How do you solve #10/12=x/18#?
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"Find the limit as x approaches infinity of #xsin(1/x)#?"
You an solve this in two ways: proportionality or just rearranging the equation
Rearranging the equation
You can multiply #10/12# by 18 to leave x to one side.
#10/12 * 18 = x#
#x = 15#
Proportionality
You can work out the proportion between 12 and 18, which will be the same for 10 and #x#.
so #18/12 = 1 1/2#
Therefore, if we multiply 10 by #1 1/2# we get the value of #x# which is #15#
Using first principles
Multiply both sides by #color(red)(18)#
#color(green)(10/12color(red)(xx18)" " =" " x/18color(red)(xx18))#
#color(green)(10/12color(red)(xx18)" " =" " x color(red)(xx)(color(red)(18))/18)#
But #18/18=1# so #x xx18/18 =x xx1=x# giving:
#(10xx18)/12=x#
Write as #" "x=(10xx18)/12" "=" "(2xx5xx2xx3xx3)/(3xx4)#
#x=(2xx2)/4xx3/3xx5xx3" " =" " 1xx1xx15#
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or if you prefer:
#x= "(cancel(2)^1xx5xxcancel(2)^1xx3xxcancel(3)^1)/(cancel(3)^1xxcancel(4)^1) = 5xx3=15#