How Can I Solve for x of y=cx+dx Using the Textbook Solution?

[Rujaxso]

$$y = cx + dx$$ solve for x textbook solution:$$x = \frac{y}{c+d}$$my steps:

$$\frac{y}{c + d}=\frac{cx + dx}{c + d}$$ here I divide to isolate x

$$\frac{y}{c + d}= x + x$$ simplifed

$$\frac{y}{c + d}=2x$$ adding the two x'sWhat am I missing here?

 

[skeeter]

$y = cx+dx$

factor $x$ from the two terms on the right side of the equation ...

$y = x(c+d)$

divide both sides by $(c+d)$ ...

$\dfrac{y}{c+d} = x$

 

[Rujaxso]

Thanks Skeeter,

basically reverse distribution of multiplication over addition I guess.

 

[skeeter]

$$y = cx + dx$$ solve for x textbook solution:$$x = \frac{y}{c+d}$$my steps:

$$\frac{y}{c + d}=\frac{cx + dx}{c + d}$$ here I divide to isolate x

$$\color{red}{\frac{y}{c + d}=\frac{x(c+d)}{c + d}}$$

$$\color{red}{\frac{y}{c + d}=\frac{x(\cancel{c+d})}{\cancel{c + d}}}$$

see above

 

[Rujaxso]

Thanks, got you the first time, I was just trying to confirm the rule/move in English by my 2nd post.

 

[Ackbach]

And if you want to wow your teacher, add the proviso, "for $c+d\not=0$."

 

[Rujaxso]

Should I be more impressed by the use of the word proviso? =)

btw you guys and this textbook is the closest thing to a teacher I have right now. Prepare for an onslaught and flurry
of annoying questions in the future.