3<5-3x<9 then I added three to everything 6<8-3x<12 now what? can u help.
V vinceman New member Joined Oct 17, 2005 Messages 13 Nov 22, 2005 #1 3<5-3x<9 then I added three to everything 6<8-3x<12 now what? can u help.
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Nov 22, 2005 #2 Re: Combined Inequalities Hello, vinceman! \(\displaystyle 3\:<\:5\,-\,3x\:<\:9\) then I added three to everything . . . . but what good did that do? Click to expand... Remember: you are solving for \(\displaystyle x\). We have: . . . \(\displaystyle 3\:<\:5\,-\,3x\:<\:9\) Subtract 5 from everything: . . . -\(\displaystyle 2\:<\:-3x\:<\:4\) Divide by -3 . . . and reverse the inequalities! . . . \(\displaystyle \frac{2}{3}\:>\:x\:>\:-\frac{4}{3}\) Therefore: .-\(\displaystyle \frac{4}{3}\:<\:x\:<\:\frac{2}{3}\)
Re: Combined Inequalities Hello, vinceman! \(\displaystyle 3\:<\:5\,-\,3x\:<\:9\) then I added three to everything . . . . but what good did that do? Click to expand... Remember: you are solving for \(\displaystyle x\). We have: . . . \(\displaystyle 3\:<\:5\,-\,3x\:<\:9\) Subtract 5 from everything: . . . -\(\displaystyle 2\:<\:-3x\:<\:4\) Divide by -3 . . . and reverse the inequalities! . . . \(\displaystyle \frac{2}{3}\:>\:x\:>\:-\frac{4}{3}\) Therefore: .-\(\displaystyle \frac{4}{3}\:<\:x\:<\:\frac{2}{3}\)
D Denis Senior Member Joined Feb 17, 2004 Messages 1,707 Nov 23, 2005 #3 3 < 5-3x < 9 Want to check if Soroban is correct? 5 - 3x = 4 to 8 : right? 3x = 5 - (4 to 8) 3x = 1, 0, -1, -2, -3 x = 1/3, 0, -1/3, -2/3, -1 so x < 2/3 and > -4/3 : Soroban gets an A+ :shock:
3 < 5-3x < 9 Want to check if Soroban is correct? 5 - 3x = 4 to 8 : right? 3x = 5 - (4 to 8) 3x = 1, 0, -1, -2, -3 x = 1/3, 0, -1/3, -2/3, -1 so x < 2/3 and > -4/3 : Soroban gets an A+ :shock:
