At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
you have to make them become corresponding 5x-2y= 28 (multiply by 8 ) = 80x-16y =224
How are those corresponding :s
and you have to make this for the second one -20x +8y=-100 (muliply by 2) = -40x +16y = -200 and solve by elimination now
why did you multoply by 8 and 2 sorry
If I was given that problem I would eliminate x... seems easier if I distribute the 4 all over the first equation
and for that reason. I have to distribute the 4 all over the first equation because I want to eliminate a variable and solve for the other. In this case, 4(5x-2y=28) 20x-8y=112 (this is my new first equation) so I add my new first equation to the second equation so I can get rid of the x
to make them corresponding...
@Michele_Laino please help
i got 3/5...
oh geez we are forced to take the y's out and then solve for x because by distributing the 4 all over the first equation, I ended up destroying x and y., and we don't want that.
\[8(5x-2y=28)\] \[2(-20x+8y=-100)\] \[40x-16y=224\] \[-40x+16y=-200\] HUH? Am I reading this wrong? even I got the same problem.
i was right @icalibear
i meant to type 40x
i will fix this
we are supposed to distribute 8 and 2 THROUGHOUT THE equation. We can't have it one sided
your x portion is off and when you fix it... x and y will cancel
i made a mistake on the multiplication.. sorry
40x - 40x = 0 -16y +16y = 0 224 -200 =24 so you get 0 = 24 in the end. Is that even possible?
is there even a solution if both x and y cancel because that leaves nothing
it's an inconsistent system which leads to no solutions at all
0 =24 true or false?
so there is no solution
right, so there is NO value of x or y which can satisfy those 2 equations. correct, no solution!
thanks so much guys.. i do not remember this from 9th grade..
yeah. I did it both ways... when I tried to eliminate x and I ended up with x and y going away similarly when I tried to eliminate y, I ended up and x and y going away as well leaving me with both situations having LHS doesn't equal the RHS left hand side doesn't equation the right hand side.
but I still would've done the x first... multiply that first equation by 4 makes it faster to get to the conclusion
\[4(5x-2y=28)\] \[-20x+8y=-100\] \[20x-8y=112\] \[-20x+8y=-100\] adding these two equations forces x and y to disappear.
I have one more..
I multiplied to get the middle terms the same but then i got a fraction
Nevermind made an error ! 6 x 3 is 18
is this set up right
yes it is.
so x = 9
y =.. ?
yay... that was fast x)
yes, she got it, she is now unstoppable! ;)
I didnt need to find y for the question but i know you would substitute it back in
thanks so much guys :)
right, go ahead and find it...you're doing it to learn, right? not just for completing the homework...
yes this is just practice
practice is fine....it's the questions with so many questions and multiple choices with omg this is due in an hour are frown upon on os
this isnt a class i get credit for, its just summer prep so i can get into math 128 in the fall
ive graduated and I forgot every bit of this intermediate algebra!
actually... well in my textbook... this is elementary algebra 1
either way, I need a refresher
waiting for the y value :P
sorry i have moved way on.. D:
its ok ^_^ probably next time :)