Question

Solve for r.

Answer 1

\(\color{eneter color name}{√2/3r +1 =23 } \)

Answer 2

@Agl202 ;]

Answer 3

Can you make it clearer? The r may or may not be inside the square root

Answer 4

\[\frac{ \sqrt2 }{ 3r }+1=23\]
In this case we have two objectives; the first one being getting rid of the "3r" that is in the denominator and secondly preceding the rewriting of the equation solving it as we normally would.
So, we can operate that fraction, but, as some hokus pokus, we can multiply both sides by "3r" in order to simplify the first term:
\[\frac{ \sqrt 2 }{ 3r }+1=23 \iff (3r)\left( \frac{ \sqrt 2 }{ 3r }+1 \right)=(3r)(23)\]
Of course, distributive axiom will allow me to distributie that "3r" to all the terms, thus completing our first objective, geting rid of that 3r in the denominator:
\[(3r)\left( \frac{ \sqrt 2 }{ 3r }+1 \right)=(3r)(23) \iff (3r)\left( \frac{ \sqrt2 }{ 3r } \right)+(3r)(1)=(23)(3r)\]
Ending up with:
\[\sqrt 2 + 3r=69r\]
And this you can solve as you normally would.

Answer 5

Okay @agent0smith here