Question

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

Answer 1

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!