Absolute Value Inequalities Analytical Approach
$begingroup$
For $|x-3|-|2x+1|<0$, I considered adding $|2x+1|$ on both sides and solving it.
$$|x-3|<|2x+1|$$
$$x-3< |2x+1|$$
$$x-3 < 2x+1$$
However, I keep getting $x>-4$ and $x<2/3$ instead of $x<-4$ and $x>2/3$. I know I can do a graphical approach and find the values, but why do I keep getting the wrong answer?
algebra-precalculus absolute-value
edited Dec 19 '18 at 1:15
Will Fisher
4,05311132
asked Dec 19 '18 at 1:09
anonymousanonymous
145
$endgroup$
add a comment |
$begingroup$
For $|x-3|-|2x+1|<0$, I considered adding $|2x+1|$ on both sides and solving it.
$$|x-3|<|2x+1|$$
$$x-3< |2x+1|$$
$$x-3 < 2x+1$$
However, I keep getting $x>-4$ and $x<2/3$ instead of $x<-4$ and $x>2/3$. I know I can do a graphical approach and find the values, but why do I keep getting the wrong answer?
algebra-precalculus absolute-value
edited Dec 19 '18 at 1:15
Will Fisher
4,05311132
asked Dec 19 '18 at 1:09
anonymousanonymous
145
$endgroup$
1
$begingroup$
How do you justify simply erasing the absolute-value signs?
$endgroup$
– Henning Makholm
Dec 19 '18 at 1:19
add a comment |
$begingroup$
For $|x-3|-|2x+1|<0$, I considered adding $|2x+1|$ on both sides and solving it.
$$|x-3|<|2x+1|$$
$$x-3< |2x+1|$$
$$x-3 < 2x+1$$
However, I keep getting $x>-4$ and $x<2/3$ instead of $x<-4$ and $x>2/3$. I know I can do a graphical approach and find the values, but why do I keep getting the wrong answer?
algebra-precalculus absolute-value
edited Dec 19 '18 at 1:15
Will Fisher
4,05311132
asked Dec 19 '18 at 1:09
anonymousanonymous
145
$endgroup$
For $|x-3|-|2x+1|<0$, I considered adding $|2x+1|$ on both sides and solving it.
$$|x-3|<|2x+1|$$
$$x-3< |2x+1|$$
$$x-3 < 2x+1$$
However, I keep getting $x>-4$ and $x<2/3$ instead of $x<-4$ and $x>2/3$. I know I can do a graphical approach and find the values, but why do I keep getting the wrong answer?
algebra-precalculus absolute-value
algebra-precalculus absolute-value
edited Dec 19 '18 at 1:15
Will Fisher
4,05311132
asked Dec 19 '18 at 1:09
anonymousanonymous
145
edited Dec 19 '18 at 1:15
Will Fisher
4,05311132
asked Dec 19 '18 at 1:09
anonymousanonymous
145
edited Dec 19 '18 at 1:15
Will Fisher
4,05311132
edited Dec 19 '18 at 1:15
Will Fisher
4,05311132
edited Dec 19 '18 at 1:15
Will Fisher
4,05311132
4,05311132
asked Dec 19 '18 at 1:09
anonymousanonymous
145
asked Dec 19 '18 at 1:09
anonymousanonymous
145
asked Dec 19 '18 at 1:09
anonymousanonymous
145
145
1
$begingroup$
How do you justify simply erasing the absolute-value signs?
$endgroup$
– Henning Makholm
Dec 19 '18 at 1:19
add a comment |
1
$begingroup$
How do you justify simply erasing the absolute-value signs?
$endgroup$
– Henning Makholm
Dec 19 '18 at 1:19
1
1
$begingroup$
How do you justify simply erasing the absolute-value signs?
$endgroup$
– Henning Makholm
Dec 19 '18 at 1:19
$begingroup$
How do you justify simply erasing the absolute-value signs?
$endgroup$
– Henning Makholm
Dec 19 '18 at 1:19
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Analytically you may also proceed as follows:
begin{eqnarray*} |x-3| < |2x+1|
& Leftrightarrow & (x-3)^2 < (2x+1)^2 \
& Leftrightarrow & 0 < (2x+1)^2 -(x-3)^2 stackrel{a^2-b^2 =(a-b)(a+b)}{=} (2x+1 - (x-3))(2x+1 + (x-3))\
& Leftrightarrow & 0 < (x+4)(3x-2) \
& Leftrightarrow & boxed{x< -4 mbox{ or } x>frac{2}{3}}\
end{eqnarray*}
answered Dec 19 '18 at 2:59
trancelocationtrancelocation
13.4k1827
$endgroup$
add a comment |
$begingroup$
You have $|x-3| < |2x+1|$. Now there are 4 possible cases:
1) $x-3ge 0$ and $2x + 1 ge 0$.
This means:
a) $x - 3 ge 0$ so $x ge 3$.
b) $2x + 1 ge 0$ so $x ge -frac 12$
c) $x - 3 < 2x + 1$ so $-4 < x$ or $x > -4$.
Combining we get $x ge 3$.
2) $x-3 < 0$ and $2x +1 ge 0$.
This means
a) $x - 3 < 0$ so $x < 3$
b) $2x + 1 ge 0$ so $x ge -frac 12$
c) $3-x < 2x + 1$ so $2 < 3x$ and $x > frac 23$
Combining we get $frac 23 < x < 3$.
Case 3) $x-3 ge 0$ and $2x + 1 < 0$ then
a) $x -3 ge 0$ and $x > 3$
b) $2x + 1 < 0$ and $x < -frac 12$
c) $x-3 < -1-2x$ and $3x < 2$ and $x < frac 23$.
Combining we get contradictions. This is not possible.
Case 4) $x-3 < 0$ and $2x + 1 < 0$ then
a) $x - 3 < 0 $ so x < 3$
b) $2x +1< 0$ so x < -frac 12$.
c) $3-x < -2x - 1$ so $x < -4$
Combining we get $x < -4$.
So we have 3 possible results $x ge 3$ or $frac 23 < x < 3$ or $x < -4$. Combining we get either $x < -4$ of $x > frac 23$.
answered Dec 19 '18 at 1:31
fleabloodfleablood
73.7k22891
$endgroup$
$begingroup$
If $3 - x < 2x + 1$, then $2 < 3x implies x > frac{2}{3}$.
$endgroup$
– N. F. Taussig
Dec 19 '18 at 10:24
$begingroup$
Dang. Arithmetic in your head is hard .....
$endgroup$
– fleablood
Dec 19 '18 at 17:18
$begingroup$
Yee gods! I made at least 3 errors and a totally wrong answer! Still, I think cases are the most straight forward way to learn it.
$endgroup$
– fleablood
Dec 19 '18 at 17:43
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
Analytically you may also proceed as follows:
begin{eqnarray*} |x-3| < |2x+1|
& Leftrightarrow & (x-3)^2 < (2x+1)^2 \
& Leftrightarrow & 0 < (2x+1)^2 -(x-3)^2 stackrel{a^2-b^2 =(a-b)(a+b)}{=} (2x+1 - (x-3))(2x+1 + (x-3))\
& Leftrightarrow & 0 < (x+4)(3x-2) \
& Leftrightarrow & boxed{x< -4 mbox{ or } x>frac{2}{3}}\
end{eqnarray*}
answered Dec 19 '18 at 2:59
trancelocationtrancelocation
13.4k1827
$endgroup$
add a comment |
$begingroup$
Analytically you may also proceed as follows:
begin{eqnarray*} |x-3| < |2x+1|
& Leftrightarrow & (x-3)^2 < (2x+1)^2 \
& Leftrightarrow & 0 < (2x+1)^2 -(x-3)^2 stackrel{a^2-b^2 =(a-b)(a+b)}{=} (2x+1 - (x-3))(2x+1 + (x-3))\
& Leftrightarrow & 0 < (x+4)(3x-2) \
& Leftrightarrow & boxed{x< -4 mbox{ or } x>frac{2}{3}}\
end{eqnarray*}
answered Dec 19 '18 at 2:59
trancelocationtrancelocation
13.4k1827
$endgroup$
add a comment |
$begingroup$
Analytically you may also proceed as follows:
begin{eqnarray*} |x-3| < |2x+1|
& Leftrightarrow & (x-3)^2 < (2x+1)^2 \
& Leftrightarrow & 0 < (2x+1)^2 -(x-3)^2 stackrel{a^2-b^2 =(a-b)(a+b)}{=} (2x+1 - (x-3))(2x+1 + (x-3))\
& Leftrightarrow & 0 < (x+4)(3x-2) \
& Leftrightarrow & boxed{x< -4 mbox{ or } x>frac{2}{3}}\
end{eqnarray*}
answered Dec 19 '18 at 2:59
trancelocationtrancelocation
13.4k1827
$endgroup$
Analytically you may also proceed as follows:
begin{eqnarray*} |x-3| < |2x+1|
& Leftrightarrow & (x-3)^2 < (2x+1)^2 \
& Leftrightarrow & 0 < (2x+1)^2 -(x-3)^2 stackrel{a^2-b^2 =(a-b)(a+b)}{=} (2x+1 - (x-3))(2x+1 + (x-3))\
& Leftrightarrow & 0 < (x+4)(3x-2) \
& Leftrightarrow & boxed{x< -4 mbox{ or } x>frac{2}{3}}\
end{eqnarray*}
answered Dec 19 '18 at 2:59
trancelocationtrancelocation
13.4k1827
answered Dec 19 '18 at 2:59
trancelocationtrancelocation
13.4k1827
answered Dec 19 '18 at 2:59
trancelocationtrancelocation
13.4k1827
answered Dec 19 '18 at 2:59
trancelocationtrancelocation
13.4k1827
13.4k1827
add a comment |
add a comment |
$begingroup$
You have $|x-3| < |2x+1|$. Now there are 4 possible cases:
1) $x-3ge 0$ and $2x + 1 ge 0$.
This means:
a) $x - 3 ge 0$ so $x ge 3$.
b) $2x + 1 ge 0$ so $x ge -frac 12$
c) $x - 3 < 2x + 1$ so $-4 < x$ or $x > -4$.
Combining we get $x ge 3$.
2) $x-3 < 0$ and $2x +1 ge 0$.
This means
a) $x - 3 < 0$ so $x < 3$
b) $2x + 1 ge 0$ so $x ge -frac 12$
c) $3-x < 2x + 1$ so $2 < 3x$ and $x > frac 23$
Combining we get $frac 23 < x < 3$.
Case 3) $x-3 ge 0$ and $2x + 1 < 0$ then
a) $x -3 ge 0$ and $x > 3$
b) $2x + 1 < 0$ and $x < -frac 12$
c) $x-3 < -1-2x$ and $3x < 2$ and $x < frac 23$.
Combining we get contradictions. This is not possible.
Case 4) $x-3 < 0$ and $2x + 1 < 0$ then
a) $x - 3 < 0 $ so x < 3$
b) $2x +1< 0$ so x < -frac 12$.
c) $3-x < -2x - 1$ so $x < -4$
Combining we get $x < -4$.
So we have 3 possible results $x ge 3$ or $frac 23 < x < 3$ or $x < -4$. Combining we get either $x < -4$ of $x > frac 23$.
answered Dec 19 '18 at 1:31
fleabloodfleablood
73.7k22891
$endgroup$
$begingroup$
If $3 - x < 2x + 1$, then $2 < 3x implies x > frac{2}{3}$.
$endgroup$
– N. F. Taussig
Dec 19 '18 at 10:24
$begingroup$
Dang. Arithmetic in your head is hard .....
$endgroup$
– fleablood
Dec 19 '18 at 17:18
$begingroup$
Yee gods! I made at least 3 errors and a totally wrong answer! Still, I think cases are the most straight forward way to learn it.
$endgroup$
– fleablood
Dec 19 '18 at 17:43
add a comment |
$begingroup$
You have $|x-3| < |2x+1|$. Now there are 4 possible cases:
1) $x-3ge 0$ and $2x + 1 ge 0$.
This means:
a) $x - 3 ge 0$ so $x ge 3$.
b) $2x + 1 ge 0$ so $x ge -frac 12$
c) $x - 3 < 2x + 1$ so $-4 < x$ or $x > -4$.
Combining we get $x ge 3$.
2) $x-3 < 0$ and $2x +1 ge 0$.
This means
a) $x - 3 < 0$ so $x < 3$
b) $2x + 1 ge 0$ so $x ge -frac 12$
c) $3-x < 2x + 1$ so $2 < 3x$ and $x > frac 23$
Combining we get $frac 23 < x < 3$.
Case 3) $x-3 ge 0$ and $2x + 1 < 0$ then
a) $x -3 ge 0$ and $x > 3$
b) $2x + 1 < 0$ and $x < -frac 12$
c) $x-3 < -1-2x$ and $3x < 2$ and $x < frac 23$.
Combining we get contradictions. This is not possible.
Case 4) $x-3 < 0$ and $2x + 1 < 0$ then
a) $x - 3 < 0 $ so x < 3$
b) $2x +1< 0$ so x < -frac 12$.
c) $3-x < -2x - 1$ so $x < -4$
Combining we get $x < -4$.
So we have 3 possible results $x ge 3$ or $frac 23 < x < 3$ or $x < -4$. Combining we get either $x < -4$ of $x > frac 23$.
answered Dec 19 '18 at 1:31
fleabloodfleablood
73.7k22891
$endgroup$
$begingroup$
If $3 - x < 2x + 1$, then $2 < 3x implies x > frac{2}{3}$.
$endgroup$
– N. F. Taussig
Dec 19 '18 at 10:24
$begingroup$
Dang. Arithmetic in your head is hard .....
$endgroup$
– fleablood
Dec 19 '18 at 17:18
$begingroup$
Yee gods! I made at least 3 errors and a totally wrong answer! Still, I think cases are the most straight forward way to learn it.
$endgroup$
– fleablood
Dec 19 '18 at 17:43
add a comment |
$begingroup$
You have $|x-3| < |2x+1|$. Now there are 4 possible cases:
1) $x-3ge 0$ and $2x + 1 ge 0$.
This means:
a) $x - 3 ge 0$ so $x ge 3$.
b) $2x + 1 ge 0$ so $x ge -frac 12$
c) $x - 3 < 2x + 1$ so $-4 < x$ or $x > -4$.
Combining we get $x ge 3$.
2) $x-3 < 0$ and $2x +1 ge 0$.
This means
a) $x - 3 < 0$ so $x < 3$
b) $2x + 1 ge 0$ so $x ge -frac 12$
c) $3-x < 2x + 1$ so $2 < 3x$ and $x > frac 23$
Combining we get $frac 23 < x < 3$.
Case 3) $x-3 ge 0$ and $2x + 1 < 0$ then
a) $x -3 ge 0$ and $x > 3$
b) $2x + 1 < 0$ and $x < -frac 12$
c) $x-3 < -1-2x$ and $3x < 2$ and $x < frac 23$.
Combining we get contradictions. This is not possible.
Case 4) $x-3 < 0$ and $2x + 1 < 0$ then
a) $x - 3 < 0 $ so x < 3$
b) $2x +1< 0$ so x < -frac 12$.
c) $3-x < -2x - 1$ so $x < -4$
Combining we get $x < -4$.
So we have 3 possible results $x ge 3$ or $frac 23 < x < 3$ or $x < -4$. Combining we get either $x < -4$ of $x > frac 23$.
answered Dec 19 '18 at 1:31
fleabloodfleablood
73.7k22891
$endgroup$
You have $|x-3| < |2x+1|$. Now there are 4 possible cases:
1) $x-3ge 0$ and $2x + 1 ge 0$.
This means:
a) $x - 3 ge 0$ so $x ge 3$.
b) $2x + 1 ge 0$ so $x ge -frac 12$
c) $x - 3 < 2x + 1$ so $-4 < x$ or $x > -4$.
Combining we get $x ge 3$.
2) $x-3 < 0$ and $2x +1 ge 0$.
This means
a) $x - 3 < 0$ so $x < 3$
b) $2x + 1 ge 0$ so $x ge -frac 12$
c) $3-x < 2x + 1$ so $2 < 3x$ and $x > frac 23$
Combining we get $frac 23 < x < 3$.
Case 3) $x-3 ge 0$ and $2x + 1 < 0$ then
a) $x -3 ge 0$ and $x > 3$
b) $2x + 1 < 0$ and $x < -frac 12$
c) $x-3 < -1-2x$ and $3x < 2$ and $x < frac 23$.
Combining we get contradictions. This is not possible.
Case 4) $x-3 < 0$ and $2x + 1 < 0$ then
a) $x - 3 < 0 $ so x < 3$
b) $2x +1< 0$ so x < -frac 12$.
c) $3-x < -2x - 1$ so $x < -4$
Combining we get $x < -4$.
So we have 3 possible results $x ge 3$ or $frac 23 < x < 3$ or $x < -4$. Combining we get either $x < -4$ of $x > frac 23$.
answered Dec 19 '18 at 1:31
fleabloodfleablood
73.7k22891
edited Dec 19 '18 at 17:42
answered Dec 19 '18 at 1:31
fleabloodfleablood
73.7k22891
answered Dec 19 '18 at 1:31
fleabloodfleablood
73.7k22891
answered Dec 19 '18 at 1:31
fleabloodfleablood
73.7k22891
73.7k22891
$begingroup$
If $3 - x < 2x + 1$, then $2 < 3x implies x > frac{2}{3}$.
$endgroup$
– N. F. Taussig
Dec 19 '18 at 10:24
$begingroup$
Dang. Arithmetic in your head is hard .....
$endgroup$
– fleablood
Dec 19 '18 at 17:18
$begingroup$
Yee gods! I made at least 3 errors and a totally wrong answer! Still, I think cases are the most straight forward way to learn it.
$endgroup$
– fleablood
Dec 19 '18 at 17:43
add a comment |
$begingroup$
If $3 - x < 2x + 1$, then $2 < 3x implies x > frac{2}{3}$.
$endgroup$
– N. F. Taussig
Dec 19 '18 at 10:24
$begingroup$
Dang. Arithmetic in your head is hard .....
$endgroup$
– fleablood
Dec 19 '18 at 17:18
$begingroup$
Yee gods! I made at least 3 errors and a totally wrong answer! Still, I think cases are the most straight forward way to learn it.
$endgroup$
– fleablood
Dec 19 '18 at 17:43
$begingroup$
If $3 - x < 2x + 1$, then $2 < 3x implies x > frac{2}{3}$.
$endgroup$
– N. F. Taussig
Dec 19 '18 at 10:24
$begingroup$
If $3 - x < 2x + 1$, then $2 < 3x implies x > frac{2}{3}$.
$endgroup$
– N. F. Taussig
Dec 19 '18 at 10:24
$begingroup$
Dang. Arithmetic in your head is hard .....
$endgroup$
– fleablood
Dec 19 '18 at 17:18
$begingroup$
Dang. Arithmetic in your head is hard .....
$endgroup$
– fleablood
Dec 19 '18 at 17:18
$begingroup$
Yee gods! I made at least 3 errors and a totally wrong answer! Still, I think cases are the most straight forward way to learn it.
$endgroup$
– fleablood
Dec 19 '18 at 17:43
$begingroup$
Yee gods! I made at least 3 errors and a totally wrong answer! Still, I think cases are the most straight forward way to learn it.
$endgroup$
– fleablood
Dec 19 '18 at 17:43
add a comment |
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$begingroup$
How do you justify simply erasing the absolute-value signs?
$endgroup$
– Henning Makholm
Dec 19 '18 at 1:19