Create a graph and a table to solve the system. Check your answer. Enter your answer as the following example: (5, -6) x+y=2 x-y=4
easier to add them up and get \[2x=6\] then sole for \(x\)
*solve
x is 3 ?
a graph is a long and time consuming way to solve this, but it will work yes \(x=3\) http://www.wolframalpha.com/input/?i=x%2By%3D2%2Cx-y%3D4
and since \(x=3\) and \(x+y=2\) you have \(3+y=2\) and you can solve for \(y\)
is y 5
no
im so bad in alg 2 :(
\[3+y=2\] subtract \(3\) from both sides to get \(y\)
i think you added 3 but you need to subtract 3 a little practice is all you need, it is not like you are born knowing this
Oh yeah i know how to do that, i was good ion alg 1 & geometry when i was in public school
so y is -1
yes, \(y=-1\)
you can check that this is right, because \[3-1=2\\3+1=4\]
ok i got you, & so i write the answer like this? (3,-1)
yes
Awesome! Thank you :)
yw
I got it wrong!?
no, it is right, the solution is \((3,-1)\)
It told me it was wrong..
wait i typed it on the wrong one :(
lol
damn on line systems all ways messing with you
I know! If i wouldnt have done that i woulda passed that damn quiz!
what? you have to do it again??
Yea i gotta retake it, im doing online school :/
ick
good luck
Use a graph and a table to solve each system. Check your answer. y+x=5 3x-5y=-1
can u explain step by step how to do these?
do you have to use a graph? or can we just solve it?
it would be much easier and quicker to solve using algebra rather than a graph
yea we can just sole its multi choice no graph stuff
ok then lets knock it out
\[y+x=5\\ 3x-5y=-1\] first equation is the same as \(y=5-x\) so we can replace the \(y\) in the second equation by \(5-x\) with careful use a parentheses and get \[3x-5(5-x)=-1\]
this has only one variable, so we can solve for \(x\) via \[3x-5(5-x)=-1\\ 3x-25+5x=-1\\ 8x-25=-1\]
now add \(25\) to both sides and get \[8x=24\] and solve for \(x\) by dividing by \(8\) to get \(x=3\)
now since \(x=3\) and \(x+y=5\) we get \(3+y=5\) making \(y=2\) and the solution is \((3,2)\)
more or less clear?
still so confusing but i can use that as a guide! Thank you
yw
Do you know this? I have got it wrong 2 times already! Classify the system and determine the number of solutions. Check all that apply. 7x + y = 13 28x + 4y = -12 infinitely many solutions inconsistent consistent no solution
this is NO Solution for this also known as "inconsistent"
i think you are supposed to select them both
\[7x + y = 13 \\28x + 4y = -12\]if yo multiply the first equation by \(4\) you get \[28x + 4y = 52\\ 28x + 4y = -12\] and there is no way to make \(52=-12\)
what about this one?? Classify the system and determine the number of solutions. (Check all that apply.) $7x - y = -11$ $3y = 21x + 33$ dependent inconsistent infinitely many solutions no solution consistent independent
what's with the dollar signs?
Oh it just put that in there ignore those, i just copied it from my page
ok well they are the same line, so one answer is "infinitely many solutions" also i believe this can be called "consistent"
oh, also add "dependent"
Thank you so so much!!!
yw
did you pass the quiz?
im trying to do these last 2 by going with what u explained im so freaking lost!
i will help you if you like
Use a graph and a table to solve the system. Check your answer. Enter your answer as the following example: (5, -6) x - y = -1 4x - 2y = 2
See i had a baby and thats the only reason im finishing my senior year online. & its so hard with the math, i got behind so i hvae 3 units to do by the 9th!
\[x - y = -1 \\4x - 2y = 2\] multiply the first equation by \(-2\) and get \[-2x+2y=2\\ 4x-2y=2\]
now add them up and get \(2x=4\) and so \(x=2\)
since \(x=2\) and \(x-y=-1\) you have \(2-y=-1\) making \(y=3\) and the solution is \((2,3)\) how old is your baby?
That was much easier to understand, now i just gotta get it to stick with me, & he will be 7 months on the 6th
must keep you pretty busy i am sure hard to do math, on line or otherwise
you done?
Yes i passed that time!!
whew!!
good luck with the rest of the class
Thank you!
yw
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