SOLUTION: 4+x<-4 3x+1<2x+5 2(x+1)>x-7 7<15-x 9>12-6

Question 22920: 4+x<-4
3x+1<2x+5
2(x+1)>x-7
7<15-x
9>12-6

Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE THE FOLLOWING TO KNOW THE METHOD OF SOLUTION AND TRY..IF YOU STILL HAVE PROBLEM PLEASE COME BACK
2<5-2x/3<71 solutions --------------------------------------------------------------------------------Answer 8553 by venugopalramana(298) on 2005-10-30 11:25:19 (Show Source): 2<5-2x/3<7..take one by one to solve2<5-2x/32x/3<5-2=3x<3*3/2x<9/25-2x/3<75-7<2x/3-2<2x/3...or..-2*3/2<1*x...or.....-3<1*x ....so the answer is -3<1*x<9/2********************************Graphs/17965: how do you solve eniqualities1 solutions --------------------------------------------------------------------------------Answer 8675 by venugopalramana(298) on 2005-11-01 11:13:46 (Show Source): the rules of transformations between equalities and inequalities are same but for few exceptions..they are given below..use them to solve the inequalities...........property..............equality....inequality.......conclusion...example........................3+2=5.......5<81.reverse sides.left to right ...5=3+2......8>5..inequality changes from < to > 2.multiply/divide with -ve number.-5=-(3+2)..-5>-8 same as above.........3.take reciprocal.................1/(3+2)=1/5..1/5>1/8..same as above..so take care of these and solve1example...x+5>8..or x>8-5=3..or x>32.3>5+x..or x+5<3..or x<3-5=-2..or x<-2..(exception rule no.1 above)2.example...-x+5>8..or-x>8-5=3..or -x>3..or x<3(exception rule no2. above)3.example...1/x>1/8...or x<8....(exception rule no.3 above)SO BE CAREFULL IN THESE OPERATIONS..PARTICULARLY MULTIPLICATION/DIVISION WITH UNKNOWNS..YOU HAVE TO BE SURE WHETHER IT IS A +VE NUMBER OR _VE NUMBER

Inequalities/18942: What is the solution to -4x <= -16 -4x is less than or equal to -16. how do i isolate the x from the -4x?1 solutions Answer 9138 by venugopalramana(345) on 2005-11-09 05:54:34 (Show Source): -4x <= -16let us cosider the 2 cases included here seperately1st.case is -4x=-16...............2nd.case is -4x<-16...........................1st.case is -4x=-16here on lhs we have - 4 multiplying x so when it is transfered to rhs it divides the number there.so we get x=-16/-4 =+42nd.case is -4x<-16here also on lhs we have - 4 multiplying x so when it is transfered to rhs it divides the number there.BUT IN CASE OF INEQUALITIES WHEN A -VE NUMBER IS USED TO MULTIPLY OR DIVIDE THEN THE INEQUALITY GETS REVERSED...FOR EXAMPLE WE KNOW5>0...BUT WHEN WE MULTIPLY BOTH SIDES WITH -1 SAY WE GET -5<0.THIS IS ONE OF THE FEW MAJOR DIFFERENCES WE HAVE BETWEEN INEQUALITIES AND EQUALITIESso we get x>-16/-4 ....or...x>+4hope you understood ..if you want to see more refer my solutions for similar problems posted in the site

Quadratic_Equations/18722: Could some one help me understand this problem? The instructions are to solve the inequality. State the solution set using interval notation and graph it:x - 2/x + 3 is less than 1 Thanks for the help1 solutions Answer 8977 by venugopalramana(345) on 2005-11-07 04:29:05 (Show Source): x - 2/x + 3 is less than 1 let y=(x - 2/x + 3)<1 x - 2/x + 3-1<0 ....or....x - 2/x + 2<0 (x^2-2+2x)/x<0...or ...(x^2+2x-2)/x<0..now a fraction will be -ve (<0)if n.r and d.r are of different signs..let us take the 2 cases dr=x is +ve...then nr should be -vedr=x is -ve...then nr should be +ve.now solve the nr using quadratic formula hence x=(-1+sqrt3)/2...and....(-1-sqrt3)/2..for conveinience if we call these 2 values as p(approximately=-1.37)and q(approximately=0.37),we find that if x lies between p and q (-1.37 and 0.37 )nr is -ve and when x is less than p(-1.37) or greater than q(0.37) ,nr is +ve.now we have to combine this with the above assumption on drdr=x is +ve...then nr should be -ve..so x should be between 0 and (-1+sqrt3)/2(not -1.37 to zero as x is already taken as positive)dr=x is -ve...then nr should be +ve.so x should be less than ....(-1-sqrt3)/2..(not 0.37 to zero as x is already taken as negative)

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