Mixing
Computing the product $xy$
$begingroup$
$$dfrac{|x|}{2}=y$$
$$|y|-x=4$$
- Compute $xy$
Here I would have been able to solve this question in such case there was only one unknown. I, however, cannot solve this type of questions when there are more than one unknown so let me at least share my thoughts with you.
$$|x| = 2y implies dfrac{1}{2}|x| = y $$
Plugging into the second equation and we get that
$$biggr |dfrac{1}{2}|x|biggr |-x = 4 $$
This is where I'm stuck.
Regards
algebra-precalculus systems-of-equations absolute-value
asked Jan 8 at 14:04
EnzoEnzo
1737
$endgroup$
add a comment |
$begingroup$
$$dfrac{|x|}{2}=y$$
$$|y|-x=4$$
- Compute $xy$
Here I would have been able to solve this question in such case there was only one unknown. I, however, cannot solve this type of questions when there are more than one unknown so let me at least share my thoughts with you.
$$|x| = 2y implies dfrac{1}{2}|x| = y $$
Plugging into the second equation and we get that
$$biggr |dfrac{1}{2}|x|biggr |-x = 4 $$
This is where I'm stuck.
Regards
algebra-precalculus systems-of-equations absolute-value
asked Jan 8 at 14:04
EnzoEnzo
1737
$endgroup$
add a comment |
$begingroup$
$$dfrac{|x|}{2}=y$$
$$|y|-x=4$$
- Compute $xy$
Here I would have been able to solve this question in such case there was only one unknown. I, however, cannot solve this type of questions when there are more than one unknown so let me at least share my thoughts with you.
$$|x| = 2y implies dfrac{1}{2}|x| = y $$
Plugging into the second equation and we get that
$$biggr |dfrac{1}{2}|x|biggr |-x = 4 $$
This is where I'm stuck.
Regards
algebra-precalculus systems-of-equations absolute-value
asked Jan 8 at 14:04
EnzoEnzo
1737
$endgroup$
$$dfrac{|x|}{2}=y$$
$$|y|-x=4$$
- Compute $xy$
Here I would have been able to solve this question in such case there was only one unknown. I, however, cannot solve this type of questions when there are more than one unknown so let me at least share my thoughts with you.
$$|x| = 2y implies dfrac{1}{2}|x| = y $$
Plugging into the second equation and we get that
$$biggr |dfrac{1}{2}|x|biggr |-x = 4 $$
This is where I'm stuck.
Regards
algebra-precalculus systems-of-equations absolute-value
algebra-precalculus systems-of-equations absolute-value
asked Jan 8 at 14:04
EnzoEnzo
1737
asked Jan 8 at 14:04
EnzoEnzo
1737
edited Jan 8 at 14:16
user593746
asked Jan 8 at 14:04
EnzoEnzo
1737
asked Jan 8 at 14:04
EnzoEnzo
1737
asked Jan 8 at 14:04
EnzoEnzo
1737
1737
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Since $y=|x|/2$ you know $yge 0,$ so in the second equation $|y|$ can be replaced by simply $y.$
answered Jan 8 at 14:07
coffeemathcoffeemath
2,7501415
$endgroup$
add a comment |
$begingroup$
Hint: Since $$y=|x|/2implies ygeq 0implies |y|=y$$
so $$|x|= 2(x+4)implies ...$$
answered Jan 8 at 14:07
greedoidgreedoid
40.7k1149100
$endgroup$
add a comment |
$begingroup$
$$y=|x|/2implies ygeq 0implies |y|=y$$
Also
$$|x|=2y, |y|-x=y-x=4$$
If $x$ is positive, then
$$x=2y$$
$$y-x=y-2y=-y=4implies y=-4$$
which contradicts the claim that $$y geq 0$$
So $xle0$, and
$$x=-2y$$
$$y-x=y-(-2y)=3y=4implies y=4/3$$
$$x=-2y=-8/3$$
$$xy=-32/9$$
answered Jan 8 at 14:19
LarryLarry
2,35131029
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $y=|x|/2$ you know $yge 0,$ so in the second equation $|y|$ can be replaced by simply $y.$
answered Jan 8 at 14:07
coffeemathcoffeemath
2,7501415
$endgroup$
add a comment |
$begingroup$
Since $y=|x|/2$ you know $yge 0,$ so in the second equation $|y|$ can be replaced by simply $y.$
answered Jan 8 at 14:07
coffeemathcoffeemath
2,7501415
$endgroup$
add a comment |
$begingroup$
Since $y=|x|/2$ you know $yge 0,$ so in the second equation $|y|$ can be replaced by simply $y.$
answered Jan 8 at 14:07
coffeemathcoffeemath
2,7501415
$endgroup$
Since $y=|x|/2$ you know $yge 0,$ so in the second equation $|y|$ can be replaced by simply $y.$
answered Jan 8 at 14:07
coffeemathcoffeemath
2,7501415
answered Jan 8 at 14:07
coffeemathcoffeemath
2,7501415
answered Jan 8 at 14:07
coffeemathcoffeemath
2,7501415
answered Jan 8 at 14:07
coffeemathcoffeemath
2,7501415
2,7501415
add a comment |
add a comment |
$begingroup$
Hint: Since $$y=|x|/2implies ygeq 0implies |y|=y$$
so $$|x|= 2(x+4)implies ...$$
answered Jan 8 at 14:07
greedoidgreedoid
40.7k1149100
$endgroup$
add a comment |
$begingroup$
Hint: Since $$y=|x|/2implies ygeq 0implies |y|=y$$
so $$|x|= 2(x+4)implies ...$$
answered Jan 8 at 14:07
greedoidgreedoid
40.7k1149100
$endgroup$
add a comment |
$begingroup$
Hint: Since $$y=|x|/2implies ygeq 0implies |y|=y$$
so $$|x|= 2(x+4)implies ...$$
answered Jan 8 at 14:07
greedoidgreedoid
40.7k1149100
$endgroup$
Hint: Since $$y=|x|/2implies ygeq 0implies |y|=y$$
so $$|x|= 2(x+4)implies ...$$
answered Jan 8 at 14:07
greedoidgreedoid
40.7k1149100
answered Jan 8 at 14:07
greedoidgreedoid
40.7k1149100
answered Jan 8 at 14:07
greedoidgreedoid
40.7k1149100
answered Jan 8 at 14:07
greedoidgreedoid
40.7k1149100
40.7k1149100
add a comment |
add a comment |
$begingroup$
$$y=|x|/2implies ygeq 0implies |y|=y$$
Also
$$|x|=2y, |y|-x=y-x=4$$
If $x$ is positive, then
$$x=2y$$
$$y-x=y-2y=-y=4implies y=-4$$
which contradicts the claim that $$y geq 0$$
So $xle0$, and
$$x=-2y$$
$$y-x=y-(-2y)=3y=4implies y=4/3$$
$$x=-2y=-8/3$$
$$xy=-32/9$$
answered Jan 8 at 14:19
LarryLarry
2,35131029
$endgroup$
add a comment |
$begingroup$
$$y=|x|/2implies ygeq 0implies |y|=y$$
Also
$$|x|=2y, |y|-x=y-x=4$$
If $x$ is positive, then
$$x=2y$$
$$y-x=y-2y=-y=4implies y=-4$$
which contradicts the claim that $$y geq 0$$
So $xle0$, and
$$x=-2y$$
$$y-x=y-(-2y)=3y=4implies y=4/3$$
$$x=-2y=-8/3$$
$$xy=-32/9$$
answered Jan 8 at 14:19
LarryLarry
2,35131029
$endgroup$
add a comment |
$begingroup$
$$y=|x|/2implies ygeq 0implies |y|=y$$
Also
$$|x|=2y, |y|-x=y-x=4$$
If $x$ is positive, then
$$x=2y$$
$$y-x=y-2y=-y=4implies y=-4$$
which contradicts the claim that $$y geq 0$$
So $xle0$, and
$$x=-2y$$
$$y-x=y-(-2y)=3y=4implies y=4/3$$
$$x=-2y=-8/3$$
$$xy=-32/9$$
answered Jan 8 at 14:19
LarryLarry
2,35131029
$endgroup$
$$y=|x|/2implies ygeq 0implies |y|=y$$
Also
$$|x|=2y, |y|-x=y-x=4$$
If $x$ is positive, then
$$x=2y$$
$$y-x=y-2y=-y=4implies y=-4$$
which contradicts the claim that $$y geq 0$$
So $xle0$, and
$$x=-2y$$
$$y-x=y-(-2y)=3y=4implies y=4/3$$
$$x=-2y=-8/3$$
$$xy=-32/9$$
answered Jan 8 at 14:19
LarryLarry
2,35131029
answered Jan 8 at 14:19
LarryLarry
2,35131029
answered Jan 8 at 14:19
LarryLarry
2,35131029
answered Jan 8 at 14:19
LarryLarry
2,35131029
2,35131029
add a comment |
add a comment |
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