Solve for $X$: $4+2X > 7$












1












$begingroup$


Solve for $$4+2x>7$$.



My answer was 1 and 1/2 but I am not so sure it is right, could anyone confirm it for me. i need someone to solve the inequality. the first step would be to subtract $4$ from both sides.






algebra-precalculus







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$endgroup$








  • 1




    $begingroup$
    It is an inequality, so you cannot have just one answer. You would have to have a spectrum of answers. Yiur answer is correct - just divide the answer that you get by subtracting 4 from each side to get tbe highest bound on the value of x.
    $endgroup$
    – user122283
    Jan 20 '14 at 15:28










  • $begingroup$
    That being said, 1/2 is not correct since $4+2(1/2) = 5 not >7$
    $endgroup$
    – Squirtle
    Jan 20 '14 at 18:05
















1












$begingroup$


Solve for $$4+2x>7$$.



My answer was 1 and 1/2 but I am not so sure it is right, could anyone confirm it for me. i need someone to solve the inequality. the first step would be to subtract $4$ from both sides.






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It is an inequality, so you cannot have just one answer. You would have to have a spectrum of answers. Yiur answer is correct - just divide the answer that you get by subtracting 4 from each side to get tbe highest bound on the value of x.
    $endgroup$
    – user122283
    Jan 20 '14 at 15:28










  • $begingroup$
    That being said, 1/2 is not correct since $4+2(1/2) = 5 not >7$
    $endgroup$
    – Squirtle
    Jan 20 '14 at 18:05














1












1








1





$begingroup$


Solve for $$4+2x>7$$.



My answer was 1 and 1/2 but I am not so sure it is right, could anyone confirm it for me. i need someone to solve the inequality. the first step would be to subtract $4$ from both sides.






algebra-precalculus







share|cite|improve this question











$endgroup$




Solve for $$4+2x>7$$.



My answer was 1 and 1/2 but I am not so sure it is right, could anyone confirm it for me. i need someone to solve the inequality. the first step would be to subtract $4$ from both sides.






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 20 '14 at 15:35









Yiorgos S. Smyrlis

63k1384163




63k1384163










asked Jan 20 '14 at 15:14









HasanHasan

111




111








  • 1




    $begingroup$
    It is an inequality, so you cannot have just one answer. You would have to have a spectrum of answers. Yiur answer is correct - just divide the answer that you get by subtracting 4 from each side to get tbe highest bound on the value of x.
    $endgroup$
    – user122283
    Jan 20 '14 at 15:28










  • $begingroup$
    That being said, 1/2 is not correct since $4+2(1/2) = 5 not >7$
    $endgroup$
    – Squirtle
    Jan 20 '14 at 18:05














  • 1




    $begingroup$
    It is an inequality, so you cannot have just one answer. You would have to have a spectrum of answers. Yiur answer is correct - just divide the answer that you get by subtracting 4 from each side to get tbe highest bound on the value of x.
    $endgroup$
    – user122283
    Jan 20 '14 at 15:28










  • $begingroup$
    That being said, 1/2 is not correct since $4+2(1/2) = 5 not >7$
    $endgroup$
    – Squirtle
    Jan 20 '14 at 18:05








1




1




$begingroup$
It is an inequality, so you cannot have just one answer. You would have to have a spectrum of answers. Yiur answer is correct - just divide the answer that you get by subtracting 4 from each side to get tbe highest bound on the value of x.
$endgroup$
– user122283
Jan 20 '14 at 15:28




$begingroup$
It is an inequality, so you cannot have just one answer. You would have to have a spectrum of answers. Yiur answer is correct - just divide the answer that you get by subtracting 4 from each side to get tbe highest bound on the value of x.
$endgroup$
– user122283
Jan 20 '14 at 15:28












$begingroup$
That being said, 1/2 is not correct since $4+2(1/2) = 5 not >7$
$endgroup$
– Squirtle
Jan 20 '14 at 18:05




$begingroup$
That being said, 1/2 is not correct since $4+2(1/2) = 5 not >7$
$endgroup$
– Squirtle
Jan 20 '14 at 18:05










3 Answers
3






active

oldest

votes


















1












$begingroup$

Yes. You'll have



$$begin{align}4+2xgt 7&iff 4+2x-4gt 7-4\&iff 2xgt 3\&iff 2x/2gt 3/2 (because 2gt 0)\&iff xgt 3/2=1.5.end{align}$$



Note that if $agt 0$, then
$$bgt ciff b/agt c/a.$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Property $bf1$: If $x,y$ are two real numbers such that: $x>y$, we have that $x+mathrm c>y+mathrm c$ for every $mathrm c$ in $Bbb R$.




    Property $bf2$: If $x,y$ are two real numbers such that: $x>y$, we have that $xcdotmathrm c>ycdotmathrm c$ for every positive $mathrm c$ in $Bbb R$.




    $$require{cancel}begin{align}
    4+2xgt 7&overset{mathbf{Pr. 1}}{iff} 4+2xcolor{red}{-4}gt 7color{red}{-4}\,\
    & iff 2x gt 3 \ ,\
    & overset{mathbf{Pr. 2}}{iff}2xcdotdfrac12 gt 3dfrac 12 \ , \
    & iff xgt 3/2=1.5
    end{align}$$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Yes, after subtracting $ 4 $ to both sides



      $$ 2x > 3 $$



      Then divide $ 2 $ to both sides to solve for x:



      $$ x > frac{3}{2} $$ or $$ x > 1.5 $$






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        Yes. You'll have



        $$begin{align}4+2xgt 7&iff 4+2x-4gt 7-4\&iff 2xgt 3\&iff 2x/2gt 3/2 (because 2gt 0)\&iff xgt 3/2=1.5.end{align}$$



        Note that if $agt 0$, then
        $$bgt ciff b/agt c/a.$$






        share|cite|improve this answer











        $endgroup$


















          1












          $begingroup$

          Yes. You'll have



          $$begin{align}4+2xgt 7&iff 4+2x-4gt 7-4\&iff 2xgt 3\&iff 2x/2gt 3/2 (because 2gt 0)\&iff xgt 3/2=1.5.end{align}$$



          Note that if $agt 0$, then
          $$bgt ciff b/agt c/a.$$






          share|cite|improve this answer











          $endgroup$
















            1












            1








            1





            $begingroup$

            Yes. You'll have



            $$begin{align}4+2xgt 7&iff 4+2x-4gt 7-4\&iff 2xgt 3\&iff 2x/2gt 3/2 (because 2gt 0)\&iff xgt 3/2=1.5.end{align}$$



            Note that if $agt 0$, then
            $$bgt ciff b/agt c/a.$$






            share|cite|improve this answer











            $endgroup$



            Yes. You'll have



            $$begin{align}4+2xgt 7&iff 4+2x-4gt 7-4\&iff 2xgt 3\&iff 2x/2gt 3/2 (because 2gt 0)\&iff xgt 3/2=1.5.end{align}$$



            Note that if $agt 0$, then
            $$bgt ciff b/agt c/a.$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 20 '14 at 15:23

























            answered Jan 20 '14 at 15:16









            mathlovemathlove

            91.7k881215




            91.7k881215























                1












                $begingroup$

                Property $bf1$: If $x,y$ are two real numbers such that: $x>y$, we have that $x+mathrm c>y+mathrm c$ for every $mathrm c$ in $Bbb R$.




                Property $bf2$: If $x,y$ are two real numbers such that: $x>y$, we have that $xcdotmathrm c>ycdotmathrm c$ for every positive $mathrm c$ in $Bbb R$.




                $$require{cancel}begin{align}
                4+2xgt 7&overset{mathbf{Pr. 1}}{iff} 4+2xcolor{red}{-4}gt 7color{red}{-4}\,\
                & iff 2x gt 3 \ ,\
                & overset{mathbf{Pr. 2}}{iff}2xcdotdfrac12 gt 3dfrac 12 \ , \
                & iff xgt 3/2=1.5
                end{align}$$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Property $bf1$: If $x,y$ are two real numbers such that: $x>y$, we have that $x+mathrm c>y+mathrm c$ for every $mathrm c$ in $Bbb R$.




                  Property $bf2$: If $x,y$ are two real numbers such that: $x>y$, we have that $xcdotmathrm c>ycdotmathrm c$ for every positive $mathrm c$ in $Bbb R$.




                  $$require{cancel}begin{align}
                  4+2xgt 7&overset{mathbf{Pr. 1}}{iff} 4+2xcolor{red}{-4}gt 7color{red}{-4}\,\
                  & iff 2x gt 3 \ ,\
                  & overset{mathbf{Pr. 2}}{iff}2xcdotdfrac12 gt 3dfrac 12 \ , \
                  & iff xgt 3/2=1.5
                  end{align}$$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Property $bf1$: If $x,y$ are two real numbers such that: $x>y$, we have that $x+mathrm c>y+mathrm c$ for every $mathrm c$ in $Bbb R$.




                    Property $bf2$: If $x,y$ are two real numbers such that: $x>y$, we have that $xcdotmathrm c>ycdotmathrm c$ for every positive $mathrm c$ in $Bbb R$.




                    $$require{cancel}begin{align}
                    4+2xgt 7&overset{mathbf{Pr. 1}}{iff} 4+2xcolor{red}{-4}gt 7color{red}{-4}\,\
                    & iff 2x gt 3 \ ,\
                    & overset{mathbf{Pr. 2}}{iff}2xcdotdfrac12 gt 3dfrac 12 \ , \
                    & iff xgt 3/2=1.5
                    end{align}$$






                    share|cite|improve this answer









                    $endgroup$



                    Property $bf1$: If $x,y$ are two real numbers such that: $x>y$, we have that $x+mathrm c>y+mathrm c$ for every $mathrm c$ in $Bbb R$.




                    Property $bf2$: If $x,y$ are two real numbers such that: $x>y$, we have that $xcdotmathrm c>ycdotmathrm c$ for every positive $mathrm c$ in $Bbb R$.




                    $$require{cancel}begin{align}
                    4+2xgt 7&overset{mathbf{Pr. 1}}{iff} 4+2xcolor{red}{-4}gt 7color{red}{-4}\,\
                    & iff 2x gt 3 \ ,\
                    & overset{mathbf{Pr. 2}}{iff}2xcdotdfrac12 gt 3dfrac 12 \ , \
                    & iff xgt 3/2=1.5
                    end{align}$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 31 '14 at 22:25







                    user93957






























                        0












                        $begingroup$

                        Yes, after subtracting $ 4 $ to both sides



                        $$ 2x > 3 $$



                        Then divide $ 2 $ to both sides to solve for x:



                        $$ x > frac{3}{2} $$ or $$ x > 1.5 $$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Yes, after subtracting $ 4 $ to both sides



                          $$ 2x > 3 $$



                          Then divide $ 2 $ to both sides to solve for x:



                          $$ x > frac{3}{2} $$ or $$ x > 1.5 $$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Yes, after subtracting $ 4 $ to both sides



                            $$ 2x > 3 $$



                            Then divide $ 2 $ to both sides to solve for x:



                            $$ x > frac{3}{2} $$ or $$ x > 1.5 $$






                            share|cite|improve this answer









                            $endgroup$



                            Yes, after subtracting $ 4 $ to both sides



                            $$ 2x > 3 $$



                            Then divide $ 2 $ to both sides to solve for x:



                            $$ x > frac{3}{2} $$ or $$ x > 1.5 $$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 8 at 3:57









                            Marvin CohenMarvin Cohen

                            112117




                            112117






























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