Solving inqualities by Using addition and subtract |
 May 11 2005, 12:06 AM
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#1
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 Senior Member  Group: Posts: 8,274 Joined: Mar 2004 Member No: 8,001  |
Solving inqualities by Using addition and subtraction
PROBLEM 1 6r > 10r - r -3r 6r > 9r - 3r 6r > 6r < -- i'm stuck .. what do i do ? do i cancel them out w/ 6, it will end up r > r, if i cancel out w/ 1, it would be 0 > 0 i'm so confused. So, am i right or wrong ... if wrong, help me please. PROBLEM 2 3.2x < 2x-(9 -1.2x) do i distribute ... 2x-9 and 2x - -1.3x ? PROBLEM 3 3 over 2 q minus 25 over 5 exact greater then 2 q over 4 For this problem, is the answer 5 exact greater then q ? if not, dont tell me the answer. |
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May 12 2005, 07:39 PM
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#2
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 Quand j'étais jeune... Group: Staff Alumni Posts: 6,826 Joined: Jan 2004 Member No: 1,272 |
QUOTE(aznxboredxperson @ May 11 2005, 12:06 AM) Solving inqualities by Using addition and subtraction PROBLEM 1 6r > 10r - r -3r 6r > 9r - 3r 6r > 6r < -- i'm stuck .. what do i do ? do i cancel them out w/ 6, it will end up r > r, if i cancel out w/ 1, it would be 0 > 0 i'm so confused. So, am i right or wrong ... if wrong, help me please. PROBLEM 2 3.2x < 2x-(9 -1.2x) do i distribute ... 2x-9 and 2x - -1.3x ? PROBLEM 3 3 over 2 q minus 25 over 5 exact greater then 2 q over 4 For this problem, is the answer 5 exact greater then q ? if not, dont tell me the answer.  I'm math stupid. Problem number 1 isn't possible, for me that is. But anyway, problem two requires that you distribute the negative sign. So, it's 3.2x < 2x-9+1.2x You always distribute the sign before the paranthesis I think this the equation you need to get to the answer: 6-20q-2q^2>0 But then I'm stuck there, too. Yes, did I say I'm math stupid? |
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May 12 2005, 11:05 PM
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#3
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 hi. call me linda. Group: Official Member Posts: 8,187 Joined: Feb 2004 Member No: 3,475 |
Hmm, are you sure that the first problem is written correctly? Because that's not exactly possible.
For the second problem, distribute the negative so you get: 3.2x < 2-9+1.2x Then you subtract 1.2x from both sides and subtrack 9 from 2 getting: 2x<-7 Then you divide by 2, getting x < 7/2 For the third problem, is this the inequality you're looking for? (I'm not sure by your wording, and what do you mean by exact greater? greater than or equal to?): 3/2 q - 25/5 >= 2/4 q ? If it is, then your answer isn't right. |
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May 13 2005, 08:48 PM
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#4
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Senior Member Group: Posts: 8,274 Joined: Mar 2004 Member No: 8,001 |
hehehe, i'm already done and i know how to do it now. thanks.
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May 13 2005, 08:50 PM
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#5
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Quand j'étais jeune... Group: Staff Alumni Posts: 6,826 Joined: Jan 2004 Member No: 1,272 |
QUOTE(aznxboredxperson @ May 13 2005, 8:48 PM) hehehe, i'm already done and i know how to do it now. thanks. Tell us! Tell us! |
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