# Some algebra I don't understand: Solve for R

• Jaccobtw
In summary, the conversation is discussing how to solve for the variable R in the equation $\frac{5}{3} = \frac{0.28 + R}{R}$ using algebra. Suggestions include eliminating fractions, isolating the terms with R on one side, and simplifying the right-hand-side before solving.

#### Jaccobtw

Homework Statement
Solve for R. $$\frac{5}{3} = \frac{0.28 + R}{R}$$
Relevant Equations
Use algebra I guess
The variable is already isolated on one side. I dont' know how to solve for R though. Any help? Thank you.

Jaccobtw said:
Homework Statement:: Solve for R. $$\frac{5}{3} = \frac{0.28 + R}{R}$$
Relevant Equations:: Use algebra I guess

The variable is already isolated on one side. I dont' know how to solve for R though. Any help? Thank you.
Fractions are always a problem. Get rid of them!

• FactChecker and Jaccobtw
PeroK said:
Fractions are always a problem. Get rid of them!
Ah ok. Multiply both sides by R and then subtract R from both sides. The rest is cake

• berkeman, FactChecker, PhDeezNutz and 1 other person
Jaccobtw said:
Ah ok. Multiply both sides by R and then subtract R from both sides. The rest is cake
Better yet, multiply both sides of the equation by 3R, and then isolate the terms with R on one side.

the difficulty here is R appears twice
we can either merge the R's or eliminate one
to do the later we can...

subtract one from both sides
$$\frac{5}{3} = \frac{0.28 + R}{R}$$
$$\frac{5}{3} -\frac{3}{3}= \frac{0.28 + R}{R}-\frac{R}{R}$$

Jaccobtw said:
Homework Statement:: Solve for R. $$\frac{5}{3} = \frac{0.28 + R}{R}$$
Relevant Equations:: Use algebra I guess

The variable is already isolated on one side. I dont' know how to solve for R though. Any help? Thank you.
Although R is only on the right-side, I would not describe that R as "isolated".

lurflurf said:
the difficulty here is R appears twice
we can either merge the R's or eliminate one
to do the later we can...

subtract one from both sides
$$\frac{5}{3} = \frac{0.28 + R}{R}$$
$$\frac{5}{3} -\frac{3}{3}= \frac{0.28 + R}{R}-\frac{R}{R}$$
While true, it may be better to simplify [to reduce the number of R's one sees on]
the right-hand-side by first distributing the [common] denominator, then simplifying further
\begin{align*}\frac{5}{3} &= \frac{0.28 + R}{R}\\ &= \frac{0.28}{R} + 1\\ \end{align*}

• SammyS