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simplify 4x - 8y = 48 to x - 2y = 12 (1) and 11x + 3y = -105.5 (2) then use (1) x 11 - (2) which will eliminate x and solve for y
yikes another one with decimals!
you got this?
no not really
hold on we can do it
\[x - 2y = 12 \] \[11x+3y=-105.5\]
multiply the first equation by -11 and add to the second one
ok is it 2y or 3y
is it suppose to be x-2y or x-3y?
first equation simplifies by dividing by 4, giving you \[x-2y=12\]
now add the two equations together
the x terms add up to zero and you get \[25y=26.5\] if my arithmetic is correct
damn i screwed up again!
ok lol we can fix it
you multiply the first one by -11
i forgot the negative on the right hand side
ok so where does that come in
ok lets go slow
you have \[x-2y=12\] \[11x+3y=-105.5\] right?
and you want to eliminate one of the variables so you can solve on equation with one variable
i am confused on what comes first can you lay it out the way it suppose to be
yes it looks like this
\[x-2y=12\] \[11x+3y=-105.5\] then \[(-11)\times (x-2y)=(-11)\times 12\] \[11x+3y=-105.5\] then \[-11x+22y=-132\] \[11x+3y=-105.5\]
you have to figure out for yourself that if you multiply the first equation across by -11 and then add, the x terms will drop out when you add the two equations. finding what to multiply by is up to you.
i am lost
now add and get \[25y=-235.5\] so now \[x=-237.5\div 25=-9.5\]
why don't we try an easy one first. suppose i see \[x+y=1\] \[x-y=7\]how can i find x and y?
if you add the two equations together you get \[2x=8\] because you get \[x+y+x-y=1+7\] \[2x=8\]
now i have one variable and one equation, so easy to solve, if \[2x=8\] then \[x=4\] and if i know \[x=4\] then i know \[y=3\] because the first equation says \[x+y=7\] so \[4+y=7\] and therefore \[y=3\]
unfortunately in your problem if i simply add, nothing drops out. so you have to arrange it so that one of the variables will go away when you add
ok when it comes to my problem i get really lost
so for example if i see \[2x+3y=7\] \[x-y=1\] have to get rid of one variable by multiplying one equation all the way across by a number that will make one of the variables go away when i add
so i could multiply the second equation by 3 and get \[2x+3y=7\] \[3x-3y=3\] and now it is easy add to get \[5x=10\] so \[x=2\]
in your problem you have \[x-2y=12\] \[11x+3y=-105.5\]
to make one of the variables go, multiply the top one by -11 so you will have \[-11x\] in the first equation and \[11x\] in the second one, and they will add up to zero
we have to multiply EVERYTHING by -11 in the first equation and we get \[-11x+22y=-132\] \[11x+3y=-105.5\]
now when you add there will be no more x's
ok is that the first part or the second part
i am not sure what you mean
step one was to turn \[4x-8y=48\] into \[x-2y=4\]
step two was to multiply by -11 to get \[-11x+22y=-132\]
step 3 is to add it to the second equation. the x's add to zero and you get \[25y=-137.5\]
step 4 is to solve for y via \[y=-237.5\div 25\] \[y=-9.5\]
typo in step 3, should have been \[25y=-237.5\]
and finally to find x, replace y by -9.5 in either equation
ok with x would you divide by 2 for the first step as well
\[x-2y=12\] \[x-2\times (-9.5)=12\] \[x+19=12\] \[x=12-19\] \[x=-7\]
you can divide the first equation by 4, because each term has a common factor of 4 just makes it easier to work with
in general no, you do not divide as a first step. in general you see what you can multiply one equation by to make one variable drop out when you add the two equations
ok still a little lost but getting the hang of it