I need help solving for {X|X ?}
3/8 (5+6x)+3 greater than or equal to 21
I appreciate help working through this!
If I understand what you're asking, you want:
$\displaystyle \{x\ :\ \frac{3}{8}(5+6x) + 3 \geq 21\}$
Technically that set definition works (though, we might put a condition on the left-hand side by $\displaystyle x\in \mathbb{R}$, the real numbers). You should, however, simplify the specification on the right-hand side. That is just algebra, though. You should be able to do that. It's straight-foward. The answer is $\displaystyle 6x \geq 43$.
Your goal is to get x by itself. To do this first you subtract the constant '1' from each side to get
$\displaystyle \frac{3}{8}(3 + 2x) \geq 12$
Now you cross-multiply both sides by $\displaystyle \frac{8}{3}$. Do you see why this was chosen? It removes the fraction from the left side. The result is then
$\displaystyle 3+2x \geq 12\times \frac{8}{3} = 4\times 8 = 32$
Then again we subtract the free constant 3 from both sides to get
$\displaystyle 2x \geq 29$
And finally we divide both sides by 2 to get a result simplified to an inequality regarding x:
$\displaystyle x \geq \frac{29}{2} = 14.5$
Did each step make sense to you? Also, when did the equation change? Didn't we originally start with $\displaystyle \frac{3}{8}(5+6x)+3 \geq 21$? In any case, the steps would be identical to the above, just with different values, and I already reported the result almost completely simplified.