hmmm 0, 3 is not, -9,0 isn't either
x =0 x -3y=-3 0-3y=-3 y = (-3)/(-3) y =1
x-3y=-9 -3y=-9 y=1/3 x-3y=-9 x=-9
x=-9 (-9)-3y=-3 -3y=-3+9 -3y=+6 y=(6)/(-3) y=-2
wait this is confusing me dont u put 0 for them?
wel... you said it was x-3y=-3 3x+y=1 but now you seem to have changed it some
OHOHOHO I've been doing the wrong problem... :D KML
I WAS LOOKING AT THE QUESTION LOL
i got (0,1) (-3,0) and (0,1) (3,0)
but all you need to graph a line is two points \(\begin{array}{llll} x-3y=-3\implies \cfrac{x+9}{3}={\color{brown}{ y}}\\ 6x+y=3\implies {\color{brown}{ y}}=3-6x \end{array}\) http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiIoeCs5KS8zIiwiY29sb3IiOiIjM0IxRENGIn0seyJ0eXBlIjowLCJlcSI6IjMtNngiLCJjb2xvciI6IiNENDExMTEifSx7InR5cGUiOjEwMDAsIndpbmRvdyI6WyItNi41NCIsIjYuNDYiLCItMi40OCIsIjUuNTIiXX1d
D: it says my answer is wrong... i did the x|y thing
check yours PEMDAS I'd think
I'm confused... Are you saying you got 3 different solutions for x-3y=-3 3x+y=1 ? You should only get one solution, infinitely many solutions, or no solutions for system of linear equations.
OHH my friend told me to use y=mx+b
why? aren't you trying to solve the system x-3y=-3 3x+y=1 ? or is that not the question?
yes by graphing it thou but i did it and getting it wrong D:
you should just get a point for the solution of this system: you will see that when graphing these two lines that they will share a common point that point is the solution x-3y=-3 -3y=-x-3 y=1/3*x+1 --- 3x+y=1 y=-3x+1 so graphing both y=1/3*x+1 and -3x+1 we have something that looks like this: |dw:1442275696567:dw| the solution is given at the point where both lines touch each other (cross each other; intersect each other)
|dw:1442275792808:dw| so this point I darkened is the solution because is the point that two lines have in common
wait can u try thing one x-2y=-2 4x+y=1
were you trying to write in y=mx+b form to graph the lines to find the point ?
yea i was using that
ok well try writing x-2y=-2 in that form
2y=x+2 right?
so you subtracted x on both sides and multiplied -1 on both sides? x-2y=-2 subtracted x on both sides -2y=-x-2 multiplied both sides by -1 2y=x+2 this is correct so far then just divide both sides by 2
yea i got y=1/2x+1
then u graph it by going to (0,1) then rise over run
that sounds great so y=1/2x+1 says the y-intercept is 1 and the slope is 1/2 |dw:1442276083294:dw| right so you get something like this now we also need to graph 4x+y=1 on that same graph
4x+y=1 is equivalent to y=-4x+1 can you try graph that we should actually already see the solution even before graphing
yea we got 0,1 then rise over run -4/1
so before graphing you should see that (0,1) is a point on both graphs where both graphs differ in slope therefore (0,1) is the solution
yea 0,1 is the slope so we go down 4 and move 1 to the left right?
(0,1) is the y-intercept the slope is -4/1 so we move down from the y-intercept 4 units and to the right once
Oh right 1 sorry my friend told me left....
OMG your the best :D
can u help with one more but this one is substitution
sure
one more then I have to roll my croissant dough
ok :D 4x+y=24 y=2x
replace first y with (2x) solve 4x+2x=24 for x
don't you reverse it like y=4x+24 and plug into x y=2(4x+24)
the second equation says y=2x so you insert that y here: 4x+y=24 to get 4x+2x=24 anyways you can't replace x with 4x+24 because we aren't given x=4x+24 unless you already know that x=-8 which x doesn't -8 and also 4x+y=24 is not equivalent to y=4x+24
wait how did u get 4x+2x=24
\[\color{red}{y=2x} \\ 4x+\color{red}{y}=24 \\ \\ 4x+\color{red}{2x}=24\]
I replaced y with 2x because one of the equations said y=2x
Wait i understand igtg thanks for the help
k peace

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