i know it is basic but it has been awhile 3(x-5) over 2 = 2(x+5) over 3
clear the fractions by "cross multiplying"
then go in for the win
if you'd like to share your answer we'll tell you if your right
do take the 2 and mulitiply the the other side ?
yes, and the same with the 3
idea is \[\frac{a}{b}=\frac{c}{d}\iff ad=bc\]
then expand it by multiplying thru the (...)
I have already took out the ()
thats fine too :)
once youve cleared the fractions; gather like terms
do i multiply the 3x and2x?
\[\frac{3(x-5)}{2} = \frac{2(x+5)}{3}\] \[\frac{3(x-5)*(3)(2)}{2} = \frac{2(x+5)*(3)(2)}{3}\] \[3(x-5)*(3) = 2(x+5)(2)\] \[9(x-5) = 4(x+5)\] \[9x-45 = 4x+20\] this is what I get to
9x - 45 = 4x + 20 ; +45 +45 +45 ---------------- 9x = 4x + 65 ; -4x -4x -4x ---------------- 5x = 65 5x = 65
any of that ring a bell :)
yes thanks
youre welcome :)
\[3/4x\]
3/4x - 1/2=1 how would i got about this one
x=13 remove denominators by multiplying to the top of each line cancel them out and the rest is easy
9x add 4x = 20 + 45
i am on a different equation now
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