-4x+4y=-7 5x-6y=7 I cant eliminate here,right? So what do i do?
What do you mean?
First multiply each of the equations by a different factor to let either the x coefficients or y coefficients be the same. Them just add the two so the one variable would be cancelled out.
You can eliminate them as slaaibak said. think lowest common multiple between the coefficients of one of the variables.
Oh i see, so i could do this? -4x+4y=-7-------------12x+12y=-21 5x-6y=7--------------10x-12y=14 To cancel out the 12s?
bingo
So i would have -12x+12y=-21 + 10x-12y=14 = -2x=7?
yep
-2x=-7
sorry, -7 careful oh you caught it
-2x=-7 =-7/2?
yep yep! and now for y...
no, sorry positive again
negative over negative...
Substitute 5x-6y=7 getting 5*7/2-6y=7
Oh right
you caught it anyway so far so good, you're doing better than me this morning lol
35/10-6y=7?
Lol xd
where did the ten come from?
Oh, so i dont multiply 5 by both parts of the factor? 35/2-6y=7?
much better :)
in general we have this situation:\[c*\frac{a}{b}=\frac{c}{1}*\frac{a}{b}=\frac{ca}{b}\]so when a number is alone we always treat it as over one...
Ohhhh, now i get it!
try to relate it to those rules for algebra so what is y?
35/2-6y=7 Im not really sure what to do here, maybe make the two sides factors match. 35/2-6y=14/2?
that's a good start now get y on one side and the constants on another
35/2-6y=14/2 -6y=-21/2
Good so far?
yes :)
-6y=-21/2 -y=-(21/2)/6
No that cant be it
I need to make the factor divisible by six
first off take the -1 with the 6, we don't want y to be negative.. second it is right, but you need to simplify it remember that dividing by a number is the same as multiplying by it's inverse
That would mean that i have to divide both sides by -6, right?
right, and what is (-21/2)/(-6) ??? do you know the trick fro this?
y=-42/4 / -6? y=7/4
yes!!! dang you learned a lot Inopeki, nice :D many people here hate systems, it'll be nice to have you help them :P
:D Yes!!
that's very impressive doing systems like that, I can honestly say I'm impressed. Again, awesome :D
Thanks :D
Give me another one :D
well, systems can be tricky to make up off the top of your head. If you're not careful they can get messy...
ok here's one... x-2y=4 3x+6y=3 I think this is a bit easier than the last...
im sorry, OS crashed
I'm in no rush, don't worry
x-2y=4-----------3x-6y=12 3x+6y=3----------3x+6y=3
pellet
They will cancel out, i cant do that
no wait
3x-6y=12 + 3x+6y=3 = 6x=15 15/6=x?
yep, just simplify the fraction
5/2?
sweet :) and for y...?
5/2-2y=8/2 -2y=13/2?
watch the signs... made a little mistake
Ah, -2y=3/2?
there ya go !
-2y=3/2 y=6/4 / -2 y=-3/4?
very nice! do you know how to check your answers?
Yeah, put the final answers in to one of the equations to see if it is true. 5/2-2*3/4=4
Now that just doesnt make sense with fractions...
you should really put it in both equations, but yes you got the idea
it is correct: 5/2-2(-3/4)=5/2+3/2=8/2=4 again, you dropped a negative for the y-value. That happens to me a lot in systems too, just gotta be careful
try to verify that the other equation is true
The first one in this question?
the other equation in the system: 3x+6y=3
Oh
3*5/2+6*-3/4=3 15/2- -18/4=3 15/2- -9/2=6/2?
=3
so it must be right :) nicely done
Thanks :D Now what?
you should always check both equations, I'm not sure who told you that checking just one is okay, it's really not. Watch: x-2y=4 3x+6y=3 no what if I got the answer x=6, y=1 the first equation would give 6-2(1)=4 True but the second... 3(6)+6(1)=3 False so you must check both each equation by itself has infinite solutions, you need to use the other equation to just get one.
now what if*
Oh, now i see why.
because if we solve say the first one for x... x=2y+4 we have a function again: y=1 x=2(1)+4=6 y=2 x=2(2)+4=8 etc... next? I guess systems with 2 unknowns when you're comfortable with this stuff. What else are you doing? quadratics?
After this, its an introduction to ratios
Oh you mean what i HAVE been doing?
I meant the next step would be with 3 unknowns, you already are doing 2... but yeah, let's practice things you have just learned recently
I leanred averages,taking procent, slope, y-intercept,graphing lines, graphing points, integer sums.
equation of a line, equation systems
lol that's a lot... what do you mean by integer sums?
Like if the sum of 5 odd consecutive integers is 217, what is the largest odd integer?
Actually thats 7 odd consecutive integers.
nice those are fun... well for a lot of us at least. what do you feel unsure about? let's work on that but I need to get breakfast...
Yeah, i need to get something to eat too
ok we'll regroup in 10-15
Sure
Why does it say that im level 23?
ok I'm back now what to work on... let's see how you are with distribution real quick: simplify 2x(3+4x-2/x)
2x(3+4x-2/x) 2(3+4x-2) 2(1+4x)
careful, that 3 is lonely...
What?
and x(4x) is not 4x
wait!
middle step ...
2x^2(3-2)?
not quite. Middle step: 2x(3+4x-2/x)=2x(3)+2x(4x)-2x(2/x)=??? ^^^^^very important step for you, you'll see why when you do factoring
Oh right!
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