Help asap please? I'm doing Algebra 2, when you're solving a system by substitution.. For example my problem is 2x+y=-11 3x-4y=11 Could somebody walk me through it? Don't give me the answer.
to do this by substitution you first want to get one of the equations as y = so pick the one that's easiest to do that, and tell me what you get :)
Okay, I do the first one... 2x+y=-11 and i subtract y from each side and get 2x=-11-y
From first equation find value of y , that is solve for y so we will get y = -11 - 2x now just plug this value of y in second equation that is 3x - 4(-11 - 2x) = 11 now just simplify it use distributive property a*(x+y) = a*x + a*y & find value of x & then put this value of x in any equation & find y
Okay one second, tgawade
Ok
@tgawade x=-3 and y=-5 ???
yes, that is correct ^_^ good job! :D
Can you help me with one moe @jigglypuff314
Sure :)
Took my words back. Good work @tgawade and @jigglypuff314 .
Okay, on this one I have to solve the system using elimination. It's 2x+6y=-12 5x-5y=10
well 5x - 5y = 10 can be reduced by dividing through by 5 right? could you do that?
divide each side by 5 ?? under x or y
both like if you divide through by 2 for 2x+6y=-12 you would get x + 3y = -6
okay one second. i'll divide by 5
x-y=2?
correct :) so now you have x + 3y = -6 x - y = 2 and you see how the x's are lined up? you can then subtract so... x + 3y = -6 -(x - y = 2) (don't forget to distribute the minusing)
okay one minute!!
x=-9 y=1 ??? @jigglypuff314
hmmm well from x + 3y = -6 -(x - y = 2) it would be subtracted like ------------- 4y = -8 do you see how I got that? could you go on from there? :)
OH so x=0 and y = 2 !! @jigglypuff314
Correct! :D
Woot ! :) Thank you so much ! If I have any other questions , I'll let you know ! Will you be on for awhile ?
I might have to go to dinner soon, but I'll be around afterwards :)
Wait... (0,2) isn't an answer .... My answers are (2,1) (0,-2) (-2,0) (1,2)
Nevermind, it would be (0,-2) I forgot the 8 was a negative!
do you have time for one more ?
yeah sure, post it in another question :)
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