How to solve $1/x = 1/a + 1/b$ to get $x = ab/(a + b)$?
up vote
2
down vote
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I'm working through the first problem set in a text book. I have the question and the solution, however the solution gives no in between steps, only the final result.
The question is solve for $x$ in :
$$frac{1}{x} = frac{1}{a} + frac{1}{b}$$
I could only get one step in:
$$x = xleft(frac{1}{a} + frac{1}{b}right)$$
The solution by the text book answers section is:
$$x = frac{ab}{a + b}$$
How was this arrived at? Seeking the steps between to arrive at this?
algebra-precalculus
edited Nov 14 at 10:15
Dave L. Renfro
23.4k33979
asked Nov 14 at 5:36
Doug Fir
1696
add a comment |
up vote
2
down vote
favorite
I'm working through the first problem set in a text book. I have the question and the solution, however the solution gives no in between steps, only the final result.
The question is solve for $x$ in :
$$frac{1}{x} = frac{1}{a} + frac{1}{b}$$
I could only get one step in:
$$x = xleft(frac{1}{a} + frac{1}{b}right)$$
The solution by the text book answers section is:
$$x = frac{ab}{a + b}$$
How was this arrived at? Seeking the steps between to arrive at this?
algebra-precalculus
edited Nov 14 at 10:15
Dave L. Renfro
23.4k33979
asked Nov 14 at 5:36
Doug Fir
1696
4
The "one step" that you get is false. When you multiply both sides with $x$, the left hand side is $1$, not $x$. But this does not help : what helps is taking the reciprocals of both sides.
– астон вілла олоф мэллбэрг
Nov 14 at 5:39
2
You have an error in your first step: it should read $1=x(1/a+1/b)$
– N74
Nov 14 at 5:40
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm working through the first problem set in a text book. I have the question and the solution, however the solution gives no in between steps, only the final result.
The question is solve for $x$ in :
$$frac{1}{x} = frac{1}{a} + frac{1}{b}$$
I could only get one step in:
$$x = xleft(frac{1}{a} + frac{1}{b}right)$$
The solution by the text book answers section is:
$$x = frac{ab}{a + b}$$
How was this arrived at? Seeking the steps between to arrive at this?
algebra-precalculus
edited Nov 14 at 10:15
Dave L. Renfro
23.4k33979
asked Nov 14 at 5:36
Doug Fir
1696
I'm working through the first problem set in a text book. I have the question and the solution, however the solution gives no in between steps, only the final result.
The question is solve for $x$ in :
$$frac{1}{x} = frac{1}{a} + frac{1}{b}$$
I could only get one step in:
$$x = xleft(frac{1}{a} + frac{1}{b}right)$$
The solution by the text book answers section is:
$$x = frac{ab}{a + b}$$
How was this arrived at? Seeking the steps between to arrive at this?
algebra-precalculus
algebra-precalculus
edited Nov 14 at 10:15
Dave L. Renfro
23.4k33979
asked Nov 14 at 5:36
Doug Fir
1696
edited Nov 14 at 10:15
Dave L. Renfro
23.4k33979
asked Nov 14 at 5:36
Doug Fir
1696
edited Nov 14 at 10:15
Dave L. Renfro
23.4k33979
edited Nov 14 at 10:15
Dave L. Renfro
23.4k33979
edited Nov 14 at 10:15
Dave L. Renfro
23.4k33979
23.4k33979
asked Nov 14 at 5:36
Doug Fir
1696
asked Nov 14 at 5:36
Doug Fir
1696
asked Nov 14 at 5:36
Doug Fir
1696
1696
4
The "one step" that you get is false. When you multiply both sides with $x$, the left hand side is $1$, not $x$. But this does not help : what helps is taking the reciprocals of both sides.
– астон вілла олоф мэллбэрг
Nov 14 at 5:39
2
You have an error in your first step: it should read $1=x(1/a+1/b)$
– N74
Nov 14 at 5:40
add a comment |
4
The "one step" that you get is false. When you multiply both sides with $x$, the left hand side is $1$, not $x$. But this does not help : what helps is taking the reciprocals of both sides.
– астон вілла олоф мэллбэрг
Nov 14 at 5:39
2
You have an error in your first step: it should read $1=x(1/a+1/b)$
– N74
Nov 14 at 5:40
4
4
The "one step" that you get is false. When you multiply both sides with $x$, the left hand side is $1$, not $x$. But this does not help : what helps is taking the reciprocals of both sides.
– астон вілла олоф мэллбэрг
Nov 14 at 5:39
The "one step" that you get is false. When you multiply both sides with $x$, the left hand side is $1$, not $x$. But this does not help : what helps is taking the reciprocals of both sides.
– астон вілла олоф мэллбэрг
Nov 14 at 5:39
2
2
You have an error in your first step: it should read $1=x(1/a+1/b)$
– N74
Nov 14 at 5:40
You have an error in your first step: it should read $1=x(1/a+1/b)$
– N74
Nov 14 at 5:40
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
$$frac{1}{x}=frac{1}{a}+frac{1}{b}$$
$$frac{1}{x}=frac{color{red}b}{acolor{red}b}+frac{color{red}a}{bcolor{red}a}$$
$$frac{1}{x}=frac{a+b}{ba}$$
$$x=frac{ab}{a+b}$$
answered Nov 14 at 5:39
Robert Howard
1,715622
Sorry for being slow but I'm just not following the second line. How do you get tob/ab + a/ba? From the comment under my post I Googled "reciprocals" which took me to this page but I don't see how it applies? mathsisfun.com/reciprocal.html
– Doug Fir
Nov 14 at 6:08
1
That's okay! Anything divided by itself is just $1$, so $b/b=1$ and $a/a=1$. That means we can multiply other fractions by $b/b$ and $a/a$ without changing the values of the original fractions, which is just what I did. I did that in order to get a common denominator ($ab$) in both fractions, which would then make it possible to add them.
– Robert Howard
Nov 14 at 6:28
1
It clicked thanks to your explanation. I think I got it. Those parts highlighted in red are are the equivalent of 1 while keeping the original denominator a and then b keeps the equation equivalent. OK, thanks a lot!
– Doug Fir
Nov 14 at 6:33
1
I would search for something like "finding common denominators," because that's the essence of what I did. I saw that $x$ was in the denominator of the fraction on the left, so I knew that I could flip that fraction to get $x$ by itself. That would mean I would also have to flip whatever was on the other side of the equation, which works much better when the other side of the equation is a single fraction. If I hadn't combined $1/a$ and $1/b$ into a single fraction, I would have ended up with $$x=frac{1}{frac{1}{a}+frac{1}{b}},$$ which is considerably uglier than $$x=frac{ab}{a+b}$$
– Robert Howard
Nov 14 at 20:37
1
Of course, the "uglier" fraction can be simplified, as Jimmy showed, but I opted to do the simplification beforehand.
– Robert Howard
Nov 14 at 20:38
|
show 4 more comments
up vote
2
down vote
$$frac{1}{a}+frac{1}{b}=frac{b}{ab}+frac{a}{ab}=frac{a+b}{ab}$$ So $$frac{1}{frac1a+frac1b}=frac{1}{frac{a+b}{ab}}=frac{frac11}{frac{a+b}{ab}}=frac{ab}{a+b}$$
answered Nov 14 at 5:39
Jimmy R.
32.9k42157
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
$$frac{1}{x}=frac{1}{a}+frac{1}{b}$$
$$frac{1}{x}=frac{color{red}b}{acolor{red}b}+frac{color{red}a}{bcolor{red}a}$$
$$frac{1}{x}=frac{a+b}{ba}$$
$$x=frac{ab}{a+b}$$
answered Nov 14 at 5:39
Robert Howard
1,715622
Sorry for being slow but I'm just not following the second line. How do you get tob/ab + a/ba? From the comment under my post I Googled "reciprocals" which took me to this page but I don't see how it applies? mathsisfun.com/reciprocal.html
– Doug Fir
Nov 14 at 6:08
1
That's okay! Anything divided by itself is just $1$, so $b/b=1$ and $a/a=1$. That means we can multiply other fractions by $b/b$ and $a/a$ without changing the values of the original fractions, which is just what I did. I did that in order to get a common denominator ($ab$) in both fractions, which would then make it possible to add them.
– Robert Howard
Nov 14 at 6:28
1
It clicked thanks to your explanation. I think I got it. Those parts highlighted in red are are the equivalent of 1 while keeping the original denominator a and then b keeps the equation equivalent. OK, thanks a lot!
– Doug Fir
Nov 14 at 6:33
1
I would search for something like "finding common denominators," because that's the essence of what I did. I saw that $x$ was in the denominator of the fraction on the left, so I knew that I could flip that fraction to get $x$ by itself. That would mean I would also have to flip whatever was on the other side of the equation, which works much better when the other side of the equation is a single fraction. If I hadn't combined $1/a$ and $1/b$ into a single fraction, I would have ended up with $$x=frac{1}{frac{1}{a}+frac{1}{b}},$$ which is considerably uglier than $$x=frac{ab}{a+b}$$
– Robert Howard
Nov 14 at 20:37
1
Of course, the "uglier" fraction can be simplified, as Jimmy showed, but I opted to do the simplification beforehand.
– Robert Howard
Nov 14 at 20:38
|
show 4 more comments
up vote
4
down vote
accepted
$$frac{1}{x}=frac{1}{a}+frac{1}{b}$$
$$frac{1}{x}=frac{color{red}b}{acolor{red}b}+frac{color{red}a}{bcolor{red}a}$$
$$frac{1}{x}=frac{a+b}{ba}$$
$$x=frac{ab}{a+b}$$
answered Nov 14 at 5:39
Robert Howard
1,715622
Sorry for being slow but I'm just not following the second line. How do you get tob/ab + a/ba? From the comment under my post I Googled "reciprocals" which took me to this page but I don't see how it applies? mathsisfun.com/reciprocal.html
– Doug Fir
Nov 14 at 6:08
1
That's okay! Anything divided by itself is just $1$, so $b/b=1$ and $a/a=1$. That means we can multiply other fractions by $b/b$ and $a/a$ without changing the values of the original fractions, which is just what I did. I did that in order to get a common denominator ($ab$) in both fractions, which would then make it possible to add them.
– Robert Howard
Nov 14 at 6:28
1
It clicked thanks to your explanation. I think I got it. Those parts highlighted in red are are the equivalent of 1 while keeping the original denominator a and then b keeps the equation equivalent. OK, thanks a lot!
– Doug Fir
Nov 14 at 6:33
1
I would search for something like "finding common denominators," because that's the essence of what I did. I saw that $x$ was in the denominator of the fraction on the left, so I knew that I could flip that fraction to get $x$ by itself. That would mean I would also have to flip whatever was on the other side of the equation, which works much better when the other side of the equation is a single fraction. If I hadn't combined $1/a$ and $1/b$ into a single fraction, I would have ended up with $$x=frac{1}{frac{1}{a}+frac{1}{b}},$$ which is considerably uglier than $$x=frac{ab}{a+b}$$
– Robert Howard
Nov 14 at 20:37
1
Of course, the "uglier" fraction can be simplified, as Jimmy showed, but I opted to do the simplification beforehand.
– Robert Howard
Nov 14 at 20:38
|
show 4 more comments
up vote
4
down vote
accepted
up vote
4
down vote
accepted
$$frac{1}{x}=frac{1}{a}+frac{1}{b}$$
$$frac{1}{x}=frac{color{red}b}{acolor{red}b}+frac{color{red}a}{bcolor{red}a}$$
$$frac{1}{x}=frac{a+b}{ba}$$
$$x=frac{ab}{a+b}$$
answered Nov 14 at 5:39
Robert Howard
1,715622
$$frac{1}{x}=frac{1}{a}+frac{1}{b}$$
$$frac{1}{x}=frac{color{red}b}{acolor{red}b}+frac{color{red}a}{bcolor{red}a}$$
$$frac{1}{x}=frac{a+b}{ba}$$
$$x=frac{ab}{a+b}$$
answered Nov 14 at 5:39
Robert Howard
1,715622
answered Nov 14 at 5:39
Robert Howard
1,715622
answered Nov 14 at 5:39
Robert Howard
1,715622
answered Nov 14 at 5:39
Robert Howard
1,715622
1,715622
Sorry for being slow but I'm just not following the second line. How do you get tob/ab + a/ba? From the comment under my post I Googled "reciprocals" which took me to this page but I don't see how it applies? mathsisfun.com/reciprocal.html
– Doug Fir
Nov 14 at 6:08
1
That's okay! Anything divided by itself is just $1$, so $b/b=1$ and $a/a=1$. That means we can multiply other fractions by $b/b$ and $a/a$ without changing the values of the original fractions, which is just what I did. I did that in order to get a common denominator ($ab$) in both fractions, which would then make it possible to add them.
– Robert Howard
Nov 14 at 6:28
1
It clicked thanks to your explanation. I think I got it. Those parts highlighted in red are are the equivalent of 1 while keeping the original denominator a and then b keeps the equation equivalent. OK, thanks a lot!
– Doug Fir
Nov 14 at 6:33
1
I would search for something like "finding common denominators," because that's the essence of what I did. I saw that $x$ was in the denominator of the fraction on the left, so I knew that I could flip that fraction to get $x$ by itself. That would mean I would also have to flip whatever was on the other side of the equation, which works much better when the other side of the equation is a single fraction. If I hadn't combined $1/a$ and $1/b$ into a single fraction, I would have ended up with $$x=frac{1}{frac{1}{a}+frac{1}{b}},$$ which is considerably uglier than $$x=frac{ab}{a+b}$$
– Robert Howard
Nov 14 at 20:37
1
Of course, the "uglier" fraction can be simplified, as Jimmy showed, but I opted to do the simplification beforehand.
– Robert Howard
Nov 14 at 20:38
|
show 4 more comments
Sorry for being slow but I'm just not following the second line. How do you get tob/ab + a/ba? From the comment under my post I Googled "reciprocals" which took me to this page but I don't see how it applies? mathsisfun.com/reciprocal.html
– Doug Fir
Nov 14 at 6:08
1
That's okay! Anything divided by itself is just $1$, so $b/b=1$ and $a/a=1$. That means we can multiply other fractions by $b/b$ and $a/a$ without changing the values of the original fractions, which is just what I did. I did that in order to get a common denominator ($ab$) in both fractions, which would then make it possible to add them.
– Robert Howard
Nov 14 at 6:28
1
It clicked thanks to your explanation. I think I got it. Those parts highlighted in red are are the equivalent of 1 while keeping the original denominator a and then b keeps the equation equivalent. OK, thanks a lot!
– Doug Fir
Nov 14 at 6:33
1
I would search for something like "finding common denominators," because that's the essence of what I did. I saw that $x$ was in the denominator of the fraction on the left, so I knew that I could flip that fraction to get $x$ by itself. That would mean I would also have to flip whatever was on the other side of the equation, which works much better when the other side of the equation is a single fraction. If I hadn't combined $1/a$ and $1/b$ into a single fraction, I would have ended up with $$x=frac{1}{frac{1}{a}+frac{1}{b}},$$ which is considerably uglier than $$x=frac{ab}{a+b}$$
– Robert Howard
Nov 14 at 20:37
1
Of course, the "uglier" fraction can be simplified, as Jimmy showed, but I opted to do the simplification beforehand.
– Robert Howard
Nov 14 at 20:38
Sorry for being slow but I'm just not following the second line. How do you get to
b/ab + a/ba? From the comment under my post I Googled "reciprocals" which took me to this page but I don't see how it applies? mathsisfun.com/reciprocal.html– Doug Fir
Nov 14 at 6:08
Sorry for being slow but I'm just not following the second line. How do you get to
b/ab + a/ba? From the comment under my post I Googled "reciprocals" which took me to this page but I don't see how it applies? mathsisfun.com/reciprocal.html– Doug Fir
Nov 14 at 6:08
1
1
That's okay! Anything divided by itself is just $1$, so $b/b=1$ and $a/a=1$. That means we can multiply other fractions by $b/b$ and $a/a$ without changing the values of the original fractions, which is just what I did. I did that in order to get a common denominator ($ab$) in both fractions, which would then make it possible to add them.
– Robert Howard
Nov 14 at 6:28
That's okay! Anything divided by itself is just $1$, so $b/b=1$ and $a/a=1$. That means we can multiply other fractions by $b/b$ and $a/a$ without changing the values of the original fractions, which is just what I did. I did that in order to get a common denominator ($ab$) in both fractions, which would then make it possible to add them.
– Robert Howard
Nov 14 at 6:28
1
1
It clicked thanks to your explanation. I think I got it. Those parts highlighted in red are are the equivalent of 1 while keeping the original denominator a and then b keeps the equation equivalent. OK, thanks a lot!
– Doug Fir
Nov 14 at 6:33
It clicked thanks to your explanation. I think I got it. Those parts highlighted in red are are the equivalent of 1 while keeping the original denominator a and then b keeps the equation equivalent. OK, thanks a lot!
– Doug Fir
Nov 14 at 6:33
1
1
I would search for something like "finding common denominators," because that's the essence of what I did. I saw that $x$ was in the denominator of the fraction on the left, so I knew that I could flip that fraction to get $x$ by itself. That would mean I would also have to flip whatever was on the other side of the equation, which works much better when the other side of the equation is a single fraction. If I hadn't combined $1/a$ and $1/b$ into a single fraction, I would have ended up with $$x=frac{1}{frac{1}{a}+frac{1}{b}},$$ which is considerably uglier than $$x=frac{ab}{a+b}$$
– Robert Howard
Nov 14 at 20:37
I would search for something like "finding common denominators," because that's the essence of what I did. I saw that $x$ was in the denominator of the fraction on the left, so I knew that I could flip that fraction to get $x$ by itself. That would mean I would also have to flip whatever was on the other side of the equation, which works much better when the other side of the equation is a single fraction. If I hadn't combined $1/a$ and $1/b$ into a single fraction, I would have ended up with $$x=frac{1}{frac{1}{a}+frac{1}{b}},$$ which is considerably uglier than $$x=frac{ab}{a+b}$$
– Robert Howard
Nov 14 at 20:37
1
1
Of course, the "uglier" fraction can be simplified, as Jimmy showed, but I opted to do the simplification beforehand.
– Robert Howard
Nov 14 at 20:38
Of course, the "uglier" fraction can be simplified, as Jimmy showed, but I opted to do the simplification beforehand.
– Robert Howard
Nov 14 at 20:38
|
show 4 more comments
up vote
2
down vote
$$frac{1}{a}+frac{1}{b}=frac{b}{ab}+frac{a}{ab}=frac{a+b}{ab}$$ So $$frac{1}{frac1a+frac1b}=frac{1}{frac{a+b}{ab}}=frac{frac11}{frac{a+b}{ab}}=frac{ab}{a+b}$$
answered Nov 14 at 5:39
Jimmy R.
32.9k42157
add a comment |
up vote
2
down vote
$$frac{1}{a}+frac{1}{b}=frac{b}{ab}+frac{a}{ab}=frac{a+b}{ab}$$ So $$frac{1}{frac1a+frac1b}=frac{1}{frac{a+b}{ab}}=frac{frac11}{frac{a+b}{ab}}=frac{ab}{a+b}$$
answered Nov 14 at 5:39
Jimmy R.
32.9k42157
add a comment |
up vote
2
down vote
up vote
2
down vote
$$frac{1}{a}+frac{1}{b}=frac{b}{ab}+frac{a}{ab}=frac{a+b}{ab}$$ So $$frac{1}{frac1a+frac1b}=frac{1}{frac{a+b}{ab}}=frac{frac11}{frac{a+b}{ab}}=frac{ab}{a+b}$$
answered Nov 14 at 5:39
Jimmy R.
32.9k42157
$$frac{1}{a}+frac{1}{b}=frac{b}{ab}+frac{a}{ab}=frac{a+b}{ab}$$ So $$frac{1}{frac1a+frac1b}=frac{1}{frac{a+b}{ab}}=frac{frac11}{frac{a+b}{ab}}=frac{ab}{a+b}$$
answered Nov 14 at 5:39
Jimmy R.
32.9k42157
answered Nov 14 at 5:39
Jimmy R.
32.9k42157
answered Nov 14 at 5:39
Jimmy R.
32.9k42157
answered Nov 14 at 5:39
Jimmy R.
32.9k42157
32.9k42157
add a comment |
add a comment |
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4
The "one step" that you get is false. When you multiply both sides with $x$, the left hand side is $1$, not $x$. But this does not help : what helps is taking the reciprocals of both sides.
– астон вілла олоф мэллбэрг
Nov 14 at 5:39
2
You have an error in your first step: it should read $1=x(1/a+1/b)$
– N74
Nov 14 at 5:40