solve using elimination 2x+4y=-4 and 3x+5y=-3
don't be offended I am just figuring out where to start at, but do you know the general process of elimination and need help with this one or need help understanding elimination in general?
I need help solving it. I cant seem to figure it out
well first you need to choose one variable to eliminate since none of these work together nicely i would suggest eliminating x as it has the lowest coefficents
how do I do that??
so you need to find the lowest common multiple of the coefficients of x 3 and 2
so like 6
\[2x+4y=-4[1]\] \[3x+5y=-3[2]\] \[[1]\times3\] \[3(2x+4y)=3(-4)\] \[6x+12y=-12[3]\] \[[2]\times2\] \[2(3x+5y)=2(-3)\] \[6x+10y=-6[4]\] \[[3]-[4]\] \[(6x+12y)-(6x+10y)=(−12)-(-6)\] \[6x+12y-6x-10y=-12--6\] Collect like terms \[2y=-12+6\]
Do the rest.
To find x substitute the y value into any one of the 4 equations labelled above.
yes now multiply the first equation by 6/2=3 and the second equation by 6/3=2
@Azteck its better if you help them through the problem instead of just posting the work
@hannahbaker do you still want my help?
@whatisthequestion yes very much please :) the @Azteck confused me
@whatisthequestion I think you should continue helping. Sorry.
its fine just for future reference
@hannahbaker And that's the clearest step by step working you can get. There's nothing confusing about that. Once you conquer this topic, you will look back at that and think, "Wow, that is one of the most detailed and simple worked solution."
@hannahbaker can you tell me what the first equation and second equation are multiplied by 3 and 2 respectively? I explained why those numbers earlier.
@Azteck It looks so different in my head
6x+12y=-12 and -6-10y=6
@Azteck unfortunately thats not enough for alot of people otherwise schools would not have teachers just textbooks and reading them would be enough
Reread my post. I said "Once you conquer"
Once you conquer means that once you've learnt and understood the concepts from teachers/textbooks. Whatever makes you completely understand the topic.
probably just a typo but you don't have an x in your second equation
oops... I meant -6x
so the x's cancel out
okay since you already isolated multiplied the second equation by -1 we can just add the two equations to cancel the xs . . which you just posted
and now its just... 2y=-6
correct
once you solve for y you can plug that value into equation 1 or 2 and solve for x
ok then it becomes y=-3 right
yup
Well glad I could help please don't forget to award the medal to whomever helped you most and close the question.
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