Help with a system of inequalities with absolute values











up vote
1
down vote

favorite












I'm trying to solve this system on inequalities



$$
left{
begin{array}{c}
|x-3|<2x \
|2x+5|>3
end{array}
right.
$$



The steps I'm taking are:



Finding the absolute values sings, so for
$x-3 geq 0$ we have $x geq 3$ therefore
$$|x-3| =
left{
begin{array}{c}
x-3 & text{for $x geq 3$} \
-x+3 & text{for $x < 3$}
end{array}
right. $$



and



$2x+5 geq 0$ we have $x geq frac{-2}{5}$ therefore
$$|2x+5| =
left{
begin{array}{c}
2x+5 & text{for $x geq frac{-2}{5}$} \
-2x-5 & text{for $x<frac{-2}{5}$} \
end{array}
right. $$



So I build a few systems with the complete inequalities, for the first one we have:



$$
left{
begin{array}{c}
x geq 3 \
x-3<2x = x>-3
end{array}
right. $$



So the solution here would be $x>3$, then:



$$
left{
begin{array}{c}
x<3 \
-x+3<2x = x>1
end{array}
right. $$



The solution would be $1<x<3$. Then



$$
left{
begin{array}{c}
x geq frac{-2}{5} \
2x+5>3 = x>-1
end{array}
right. $$



So the solution of the system is $x>-1$, then



$$
left{
begin{array}{c}
x< frac{-2}{5} \
-2x-5>3 = x<-4
end{array}
right. $$



And the solution is $x<-4$



Now the solution my book gives is x>1 for the initial system. But I can't find that one. I can't get a solution at all. I tried finding a common point between the 4 solutions I found (as if it was a 4-inequalities system), but there isn't one really. What am I doing wrong?






calculus algebra-precalculus proof-verification







share|cite|improve this question
























  • either plot the graphs of the inequalities or square both the inequalties and solve the resulting quadratic inequalities
    – vidyarthi
    yesterday










  • Square $|x-3|<2x$ to $x^2+9-6x<4x^2$ you mean?
    – Paul
    yesterday












  • yes, exactly. Now, solve the inequality by factoring the quadratic
    – vidyarthi
    yesterday










  • Is this the only way to solve this? I don't really understand how squaring the inequalities takes away the absolute value.
    – Paul
    yesterday










  • since square of a real is always positive, so the absolute value is easily removed
    – vidyarthi
    yesterday















up vote
1
down vote

favorite












I'm trying to solve this system on inequalities



$$
left{
begin{array}{c}
|x-3|<2x \
|2x+5|>3
end{array}
right.
$$



The steps I'm taking are:



Finding the absolute values sings, so for
$x-3 geq 0$ we have $x geq 3$ therefore
$$|x-3| =
left{
begin{array}{c}
x-3 & text{for $x geq 3$} \
-x+3 & text{for $x < 3$}
end{array}
right. $$



and



$2x+5 geq 0$ we have $x geq frac{-2}{5}$ therefore
$$|2x+5| =
left{
begin{array}{c}
2x+5 & text{for $x geq frac{-2}{5}$} \
-2x-5 & text{for $x<frac{-2}{5}$} \
end{array}
right. $$



So I build a few systems with the complete inequalities, for the first one we have:



$$
left{
begin{array}{c}
x geq 3 \
x-3<2x = x>-3
end{array}
right. $$



So the solution here would be $x>3$, then:



$$
left{
begin{array}{c}
x<3 \
-x+3<2x = x>1
end{array}
right. $$



The solution would be $1<x<3$. Then



$$
left{
begin{array}{c}
x geq frac{-2}{5} \
2x+5>3 = x>-1
end{array}
right. $$



So the solution of the system is $x>-1$, then



$$
left{
begin{array}{c}
x< frac{-2}{5} \
-2x-5>3 = x<-4
end{array}
right. $$



And the solution is $x<-4$



Now the solution my book gives is x>1 for the initial system. But I can't find that one. I can't get a solution at all. I tried finding a common point between the 4 solutions I found (as if it was a 4-inequalities system), but there isn't one really. What am I doing wrong?






calculus algebra-precalculus proof-verification







share|cite|improve this question
























  • either plot the graphs of the inequalities or square both the inequalties and solve the resulting quadratic inequalities
    – vidyarthi
    yesterday










  • Square $|x-3|<2x$ to $x^2+9-6x<4x^2$ you mean?
    – Paul
    yesterday












  • yes, exactly. Now, solve the inequality by factoring the quadratic
    – vidyarthi
    yesterday










  • Is this the only way to solve this? I don't really understand how squaring the inequalities takes away the absolute value.
    – Paul
    yesterday










  • since square of a real is always positive, so the absolute value is easily removed
    – vidyarthi
    yesterday













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I'm trying to solve this system on inequalities



$$
left{
begin{array}{c}
|x-3|<2x \
|2x+5|>3
end{array}
right.
$$



The steps I'm taking are:



Finding the absolute values sings, so for
$x-3 geq 0$ we have $x geq 3$ therefore
$$|x-3| =
left{
begin{array}{c}
x-3 & text{for $x geq 3$} \
-x+3 & text{for $x < 3$}
end{array}
right. $$



and



$2x+5 geq 0$ we have $x geq frac{-2}{5}$ therefore
$$|2x+5| =
left{
begin{array}{c}
2x+5 & text{for $x geq frac{-2}{5}$} \
-2x-5 & text{for $x<frac{-2}{5}$} \
end{array}
right. $$



So I build a few systems with the complete inequalities, for the first one we have:



$$
left{
begin{array}{c}
x geq 3 \
x-3<2x = x>-3
end{array}
right. $$



So the solution here would be $x>3$, then:



$$
left{
begin{array}{c}
x<3 \
-x+3<2x = x>1
end{array}
right. $$



The solution would be $1<x<3$. Then



$$
left{
begin{array}{c}
x geq frac{-2}{5} \
2x+5>3 = x>-1
end{array}
right. $$



So the solution of the system is $x>-1$, then



$$
left{
begin{array}{c}
x< frac{-2}{5} \
-2x-5>3 = x<-4
end{array}
right. $$



And the solution is $x<-4$



Now the solution my book gives is x>1 for the initial system. But I can't find that one. I can't get a solution at all. I tried finding a common point between the 4 solutions I found (as if it was a 4-inequalities system), but there isn't one really. What am I doing wrong?






calculus algebra-precalculus proof-verification







share|cite|improve this question















I'm trying to solve this system on inequalities



$$
left{
begin{array}{c}
|x-3|<2x \
|2x+5|>3
end{array}
right.
$$



The steps I'm taking are:



Finding the absolute values sings, so for
$x-3 geq 0$ we have $x geq 3$ therefore
$$|x-3| =
left{
begin{array}{c}
x-3 & text{for $x geq 3$} \
-x+3 & text{for $x < 3$}
end{array}
right. $$



and



$2x+5 geq 0$ we have $x geq frac{-2}{5}$ therefore
$$|2x+5| =
left{
begin{array}{c}
2x+5 & text{for $x geq frac{-2}{5}$} \
-2x-5 & text{for $x<frac{-2}{5}$} \
end{array}
right. $$



So I build a few systems with the complete inequalities, for the first one we have:



$$
left{
begin{array}{c}
x geq 3 \
x-3<2x = x>-3
end{array}
right. $$



So the solution here would be $x>3$, then:



$$
left{
begin{array}{c}
x<3 \
-x+3<2x = x>1
end{array}
right. $$



The solution would be $1<x<3$. Then



$$
left{
begin{array}{c}
x geq frac{-2}{5} \
2x+5>3 = x>-1
end{array}
right. $$



So the solution of the system is $x>-1$, then



$$
left{
begin{array}{c}
x< frac{-2}{5} \
-2x-5>3 = x<-4
end{array}
right. $$



And the solution is $x<-4$



Now the solution my book gives is x>1 for the initial system. But I can't find that one. I can't get a solution at all. I tried finding a common point between the 4 solutions I found (as if it was a 4-inequalities system), but there isn't one really. What am I doing wrong?






calculus algebra-precalculus proof-verification




calculus algebra-precalculus proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 25 mins ago









N. F. Taussig

42.4k93254




42.4k93254










asked yesterday









Paul

1546




1546












  • either plot the graphs of the inequalities or square both the inequalties and solve the resulting quadratic inequalities
    – vidyarthi
    yesterday










  • Square $|x-3|<2x$ to $x^2+9-6x<4x^2$ you mean?
    – Paul
    yesterday












  • yes, exactly. Now, solve the inequality by factoring the quadratic
    – vidyarthi
    yesterday










  • Is this the only way to solve this? I don't really understand how squaring the inequalities takes away the absolute value.
    – Paul
    yesterday










  • since square of a real is always positive, so the absolute value is easily removed
    – vidyarthi
    yesterday


















  • either plot the graphs of the inequalities or square both the inequalties and solve the resulting quadratic inequalities
    – vidyarthi
    yesterday










  • Square $|x-3|<2x$ to $x^2+9-6x<4x^2$ you mean?
    – Paul
    yesterday












  • yes, exactly. Now, solve the inequality by factoring the quadratic
    – vidyarthi
    yesterday










  • Is this the only way to solve this? I don't really understand how squaring the inequalities takes away the absolute value.
    – Paul
    yesterday










  • since square of a real is always positive, so the absolute value is easily removed
    – vidyarthi
    yesterday
















either plot the graphs of the inequalities or square both the inequalties and solve the resulting quadratic inequalities
– vidyarthi
yesterday




either plot the graphs of the inequalities or square both the inequalties and solve the resulting quadratic inequalities
– vidyarthi
yesterday












Square $|x-3|<2x$ to $x^2+9-6x<4x^2$ you mean?
– Paul
yesterday






Square $|x-3|<2x$ to $x^2+9-6x<4x^2$ you mean?
– Paul
yesterday














yes, exactly. Now, solve the inequality by factoring the quadratic
– vidyarthi
yesterday




yes, exactly. Now, solve the inequality by factoring the quadratic
– vidyarthi
yesterday












Is this the only way to solve this? I don't really understand how squaring the inequalities takes away the absolute value.
– Paul
yesterday




Is this the only way to solve this? I don't really understand how squaring the inequalities takes away the absolute value.
– Paul
yesterday












since square of a real is always positive, so the absolute value is easily removed
– vidyarthi
yesterday




since square of a real is always positive, so the absolute value is easily removed
– vidyarthi
yesterday










4 Answers
4






active

oldest

votes

















up vote
1
down vote













Since $2x>|x-3|$ we get $x>0$ so $2x+5>0$ and so $2x+5>3$ so $x>-2$ which is nothing new. So $x>0$ and $|x-3|<2x$ so after squaring we get $$ x^2-6x+9<4x^2implies 3x^2+6x-9>0$$



or $$(x+3)(x-1)>0implies x-1>0$$ so $boxed{x>1}$.






share|cite|improve this answer























  • so you reduced two squarings to just one! great answer
    – vidyarthi
    yesterday


















up vote
0
down vote













We have $$|x-3|<2ximplies 3x^2+6x-9>0implies (x+3)(x-1)>0$$ and $$|2x+5|>3implies 4x^2+20x+16>0implies (x+4)(x+1)>0$$. I think you could proceed now? For a quick method on how to solve further, see here






share|cite|improve this answer























  • If I try to graph this I get x<-4 and x>1 as solutions. It should only be x>1. Where am I wrong?
    – Paul
    yesterday


















up vote
0
down vote













Since $0leleft|x-3right|$ and $left|x-3right|le2x$ we conclude $2xge0$ and then $xge0$.
From $left|x-3right|<2x$ and since $xge0$, we get $-2x< x-3<2x$ and therefore we have $left{begin{array}{c}x-3<2x \x-3>-2xend{array}right.$ and therefore $left{begin{array}{c}x>-3 \3x>3end{array}right.$ and then $left{begin{array}{c}x>-3 \x>1end{array}right.$, so $x>1$ which satisfies the initial statement $xge0$.
So far we see $x$ has to be greater than $1$ to satisfy $left|x-3right|<2x$.
Let us consider the next inequality $|2x+5|>3$ which leads to $2x+5>3$ or $2x+5<-3$. Then we get $x>-1$ or $x<-4$. And finally if we combine this with the result of first inequality, $x>1$ we get $x>1$.






share|cite|improve this answer




























    up vote
    0
    down vote













    There is actually only the first inequality to be solved:



    $$ |2x+5| stackrel{color{blue}{0}leq|x-3| color{blue}{< 2x}}{>} |x-3|+5 > 3$$
    So, you only need to solve $|x-3| < 2x$ while $x>0$:
    $$
    -2x < x-3 < 2x Leftrightarrow
    begin{cases}
    3x > 3 Leftrightarrow color{blue}{x> 1} \
    x > -3 mbox{ does not extend the solution}
    end{cases}
    $$






    share|cite|improve this answer





















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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      1
      down vote













      Since $2x>|x-3|$ we get $x>0$ so $2x+5>0$ and so $2x+5>3$ so $x>-2$ which is nothing new. So $x>0$ and $|x-3|<2x$ so after squaring we get $$ x^2-6x+9<4x^2implies 3x^2+6x-9>0$$



      or $$(x+3)(x-1)>0implies x-1>0$$ so $boxed{x>1}$.






      share|cite|improve this answer























      • so you reduced two squarings to just one! great answer
        – vidyarthi
        yesterday















      up vote
      1
      down vote













      Since $2x>|x-3|$ we get $x>0$ so $2x+5>0$ and so $2x+5>3$ so $x>-2$ which is nothing new. So $x>0$ and $|x-3|<2x$ so after squaring we get $$ x^2-6x+9<4x^2implies 3x^2+6x-9>0$$



      or $$(x+3)(x-1)>0implies x-1>0$$ so $boxed{x>1}$.






      share|cite|improve this answer























      • so you reduced two squarings to just one! great answer
        – vidyarthi
        yesterday













      up vote
      1
      down vote










      up vote
      1
      down vote









      Since $2x>|x-3|$ we get $x>0$ so $2x+5>0$ and so $2x+5>3$ so $x>-2$ which is nothing new. So $x>0$ and $|x-3|<2x$ so after squaring we get $$ x^2-6x+9<4x^2implies 3x^2+6x-9>0$$



      or $$(x+3)(x-1)>0implies x-1>0$$ so $boxed{x>1}$.






      share|cite|improve this answer














      Since $2x>|x-3|$ we get $x>0$ so $2x+5>0$ and so $2x+5>3$ so $x>-2$ which is nothing new. So $x>0$ and $|x-3|<2x$ so after squaring we get $$ x^2-6x+9<4x^2implies 3x^2+6x-9>0$$



      or $$(x+3)(x-1)>0implies x-1>0$$ so $boxed{x>1}$.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited yesterday









      vidyarthi

      2,6881831




      2,6881831










      answered yesterday









      greedoid

      34.3k114488




      34.3k114488












      • so you reduced two squarings to just one! great answer
        – vidyarthi
        yesterday


















      • so you reduced two squarings to just one! great answer
        – vidyarthi
        yesterday
















      so you reduced two squarings to just one! great answer
      – vidyarthi
      yesterday




      so you reduced two squarings to just one! great answer
      – vidyarthi
      yesterday










      up vote
      0
      down vote













      We have $$|x-3|<2ximplies 3x^2+6x-9>0implies (x+3)(x-1)>0$$ and $$|2x+5|>3implies 4x^2+20x+16>0implies (x+4)(x+1)>0$$. I think you could proceed now? For a quick method on how to solve further, see here






      share|cite|improve this answer























      • If I try to graph this I get x<-4 and x>1 as solutions. It should only be x>1. Where am I wrong?
        – Paul
        yesterday















      up vote
      0
      down vote













      We have $$|x-3|<2ximplies 3x^2+6x-9>0implies (x+3)(x-1)>0$$ and $$|2x+5|>3implies 4x^2+20x+16>0implies (x+4)(x+1)>0$$. I think you could proceed now? For a quick method on how to solve further, see here






      share|cite|improve this answer























      • If I try to graph this I get x<-4 and x>1 as solutions. It should only be x>1. Where am I wrong?
        – Paul
        yesterday













      up vote
      0
      down vote










      up vote
      0
      down vote









      We have $$|x-3|<2ximplies 3x^2+6x-9>0implies (x+3)(x-1)>0$$ and $$|2x+5|>3implies 4x^2+20x+16>0implies (x+4)(x+1)>0$$. I think you could proceed now? For a quick method on how to solve further, see here






      share|cite|improve this answer














      We have $$|x-3|<2ximplies 3x^2+6x-9>0implies (x+3)(x-1)>0$$ and $$|2x+5|>3implies 4x^2+20x+16>0implies (x+4)(x+1)>0$$. I think you could proceed now? For a quick method on how to solve further, see here







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited yesterday

























      answered yesterday









      vidyarthi

      2,6881831




      2,6881831












      • If I try to graph this I get x<-4 and x>1 as solutions. It should only be x>1. Where am I wrong?
        – Paul
        yesterday


















      • If I try to graph this I get x<-4 and x>1 as solutions. It should only be x>1. Where am I wrong?
        – Paul
        yesterday
















      If I try to graph this I get x<-4 and x>1 as solutions. It should only be x>1. Where am I wrong?
      – Paul
      yesterday




      If I try to graph this I get x<-4 and x>1 as solutions. It should only be x>1. Where am I wrong?
      – Paul
      yesterday










      up vote
      0
      down vote













      Since $0leleft|x-3right|$ and $left|x-3right|le2x$ we conclude $2xge0$ and then $xge0$.
      From $left|x-3right|<2x$ and since $xge0$, we get $-2x< x-3<2x$ and therefore we have $left{begin{array}{c}x-3<2x \x-3>-2xend{array}right.$ and therefore $left{begin{array}{c}x>-3 \3x>3end{array}right.$ and then $left{begin{array}{c}x>-3 \x>1end{array}right.$, so $x>1$ which satisfies the initial statement $xge0$.
      So far we see $x$ has to be greater than $1$ to satisfy $left|x-3right|<2x$.
      Let us consider the next inequality $|2x+5|>3$ which leads to $2x+5>3$ or $2x+5<-3$. Then we get $x>-1$ or $x<-4$. And finally if we combine this with the result of first inequality, $x>1$ we get $x>1$.






      share|cite|improve this answer

























        up vote
        0
        down vote













        Since $0leleft|x-3right|$ and $left|x-3right|le2x$ we conclude $2xge0$ and then $xge0$.
        From $left|x-3right|<2x$ and since $xge0$, we get $-2x< x-3<2x$ and therefore we have $left{begin{array}{c}x-3<2x \x-3>-2xend{array}right.$ and therefore $left{begin{array}{c}x>-3 \3x>3end{array}right.$ and then $left{begin{array}{c}x>-3 \x>1end{array}right.$, so $x>1$ which satisfies the initial statement $xge0$.
        So far we see $x$ has to be greater than $1$ to satisfy $left|x-3right|<2x$.
        Let us consider the next inequality $|2x+5|>3$ which leads to $2x+5>3$ or $2x+5<-3$. Then we get $x>-1$ or $x<-4$. And finally if we combine this with the result of first inequality, $x>1$ we get $x>1$.






        share|cite|improve this answer























          up vote
          0
          down vote










          up vote
          0
          down vote









          Since $0leleft|x-3right|$ and $left|x-3right|le2x$ we conclude $2xge0$ and then $xge0$.
          From $left|x-3right|<2x$ and since $xge0$, we get $-2x< x-3<2x$ and therefore we have $left{begin{array}{c}x-3<2x \x-3>-2xend{array}right.$ and therefore $left{begin{array}{c}x>-3 \3x>3end{array}right.$ and then $left{begin{array}{c}x>-3 \x>1end{array}right.$, so $x>1$ which satisfies the initial statement $xge0$.
          So far we see $x$ has to be greater than $1$ to satisfy $left|x-3right|<2x$.
          Let us consider the next inequality $|2x+5|>3$ which leads to $2x+5>3$ or $2x+5<-3$. Then we get $x>-1$ or $x<-4$. And finally if we combine this with the result of first inequality, $x>1$ we get $x>1$.






          share|cite|improve this answer












          Since $0leleft|x-3right|$ and $left|x-3right|le2x$ we conclude $2xge0$ and then $xge0$.
          From $left|x-3right|<2x$ and since $xge0$, we get $-2x< x-3<2x$ and therefore we have $left{begin{array}{c}x-3<2x \x-3>-2xend{array}right.$ and therefore $left{begin{array}{c}x>-3 \3x>3end{array}right.$ and then $left{begin{array}{c}x>-3 \x>1end{array}right.$, so $x>1$ which satisfies the initial statement $xge0$.
          So far we see $x$ has to be greater than $1$ to satisfy $left|x-3right|<2x$.
          Let us consider the next inequality $|2x+5|>3$ which leads to $2x+5>3$ or $2x+5<-3$. Then we get $x>-1$ or $x<-4$. And finally if we combine this with the result of first inequality, $x>1$ we get $x>1$.







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          answered yesterday









          Arash Rashidi

          558




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              up vote
              0
              down vote













              There is actually only the first inequality to be solved:



              $$ |2x+5| stackrel{color{blue}{0}leq|x-3| color{blue}{< 2x}}{>} |x-3|+5 > 3$$
              So, you only need to solve $|x-3| < 2x$ while $x>0$:
              $$
              -2x < x-3 < 2x Leftrightarrow
              begin{cases}
              3x > 3 Leftrightarrow color{blue}{x> 1} \
              x > -3 mbox{ does not extend the solution}
              end{cases}
              $$






              share|cite|improve this answer

























                up vote
                0
                down vote













                There is actually only the first inequality to be solved:



                $$ |2x+5| stackrel{color{blue}{0}leq|x-3| color{blue}{< 2x}}{>} |x-3|+5 > 3$$
                So, you only need to solve $|x-3| < 2x$ while $x>0$:
                $$
                -2x < x-3 < 2x Leftrightarrow
                begin{cases}
                3x > 3 Leftrightarrow color{blue}{x> 1} \
                x > -3 mbox{ does not extend the solution}
                end{cases}
                $$






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  There is actually only the first inequality to be solved:



                  $$ |2x+5| stackrel{color{blue}{0}leq|x-3| color{blue}{< 2x}}{>} |x-3|+5 > 3$$
                  So, you only need to solve $|x-3| < 2x$ while $x>0$:
                  $$
                  -2x < x-3 < 2x Leftrightarrow
                  begin{cases}
                  3x > 3 Leftrightarrow color{blue}{x> 1} \
                  x > -3 mbox{ does not extend the solution}
                  end{cases}
                  $$






                  share|cite|improve this answer












                  There is actually only the first inequality to be solved:



                  $$ |2x+5| stackrel{color{blue}{0}leq|x-3| color{blue}{< 2x}}{>} |x-3|+5 > 3$$
                  So, you only need to solve $|x-3| < 2x$ while $x>0$:
                  $$
                  -2x < x-3 < 2x Leftrightarrow
                  begin{cases}
                  3x > 3 Leftrightarrow color{blue}{x> 1} \
                  x > -3 mbox{ does not extend the solution}
                  end{cases}
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  trancelocation

                  8,0741519




                  8,0741519






























                       

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