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2 x(1 1/2-y)= 2 1/3+1/15
Y=?

1 Answer

4 votes
Hey there, lets solve for each side!


2x\left(1\cdot \left((1)/(2)\right)-y\right)=2\left((1)/(3)\right)+\left((1)/(15)\right)


2x\left(1\cdot \left((1)/(2)\right)-y\right) \; (a) = a\ \textgreater \ Remove\;parenthesis \ \textgreater \ 2x\left(1\cdot (1)/(2)-y\right)


\mathrm{Multiply:}\:1\cdot (1)/(2)=(1)/(2) \ \textgreater \ 2x\left((1)/(2)-y\right)


\mathrm{Distribute\:parentheses\:using}: \:a\left(b+c\right)=ab+ac

Where\; a=2x,\:b=(1)/(2),\:c=-y

2x\cdot (1)/(2)+2x\left(-y\right)


\mathrm{Apply\:minus-plus\:rules} \ \textgreater \ +\left(-a\right)=-a \ \textgreater \ 2x(1)/(2)-2xy


2x(1)/(2) \ \textgreater \ \mathrm{Multiply\:fractions}: \:a\cdot (b)/(c)=(a\:\cdot \:b)/(c) \ \textgreater \ (1\cdot \:2x)/(2) \ \textgreater \ \mathrm{Apply\:rule}\:1\cdot \:a=a

(2x)/(2) \ \textgreater \ \mathrm{Cancel\:the\:common\:factor:}\:2 \ \textgreater \ x \ \textgreater \ x-2xy

Moving on

2\left((1)/(3)\right)+\left((1)/(15)\right)

Remove parenthesis again

2\cdot (1)/(3)+(1)/(15) \ \textgreater \ 2\cdot (1)/(3) \ \textgreater \ \mathrm{Multiply\:fractions}: \:a\cdot (b)/(c)=(a\:\cdot \:b)/(c) \ \textgreater \ (1\cdot \:2)/(3)

\mathrm{Apply\:rule}\:1\cdot \:a=a \ \textgreater \ (2)/(3) \ \textgreater \ (2)/(3)+(1)/(15)

Now we want to find the LCD for
(2)/(3)+(1)/(15)

\mathrm{Factor\:each\:denominator\:into\:its\:primes} \ \textgreater \ 15=3\cdot \:5 \ \textgreater \ 15

Now adjust the fractions based on the LCD

(2\cdot \:5)/(15)+(1)/(15)

Since the denominators are equal, you can combine the fractions

(a)/(c)\pm (b)/(c)=(a\pm \:b)/(c) \ \textgreater \ (2\cdot \:5+1)/(15)

Of course no simply multiply 2 by 5 then add 1

(11)/(15)

Combine the two again

x-2xy=(11)/(15)

Hope this helps!
User Venko
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