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How to find unique solution for two variables from a single equation with those same two variables?
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I’m working on a problem - listed below - that supposedly is not solveable. I’m a bit confused because when we divide $40X$ by $3X$ we get the solution for $P$. Why can’t we just plug back P’s value of $40/3$ to get the solution for $X$.
Is it possible to find X when $40X = 3XP$
algebra-precalculus
edited Jul 14 at 18:23
Key Flex
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asked Jul 14 at 17:46
Omar
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up vote
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I’m working on a problem - listed below - that supposedly is not solveable. I’m a bit confused because when we divide $40X$ by $3X$ we get the solution for $P$. Why can’t we just plug back P’s value of $40/3$ to get the solution for $X$.
Is it possible to find X when $40X = 3XP$
algebra-precalculus
edited Jul 14 at 18:23
Key Flex
4,471525
asked Jul 14 at 17:46
Omar
324
1
Because then you just have $3x=3x$ which is true for every real number $x$. Furthermore, you can't conclude $p=40/3$ from the original equation unless you know $xne0.$
– saulspatz
Jul 14 at 17:51
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I’m working on a problem - listed below - that supposedly is not solveable. I’m a bit confused because when we divide $40X$ by $3X$ we get the solution for $P$. Why can’t we just plug back P’s value of $40/3$ to get the solution for $X$.
Is it possible to find X when $40X = 3XP$
algebra-precalculus
edited Jul 14 at 18:23
Key Flex
4,471525
asked Jul 14 at 17:46
Omar
324
I’m working on a problem - listed below - that supposedly is not solveable. I’m a bit confused because when we divide $40X$ by $3X$ we get the solution for $P$. Why can’t we just plug back P’s value of $40/3$ to get the solution for $X$.
Is it possible to find X when $40X = 3XP$
algebra-precalculus
edited Jul 14 at 18:23
Key Flex
4,471525
asked Jul 14 at 17:46
Omar
324
edited Jul 14 at 18:23
Key Flex
4,471525
edited Jul 14 at 18:23
Key Flex
4,471525
edited Jul 14 at 18:23
Key Flex
4,471525
4,471525
asked Jul 14 at 17:46
Omar
324
asked Jul 14 at 17:46
Omar
324
asked Jul 14 at 17:46
Omar
324
324
1
Because then you just have $3x=3x$ which is true for every real number $x$. Furthermore, you can't conclude $p=40/3$ from the original equation unless you know $xne0.$
– saulspatz
Jul 14 at 17:51
add a comment |Â
1
Because then you just have $3x=3x$ which is true for every real number $x$. Furthermore, you can't conclude $p=40/3$ from the original equation unless you know $xne0.$
– saulspatz
Jul 14 at 17:51
1
1
Because then you just have $3x=3x$ which is true for every real number $x$. Furthermore, you can't conclude $p=40/3$ from the original equation unless you know $xne0.$
– saulspatz
Jul 14 at 17:51
Because then you just have $3x=3x$ which is true for every real number $x$. Furthermore, you can't conclude $p=40/3$ from the original equation unless you know $xne0.$
– saulspatz
Jul 14 at 17:51
add a comment |Â
1 Answer
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When you divide both sides by $3x$, you're assuming that $x$ is nonzero - you can't do that if $x = 0$! In fact, if we take $x = 0$, the equation becomes
$$40 cdot 0 = 3 cdot 0 cdot p$$
which is true for every value of $p$. So we already have infinitely many solutions: $x = 0$, $p = $ anything.
Now, if $x neq 0$, you're right that we can divide both sides by $3x$ and get $p = 40/3$. If we plug that value of $p$ back in, here's how it goes:
$$40x = 3x cdot frac403$$
$$40x = 40x$$
$$x = x$$
This is true no matter what $x$ is - so again we have infinitely many solutions, this time with $p = 40/3$ and $x = $ anything.
This is what typically happens when you have more unknowns than you have equations - you can sort of "partially" solve it, but you're left with infinitely many solutions anyway. I'm not sure I would call this "unsolvable", though - certainly you can't solve for a unique solution, but we say equations like $x^2 + 3x + 2 = 0$ are "solvable" even though they have two solutions.
answered Jul 14 at 18:09
Reese
14.3k11135
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
When you divide both sides by $3x$, you're assuming that $x$ is nonzero - you can't do that if $x = 0$! In fact, if we take $x = 0$, the equation becomes
$$40 cdot 0 = 3 cdot 0 cdot p$$
which is true for every value of $p$. So we already have infinitely many solutions: $x = 0$, $p = $ anything.
Now, if $x neq 0$, you're right that we can divide both sides by $3x$ and get $p = 40/3$. If we plug that value of $p$ back in, here's how it goes:
$$40x = 3x cdot frac403$$
$$40x = 40x$$
$$x = x$$
This is true no matter what $x$ is - so again we have infinitely many solutions, this time with $p = 40/3$ and $x = $ anything.
This is what typically happens when you have more unknowns than you have equations - you can sort of "partially" solve it, but you're left with infinitely many solutions anyway. I'm not sure I would call this "unsolvable", though - certainly you can't solve for a unique solution, but we say equations like $x^2 + 3x + 2 = 0$ are "solvable" even though they have two solutions.
answered Jul 14 at 18:09
Reese
14.3k11135
add a comment |Â
up vote
1
down vote
accepted
When you divide both sides by $3x$, you're assuming that $x$ is nonzero - you can't do that if $x = 0$! In fact, if we take $x = 0$, the equation becomes
$$40 cdot 0 = 3 cdot 0 cdot p$$
which is true for every value of $p$. So we already have infinitely many solutions: $x = 0$, $p = $ anything.
Now, if $x neq 0$, you're right that we can divide both sides by $3x$ and get $p = 40/3$. If we plug that value of $p$ back in, here's how it goes:
$$40x = 3x cdot frac403$$
$$40x = 40x$$
$$x = x$$
This is true no matter what $x$ is - so again we have infinitely many solutions, this time with $p = 40/3$ and $x = $ anything.
This is what typically happens when you have more unknowns than you have equations - you can sort of "partially" solve it, but you're left with infinitely many solutions anyway. I'm not sure I would call this "unsolvable", though - certainly you can't solve for a unique solution, but we say equations like $x^2 + 3x + 2 = 0$ are "solvable" even though they have two solutions.
answered Jul 14 at 18:09
Reese
14.3k11135
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
When you divide both sides by $3x$, you're assuming that $x$ is nonzero - you can't do that if $x = 0$! In fact, if we take $x = 0$, the equation becomes
$$40 cdot 0 = 3 cdot 0 cdot p$$
which is true for every value of $p$. So we already have infinitely many solutions: $x = 0$, $p = $ anything.
Now, if $x neq 0$, you're right that we can divide both sides by $3x$ and get $p = 40/3$. If we plug that value of $p$ back in, here's how it goes:
$$40x = 3x cdot frac403$$
$$40x = 40x$$
$$x = x$$
This is true no matter what $x$ is - so again we have infinitely many solutions, this time with $p = 40/3$ and $x = $ anything.
This is what typically happens when you have more unknowns than you have equations - you can sort of "partially" solve it, but you're left with infinitely many solutions anyway. I'm not sure I would call this "unsolvable", though - certainly you can't solve for a unique solution, but we say equations like $x^2 + 3x + 2 = 0$ are "solvable" even though they have two solutions.
answered Jul 14 at 18:09
Reese
14.3k11135
When you divide both sides by $3x$, you're assuming that $x$ is nonzero - you can't do that if $x = 0$! In fact, if we take $x = 0$, the equation becomes
$$40 cdot 0 = 3 cdot 0 cdot p$$
which is true for every value of $p$. So we already have infinitely many solutions: $x = 0$, $p = $ anything.
Now, if $x neq 0$, you're right that we can divide both sides by $3x$ and get $p = 40/3$. If we plug that value of $p$ back in, here's how it goes:
$$40x = 3x cdot frac403$$
$$40x = 40x$$
$$x = x$$
This is true no matter what $x$ is - so again we have infinitely many solutions, this time with $p = 40/3$ and $x = $ anything.
This is what typically happens when you have more unknowns than you have equations - you can sort of "partially" solve it, but you're left with infinitely many solutions anyway. I'm not sure I would call this "unsolvable", though - certainly you can't solve for a unique solution, but we say equations like $x^2 + 3x + 2 = 0$ are "solvable" even though they have two solutions.
answered Jul 14 at 18:09
Reese
14.3k11135
answered Jul 14 at 18:09
Reese
14.3k11135
answered Jul 14 at 18:09
Reese
14.3k11135
answered Jul 14 at 18:09
Reese
14.3k11135
14.3k11135
add a comment |Â
add a comment |Â
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1
Because then you just have $3x=3x$ which is true for every real number $x$. Furthermore, you can't conclude $p=40/3$ from the original equation unless you know $xne0.$
– saulspatz
Jul 14 at 17:51