Q. -3 <= (3-5x)/3 < 3/8 Should I do BEDMAS in reverse? eg. subtract 3 from all three sides to isolate the -5xor Should I multiply all three sides to remove the denominators? Thanks
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Originally Posted by crashdummy Q. -3 <= (3-5x)/3 < 3/8 Should I do BEDMAS in reverse? eg. subtract 3 from all three sides to isolate the -5xor Should I multiply all three sides to remove the denominators? Thanks Hi crashdummy, I believe I would multiply all three parts by 24 to clear the fractions.
so does this look right? (see attachment)
hello $\displaystyle -3\leq \frac{3-5x}{3}< \frac{3}{8}\Leftrightarrow -9\leq 3-5x< \frac{9}{8}\Leftrightarrow -12\leq -5x< \frac{-15}{8}\Leftrightarrow \frac{12}{5}\geq x> \frac{3}{8}$
Originally Posted by Raoh hello $\displaystyle -3\leq \frac{3-5x}{3}< \frac{3}{8}\Leftrightarrow -9\leq 3-5x< \frac{9}{8}\Leftrightarrow -12\leq -5x< \frac{-15}{8}\Leftrightarrow \frac{12}{5}\geq x> \frac{3}{8}$ thanks, so my way is correct, just not in lowest terms as yours is.
Originally Posted by crashdummy thanks, so my way is correct, just not in lowest terms as yours is. Exactly
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