Solve a system of three inequalities in three unknowns
$begingroup$
I have been trying to solve the system of three inequalities. I could not manage to solve it. Can somebody help me?
$$ X gt 1 tag{1} $$
$$ Y lt Zcdot (1+X) tag{2} $$
$$ Z lt frac{1-Y}{X} tag{3} $$
inequality
$endgroup$
add a comment |
$begingroup$
I have been trying to solve the system of three inequalities. I could not manage to solve it. Can somebody help me?
$$ X gt 1 tag{1} $$
$$ Y lt Zcdot (1+X) tag{2} $$
$$ Z lt frac{1-Y}{X} tag{3} $$
inequality
$endgroup$
$begingroup$
Have been trying the whole day to solve it. I could not. I also tried matlab to draw in 3D to see the region that represent the solution
$endgroup$
– Yesh
Oct 14 '15 at 16:05
$begingroup$
With a bit of practice you can use mathematical notation in your posts with MathJax and $LaTeX$. I will edit your post to give you an idea how it works.
$endgroup$
– hardmath
Oct 29 '15 at 9:42
$begingroup$
Try substituting one inequality into another. Try showing that $Y<2Z$. Can you do that? It should help you get part of the way to the answer.
$endgroup$
– Cataline
Oct 29 '15 at 10:51
add a comment |
$begingroup$
I have been trying to solve the system of three inequalities. I could not manage to solve it. Can somebody help me?
$$ X gt 1 tag{1} $$
$$ Y lt Zcdot (1+X) tag{2} $$
$$ Z lt frac{1-Y}{X} tag{3} $$
inequality
$endgroup$
I have been trying to solve the system of three inequalities. I could not manage to solve it. Can somebody help me?
$$ X gt 1 tag{1} $$
$$ Y lt Zcdot (1+X) tag{2} $$
$$ Z lt frac{1-Y}{X} tag{3} $$
inequality
inequality
edited Oct 29 '15 at 9:58
hardmath
29.3k953101
29.3k953101
asked Oct 14 '15 at 14:26
YeshYesh
112
112
$begingroup$
Have been trying the whole day to solve it. I could not. I also tried matlab to draw in 3D to see the region that represent the solution
$endgroup$
– Yesh
Oct 14 '15 at 16:05
$begingroup$
With a bit of practice you can use mathematical notation in your posts with MathJax and $LaTeX$. I will edit your post to give you an idea how it works.
$endgroup$
– hardmath
Oct 29 '15 at 9:42
$begingroup$
Try substituting one inequality into another. Try showing that $Y<2Z$. Can you do that? It should help you get part of the way to the answer.
$endgroup$
– Cataline
Oct 29 '15 at 10:51
add a comment |
$begingroup$
Have been trying the whole day to solve it. I could not. I also tried matlab to draw in 3D to see the region that represent the solution
$endgroup$
– Yesh
Oct 14 '15 at 16:05
$begingroup$
With a bit of practice you can use mathematical notation in your posts with MathJax and $LaTeX$. I will edit your post to give you an idea how it works.
$endgroup$
– hardmath
Oct 29 '15 at 9:42
$begingroup$
Try substituting one inequality into another. Try showing that $Y<2Z$. Can you do that? It should help you get part of the way to the answer.
$endgroup$
– Cataline
Oct 29 '15 at 10:51
$begingroup$
Have been trying the whole day to solve it. I could not. I also tried matlab to draw in 3D to see the region that represent the solution
$endgroup$
– Yesh
Oct 14 '15 at 16:05
$begingroup$
Have been trying the whole day to solve it. I could not. I also tried matlab to draw in 3D to see the region that represent the solution
$endgroup$
– Yesh
Oct 14 '15 at 16:05
$begingroup$
With a bit of practice you can use mathematical notation in your posts with MathJax and $LaTeX$. I will edit your post to give you an idea how it works.
$endgroup$
– hardmath
Oct 29 '15 at 9:42
$begingroup$
With a bit of practice you can use mathematical notation in your posts with MathJax and $LaTeX$. I will edit your post to give you an idea how it works.
$endgroup$
– hardmath
Oct 29 '15 at 9:42
$begingroup$
Try substituting one inequality into another. Try showing that $Y<2Z$. Can you do that? It should help you get part of the way to the answer.
$endgroup$
– Cataline
Oct 29 '15 at 10:51
$begingroup$
Try substituting one inequality into another. Try showing that $Y<2Z$. Can you do that? It should help you get part of the way to the answer.
$endgroup$
– Cataline
Oct 29 '15 at 10:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Typically the set of solutions to a system of inequalities will be one or more regions in the space corresponding to the unknowns (here three-dimensional space, for three unknowns).
Working with rational expressions can be especially tricky, since our impulse to multiply both sides to clear fractions is apt to affect the direction of an inequality depending on the sign of what we multiply by.
Here we know $X gt 1$, so in particular we can multiply by $X$ without changing the direction of any inequality (since $X$ is positive). The system is then equivalent to this:
$$ X gt 1 tag{1} $$
$$ Y lt ZX + Z tag{2} $$
$$ ZX lt 1 - Y tag{3} $$
It is tempting to multiply the first inequality by $Z$, since then we would be able to replace the original system of nonlinear inequalities by one of strictly linear inequalities, the solution of which is pretty well understood (the subject of linear program feasibility, more or less).
But we don't know in advance whether $Z$ is positive, negative, or zero. We can resort to considering three separate cases.
Case: Z is zero
If $Z=0$, then the second and third inequalities simplify quite a bit:
$$ Y lt 0 tag{2'} $$
$$ 0 lt 1 - Y tag{3'} $$
Now it $Y lt 0$ (rewritten second inequality), then automatically $0 lt 1-Y$ (rewritten third inequality). So the case $Z=0$ amounts to a solution set:
$$ {(X,Y,Z) mid Xgt 1, Ylt 0, Z = 0 } tag{Z=0} $$
Case: Z is positive
If $Z gt 0$, then we could multiply the first inequality on both sides by $Z$ without changing the direction of the inequality. We would then have the equivalent system:
$$ ZX gt Z tag{1} $$
$$ Y lt ZX + Z tag{2} $$
$$ ZX lt 1 - Y tag{3} $$
If we substitute $W = ZX$, then the nature of the equivalent system as linear inequalities will be clear:
$$ W gt Z gt 0 tag{1''} $$
$$ Y lt W + Z tag{2''} $$
$$ W lt 1 - Y tag{3''} $$
The solutions of the first inequality are easy to visualize. In the $W,Z$-plane it consists of all the points in the (strict) first quadrant ($W,Z gt 0$) such that $W gt Z$, so points $(W,Z)$ falling to the right of the diagonal line $W=Z$.
The second and third inequalities then both amount to upper bounds on $Y$. That is, not only do we have $Y lt W + Z$, but also (rewriting the third inequality) $Y lt 1 - W$. So the solutions will be (in terms of $W,Y,Z$) determined by:
$$ Y lt min(W+Z,1-W) $$
Translating this result back into terms of $X = W/Z$, we have the solution set for the case that $Z$ is positive:
$$ {(X,Y,Z) mid X gt 1, Z gt 0, Y lt min(Z(1+X),1-ZX); } tag{Z>0}$$
More can be done to visualize this region. When is $Z(1+X) le 1-ZX$, and when is the reverse true? For fixed $Zgt 0$, the quantity $Z(1+X)$ increases as $X$ increases while $1-ZX$ decreases with increasing $X$. So for sufficiently large $X$, the smaller of these will be $1-ZX$.
If we work out where the two expressions are equal, $Z(1+X) = 1-ZX$, we find the crossover occurs at:
$$ X_* = frac{1-Z}{2Z} $$
Since $Xgt 1$, the value $X ge X_*$ always holds when $Z ge 1/3$ (which implies $X_* le 1$), and thus $Y lt 1 -ZX$ is the limiting constraint.
For $0 lt Z lt 1/3$, there are two cases. If $1lt X lt X_*$, then $Y lt Z(1+X)$ is the limiting constraint, and otherwise ($X ge X_*$) the limiting constraint is $Y lt 1 - ZX$.
Case: Z is negative
Left as an exercise for the Reader. Note that the first inequality becomes $ZX < Z$ because the direction of the inequality is reversed.
$endgroup$
1
$begingroup$
Thank you Hardmath. Case by case derivation is the way to solve it. Drawing the systems of inequalities in matlab I found two solution regions.
$endgroup$
– Yesh
Oct 30 '15 at 13:56
add a comment |
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$begingroup$
Typically the set of solutions to a system of inequalities will be one or more regions in the space corresponding to the unknowns (here three-dimensional space, for three unknowns).
Working with rational expressions can be especially tricky, since our impulse to multiply both sides to clear fractions is apt to affect the direction of an inequality depending on the sign of what we multiply by.
Here we know $X gt 1$, so in particular we can multiply by $X$ without changing the direction of any inequality (since $X$ is positive). The system is then equivalent to this:
$$ X gt 1 tag{1} $$
$$ Y lt ZX + Z tag{2} $$
$$ ZX lt 1 - Y tag{3} $$
It is tempting to multiply the first inequality by $Z$, since then we would be able to replace the original system of nonlinear inequalities by one of strictly linear inequalities, the solution of which is pretty well understood (the subject of linear program feasibility, more or less).
But we don't know in advance whether $Z$ is positive, negative, or zero. We can resort to considering three separate cases.
Case: Z is zero
If $Z=0$, then the second and third inequalities simplify quite a bit:
$$ Y lt 0 tag{2'} $$
$$ 0 lt 1 - Y tag{3'} $$
Now it $Y lt 0$ (rewritten second inequality), then automatically $0 lt 1-Y$ (rewritten third inequality). So the case $Z=0$ amounts to a solution set:
$$ {(X,Y,Z) mid Xgt 1, Ylt 0, Z = 0 } tag{Z=0} $$
Case: Z is positive
If $Z gt 0$, then we could multiply the first inequality on both sides by $Z$ without changing the direction of the inequality. We would then have the equivalent system:
$$ ZX gt Z tag{1} $$
$$ Y lt ZX + Z tag{2} $$
$$ ZX lt 1 - Y tag{3} $$
If we substitute $W = ZX$, then the nature of the equivalent system as linear inequalities will be clear:
$$ W gt Z gt 0 tag{1''} $$
$$ Y lt W + Z tag{2''} $$
$$ W lt 1 - Y tag{3''} $$
The solutions of the first inequality are easy to visualize. In the $W,Z$-plane it consists of all the points in the (strict) first quadrant ($W,Z gt 0$) such that $W gt Z$, so points $(W,Z)$ falling to the right of the diagonal line $W=Z$.
The second and third inequalities then both amount to upper bounds on $Y$. That is, not only do we have $Y lt W + Z$, but also (rewriting the third inequality) $Y lt 1 - W$. So the solutions will be (in terms of $W,Y,Z$) determined by:
$$ Y lt min(W+Z,1-W) $$
Translating this result back into terms of $X = W/Z$, we have the solution set for the case that $Z$ is positive:
$$ {(X,Y,Z) mid X gt 1, Z gt 0, Y lt min(Z(1+X),1-ZX); } tag{Z>0}$$
More can be done to visualize this region. When is $Z(1+X) le 1-ZX$, and when is the reverse true? For fixed $Zgt 0$, the quantity $Z(1+X)$ increases as $X$ increases while $1-ZX$ decreases with increasing $X$. So for sufficiently large $X$, the smaller of these will be $1-ZX$.
If we work out where the two expressions are equal, $Z(1+X) = 1-ZX$, we find the crossover occurs at:
$$ X_* = frac{1-Z}{2Z} $$
Since $Xgt 1$, the value $X ge X_*$ always holds when $Z ge 1/3$ (which implies $X_* le 1$), and thus $Y lt 1 -ZX$ is the limiting constraint.
For $0 lt Z lt 1/3$, there are two cases. If $1lt X lt X_*$, then $Y lt Z(1+X)$ is the limiting constraint, and otherwise ($X ge X_*$) the limiting constraint is $Y lt 1 - ZX$.
Case: Z is negative
Left as an exercise for the Reader. Note that the first inequality becomes $ZX < Z$ because the direction of the inequality is reversed.
$endgroup$
1
$begingroup$
Thank you Hardmath. Case by case derivation is the way to solve it. Drawing the systems of inequalities in matlab I found two solution regions.
$endgroup$
– Yesh
Oct 30 '15 at 13:56
add a comment |
$begingroup$
Typically the set of solutions to a system of inequalities will be one or more regions in the space corresponding to the unknowns (here three-dimensional space, for three unknowns).
Working with rational expressions can be especially tricky, since our impulse to multiply both sides to clear fractions is apt to affect the direction of an inequality depending on the sign of what we multiply by.
Here we know $X gt 1$, so in particular we can multiply by $X$ without changing the direction of any inequality (since $X$ is positive). The system is then equivalent to this:
$$ X gt 1 tag{1} $$
$$ Y lt ZX + Z tag{2} $$
$$ ZX lt 1 - Y tag{3} $$
It is tempting to multiply the first inequality by $Z$, since then we would be able to replace the original system of nonlinear inequalities by one of strictly linear inequalities, the solution of which is pretty well understood (the subject of linear program feasibility, more or less).
But we don't know in advance whether $Z$ is positive, negative, or zero. We can resort to considering three separate cases.
Case: Z is zero
If $Z=0$, then the second and third inequalities simplify quite a bit:
$$ Y lt 0 tag{2'} $$
$$ 0 lt 1 - Y tag{3'} $$
Now it $Y lt 0$ (rewritten second inequality), then automatically $0 lt 1-Y$ (rewritten third inequality). So the case $Z=0$ amounts to a solution set:
$$ {(X,Y,Z) mid Xgt 1, Ylt 0, Z = 0 } tag{Z=0} $$
Case: Z is positive
If $Z gt 0$, then we could multiply the first inequality on both sides by $Z$ without changing the direction of the inequality. We would then have the equivalent system:
$$ ZX gt Z tag{1} $$
$$ Y lt ZX + Z tag{2} $$
$$ ZX lt 1 - Y tag{3} $$
If we substitute $W = ZX$, then the nature of the equivalent system as linear inequalities will be clear:
$$ W gt Z gt 0 tag{1''} $$
$$ Y lt W + Z tag{2''} $$
$$ W lt 1 - Y tag{3''} $$
The solutions of the first inequality are easy to visualize. In the $W,Z$-plane it consists of all the points in the (strict) first quadrant ($W,Z gt 0$) such that $W gt Z$, so points $(W,Z)$ falling to the right of the diagonal line $W=Z$.
The second and third inequalities then both amount to upper bounds on $Y$. That is, not only do we have $Y lt W + Z$, but also (rewriting the third inequality) $Y lt 1 - W$. So the solutions will be (in terms of $W,Y,Z$) determined by:
$$ Y lt min(W+Z,1-W) $$
Translating this result back into terms of $X = W/Z$, we have the solution set for the case that $Z$ is positive:
$$ {(X,Y,Z) mid X gt 1, Z gt 0, Y lt min(Z(1+X),1-ZX); } tag{Z>0}$$
More can be done to visualize this region. When is $Z(1+X) le 1-ZX$, and when is the reverse true? For fixed $Zgt 0$, the quantity $Z(1+X)$ increases as $X$ increases while $1-ZX$ decreases with increasing $X$. So for sufficiently large $X$, the smaller of these will be $1-ZX$.
If we work out where the two expressions are equal, $Z(1+X) = 1-ZX$, we find the crossover occurs at:
$$ X_* = frac{1-Z}{2Z} $$
Since $Xgt 1$, the value $X ge X_*$ always holds when $Z ge 1/3$ (which implies $X_* le 1$), and thus $Y lt 1 -ZX$ is the limiting constraint.
For $0 lt Z lt 1/3$, there are two cases. If $1lt X lt X_*$, then $Y lt Z(1+X)$ is the limiting constraint, and otherwise ($X ge X_*$) the limiting constraint is $Y lt 1 - ZX$.
Case: Z is negative
Left as an exercise for the Reader. Note that the first inequality becomes $ZX < Z$ because the direction of the inequality is reversed.
$endgroup$
1
$begingroup$
Thank you Hardmath. Case by case derivation is the way to solve it. Drawing the systems of inequalities in matlab I found two solution regions.
$endgroup$
– Yesh
Oct 30 '15 at 13:56
add a comment |
$begingroup$
Typically the set of solutions to a system of inequalities will be one or more regions in the space corresponding to the unknowns (here three-dimensional space, for three unknowns).
Working with rational expressions can be especially tricky, since our impulse to multiply both sides to clear fractions is apt to affect the direction of an inequality depending on the sign of what we multiply by.
Here we know $X gt 1$, so in particular we can multiply by $X$ without changing the direction of any inequality (since $X$ is positive). The system is then equivalent to this:
$$ X gt 1 tag{1} $$
$$ Y lt ZX + Z tag{2} $$
$$ ZX lt 1 - Y tag{3} $$
It is tempting to multiply the first inequality by $Z$, since then we would be able to replace the original system of nonlinear inequalities by one of strictly linear inequalities, the solution of which is pretty well understood (the subject of linear program feasibility, more or less).
But we don't know in advance whether $Z$ is positive, negative, or zero. We can resort to considering three separate cases.
Case: Z is zero
If $Z=0$, then the second and third inequalities simplify quite a bit:
$$ Y lt 0 tag{2'} $$
$$ 0 lt 1 - Y tag{3'} $$
Now it $Y lt 0$ (rewritten second inequality), then automatically $0 lt 1-Y$ (rewritten third inequality). So the case $Z=0$ amounts to a solution set:
$$ {(X,Y,Z) mid Xgt 1, Ylt 0, Z = 0 } tag{Z=0} $$
Case: Z is positive
If $Z gt 0$, then we could multiply the first inequality on both sides by $Z$ without changing the direction of the inequality. We would then have the equivalent system:
$$ ZX gt Z tag{1} $$
$$ Y lt ZX + Z tag{2} $$
$$ ZX lt 1 - Y tag{3} $$
If we substitute $W = ZX$, then the nature of the equivalent system as linear inequalities will be clear:
$$ W gt Z gt 0 tag{1''} $$
$$ Y lt W + Z tag{2''} $$
$$ W lt 1 - Y tag{3''} $$
The solutions of the first inequality are easy to visualize. In the $W,Z$-plane it consists of all the points in the (strict) first quadrant ($W,Z gt 0$) such that $W gt Z$, so points $(W,Z)$ falling to the right of the diagonal line $W=Z$.
The second and third inequalities then both amount to upper bounds on $Y$. That is, not only do we have $Y lt W + Z$, but also (rewriting the third inequality) $Y lt 1 - W$. So the solutions will be (in terms of $W,Y,Z$) determined by:
$$ Y lt min(W+Z,1-W) $$
Translating this result back into terms of $X = W/Z$, we have the solution set for the case that $Z$ is positive:
$$ {(X,Y,Z) mid X gt 1, Z gt 0, Y lt min(Z(1+X),1-ZX); } tag{Z>0}$$
More can be done to visualize this region. When is $Z(1+X) le 1-ZX$, and when is the reverse true? For fixed $Zgt 0$, the quantity $Z(1+X)$ increases as $X$ increases while $1-ZX$ decreases with increasing $X$. So for sufficiently large $X$, the smaller of these will be $1-ZX$.
If we work out where the two expressions are equal, $Z(1+X) = 1-ZX$, we find the crossover occurs at:
$$ X_* = frac{1-Z}{2Z} $$
Since $Xgt 1$, the value $X ge X_*$ always holds when $Z ge 1/3$ (which implies $X_* le 1$), and thus $Y lt 1 -ZX$ is the limiting constraint.
For $0 lt Z lt 1/3$, there are two cases. If $1lt X lt X_*$, then $Y lt Z(1+X)$ is the limiting constraint, and otherwise ($X ge X_*$) the limiting constraint is $Y lt 1 - ZX$.
Case: Z is negative
Left as an exercise for the Reader. Note that the first inequality becomes $ZX < Z$ because the direction of the inequality is reversed.
$endgroup$
Typically the set of solutions to a system of inequalities will be one or more regions in the space corresponding to the unknowns (here three-dimensional space, for three unknowns).
Working with rational expressions can be especially tricky, since our impulse to multiply both sides to clear fractions is apt to affect the direction of an inequality depending on the sign of what we multiply by.
Here we know $X gt 1$, so in particular we can multiply by $X$ without changing the direction of any inequality (since $X$ is positive). The system is then equivalent to this:
$$ X gt 1 tag{1} $$
$$ Y lt ZX + Z tag{2} $$
$$ ZX lt 1 - Y tag{3} $$
It is tempting to multiply the first inequality by $Z$, since then we would be able to replace the original system of nonlinear inequalities by one of strictly linear inequalities, the solution of which is pretty well understood (the subject of linear program feasibility, more or less).
But we don't know in advance whether $Z$ is positive, negative, or zero. We can resort to considering three separate cases.
Case: Z is zero
If $Z=0$, then the second and third inequalities simplify quite a bit:
$$ Y lt 0 tag{2'} $$
$$ 0 lt 1 - Y tag{3'} $$
Now it $Y lt 0$ (rewritten second inequality), then automatically $0 lt 1-Y$ (rewritten third inequality). So the case $Z=0$ amounts to a solution set:
$$ {(X,Y,Z) mid Xgt 1, Ylt 0, Z = 0 } tag{Z=0} $$
Case: Z is positive
If $Z gt 0$, then we could multiply the first inequality on both sides by $Z$ without changing the direction of the inequality. We would then have the equivalent system:
$$ ZX gt Z tag{1} $$
$$ Y lt ZX + Z tag{2} $$
$$ ZX lt 1 - Y tag{3} $$
If we substitute $W = ZX$, then the nature of the equivalent system as linear inequalities will be clear:
$$ W gt Z gt 0 tag{1''} $$
$$ Y lt W + Z tag{2''} $$
$$ W lt 1 - Y tag{3''} $$
The solutions of the first inequality are easy to visualize. In the $W,Z$-plane it consists of all the points in the (strict) first quadrant ($W,Z gt 0$) such that $W gt Z$, so points $(W,Z)$ falling to the right of the diagonal line $W=Z$.
The second and third inequalities then both amount to upper bounds on $Y$. That is, not only do we have $Y lt W + Z$, but also (rewriting the third inequality) $Y lt 1 - W$. So the solutions will be (in terms of $W,Y,Z$) determined by:
$$ Y lt min(W+Z,1-W) $$
Translating this result back into terms of $X = W/Z$, we have the solution set for the case that $Z$ is positive:
$$ {(X,Y,Z) mid X gt 1, Z gt 0, Y lt min(Z(1+X),1-ZX); } tag{Z>0}$$
More can be done to visualize this region. When is $Z(1+X) le 1-ZX$, and when is the reverse true? For fixed $Zgt 0$, the quantity $Z(1+X)$ increases as $X$ increases while $1-ZX$ decreases with increasing $X$. So for sufficiently large $X$, the smaller of these will be $1-ZX$.
If we work out where the two expressions are equal, $Z(1+X) = 1-ZX$, we find the crossover occurs at:
$$ X_* = frac{1-Z}{2Z} $$
Since $Xgt 1$, the value $X ge X_*$ always holds when $Z ge 1/3$ (which implies $X_* le 1$), and thus $Y lt 1 -ZX$ is the limiting constraint.
For $0 lt Z lt 1/3$, there are two cases. If $1lt X lt X_*$, then $Y lt Z(1+X)$ is the limiting constraint, and otherwise ($X ge X_*$) the limiting constraint is $Y lt 1 - ZX$.
Case: Z is negative
Left as an exercise for the Reader. Note that the first inequality becomes $ZX < Z$ because the direction of the inequality is reversed.
answered Oct 29 '15 at 12:24
hardmathhardmath
29.3k953101
29.3k953101
1
$begingroup$
Thank you Hardmath. Case by case derivation is the way to solve it. Drawing the systems of inequalities in matlab I found two solution regions.
$endgroup$
– Yesh
Oct 30 '15 at 13:56
add a comment |
1
$begingroup$
Thank you Hardmath. Case by case derivation is the way to solve it. Drawing the systems of inequalities in matlab I found two solution regions.
$endgroup$
– Yesh
Oct 30 '15 at 13:56
1
1
$begingroup$
Thank you Hardmath. Case by case derivation is the way to solve it. Drawing the systems of inequalities in matlab I found two solution regions.
$endgroup$
– Yesh
Oct 30 '15 at 13:56
$begingroup$
Thank you Hardmath. Case by case derivation is the way to solve it. Drawing the systems of inequalities in matlab I found two solution regions.
$endgroup$
– Yesh
Oct 30 '15 at 13:56
add a comment |
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$begingroup$
Have been trying the whole day to solve it. I could not. I also tried matlab to draw in 3D to see the region that represent the solution
$endgroup$
– Yesh
Oct 14 '15 at 16:05
$begingroup$
With a bit of practice you can use mathematical notation in your posts with MathJax and $LaTeX$. I will edit your post to give you an idea how it works.
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– hardmath
Oct 29 '15 at 9:42
$begingroup$
Try substituting one inequality into another. Try showing that $Y<2Z$. Can you do that? It should help you get part of the way to the answer.
$endgroup$
– Cataline
Oct 29 '15 at 10:51