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Help with a system of inequalities with absolute values
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1
down vote
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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
edited 25 mins ago
N. F. Taussig
42.4k93254
asked yesterday
Paul
1546
|
show 4 more comments
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
edited 25 mins ago
N. F. Taussig
42.4k93254
asked yesterday
Paul
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
|
show 4 more comments
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
edited 25 mins ago
N. F. Taussig
42.4k93254
asked yesterday
Paul
1546
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
edited 25 mins ago
N. F. Taussig
42.4k93254
asked yesterday
Paul
1546
edited 25 mins ago
N. F. Taussig
42.4k93254
asked yesterday
Paul
1546
edited 25 mins ago
N. F. Taussig
42.4k93254
edited 25 mins ago
N. F. Taussig
42.4k93254
edited 25 mins ago
N. F. Taussig
42.4k93254
42.4k93254
asked yesterday
Paul
1546
asked yesterday
Paul
1546
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
|
show 4 more comments
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
|
show 4 more comments
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 $boxed{x>1}$.
edited yesterday
vidyarthi
2,6881831
answered yesterday
greedoid
34.3k114488
so you reduced two squarings to just one! great answer
– vidyarthi
yesterday
add a comment |
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
answered yesterday
vidyarthi
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
add a comment |
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 $x>1$.
answered yesterday
Arash Rashidi
558
add a comment |
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 $x>0$:
$$
-2x < x-3 < 2x Leftrightarrow
begin{cases}
3x > 3 Leftrightarrow color{blue}{x> 1} \
x > -3 mbox{ does not extend the solution}
end{cases}
$$
answered yesterday
trancelocation
8,0741519
add a comment |
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 $boxed{x>1}$.
edited yesterday
vidyarthi
2,6881831
answered yesterday
greedoid
34.3k114488
so you reduced two squarings to just one! great answer
– vidyarthi
yesterday
add a comment |
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 $boxed{x>1}$.
edited yesterday
vidyarthi
2,6881831
answered yesterday
greedoid
34.3k114488
so you reduced two squarings to just one! great answer
– vidyarthi
yesterday
add a comment |
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 $boxed{x>1}$.
edited yesterday
vidyarthi
2,6881831
answered yesterday
greedoid
34.3k114488
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 $boxed{x>1}$.
edited yesterday
vidyarthi
2,6881831
answered yesterday
greedoid
34.3k114488
edited yesterday
vidyarthi
2,6881831
edited yesterday
vidyarthi
2,6881831
edited yesterday
vidyarthi
2,6881831
2,6881831
answered yesterday
greedoid
34.3k114488
answered yesterday
greedoid
34.3k114488
answered yesterday
greedoid
34.3k114488
34.3k114488
so you reduced two squarings to just one! great answer
– vidyarthi
yesterday
add a comment |
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
add a comment |
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
answered yesterday
vidyarthi
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
add a comment |
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
answered yesterday
vidyarthi
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
add a comment |
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
answered yesterday
vidyarthi
2,6881831
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
answered yesterday
vidyarthi
2,6881831
edited yesterday
answered yesterday
vidyarthi
2,6881831
answered yesterday
vidyarthi
2,6881831
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
add a comment |
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
add a comment |
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 $x>1$.
answered yesterday
Arash Rashidi
558
add a comment |
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 $x>1$.
answered yesterday
Arash Rashidi
558
add a comment |
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 $x>1$.
answered yesterday
Arash Rashidi
558
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 $x>1$.
answered yesterday
Arash Rashidi
558
answered yesterday
Arash Rashidi
558
answered yesterday
Arash Rashidi
558
answered yesterday
Arash Rashidi
558
558
add a comment |
add a comment |
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 $x>0$:
$$
-2x < x-3 < 2x Leftrightarrow
begin{cases}
3x > 3 Leftrightarrow color{blue}{x> 1} \
x > -3 mbox{ does not extend the solution}
end{cases}
$$
answered yesterday
trancelocation
8,0741519
add a comment |
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 $x>0$:
$$
-2x < x-3 < 2x Leftrightarrow
begin{cases}
3x > 3 Leftrightarrow color{blue}{x> 1} \
x > -3 mbox{ does not extend the solution}
end{cases}
$$
answered yesterday
trancelocation
8,0741519
add a comment |
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 $x>0$:
$$
-2x < x-3 < 2x Leftrightarrow
begin{cases}
3x > 3 Leftrightarrow color{blue}{x> 1} \
x > -3 mbox{ does not extend the solution}
end{cases}
$$
answered yesterday
trancelocation
8,0741519
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 $x>0$:
$$
-2x < x-3 < 2x Leftrightarrow
begin{cases}
3x > 3 Leftrightarrow color{blue}{x> 1} \
x > -3 mbox{ does not extend the solution}
end{cases}
$$
answered yesterday
trancelocation
8,0741519
answered yesterday
trancelocation
8,0741519
answered yesterday
trancelocation
8,0741519
answered yesterday
trancelocation
8,0741519
8,0741519
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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