$a+b=21$ and $frac{1}{a}+frac{1}{b} =frac{7}{18} $. What is $ab$? [closed]
$begingroup$
$ab$ is:
A. $18$
B. $36$
C. $54$
D. $72$
E. $90$
Would like to know if there is a way other than testing out each option to get the answer.
algebra-precalculus
edited May 6 '17 at 2:47
ΘΣΦGenSan
7,93992346
asked May 6 '17 at 1:54
CuriousCurious
111
$endgroup$
closed as off-topic by user21820, Jyrki Lahtonen, Carl Schildkraut, metamorphy, Did Dec 20 '18 at 8:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Jyrki Lahtonen, Carl Schildkraut, metamorphy, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
$ab$ is:
A. $18$
B. $36$
C. $54$
D. $72$
E. $90$
Would like to know if there is a way other than testing out each option to get the answer.
algebra-precalculus
edited May 6 '17 at 2:47
ΘΣΦGenSan
7,93992346
asked May 6 '17 at 1:54
CuriousCurious
111
$endgroup$
closed as off-topic by user21820, Jyrki Lahtonen, Carl Schildkraut, metamorphy, Did Dec 20 '18 at 8:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Jyrki Lahtonen, Carl Schildkraut, metamorphy, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
You already have 2 good answers (you should accept one). This is not an answer, just a comment - but this problem illustrates how, given exactly two of the three commonly used means of two numbers - i.e. the arithmetic (AM), geometric (GM) and harmonic mean (HM) - you can derive the third mean.
$endgroup$
– Deepak
May 6 '17 at 3:54
add a comment |
$begingroup$
$ab$ is:
A. $18$
B. $36$
C. $54$
D. $72$
E. $90$
Would like to know if there is a way other than testing out each option to get the answer.
algebra-precalculus
edited May 6 '17 at 2:47
ΘΣΦGenSan
7,93992346
asked May 6 '17 at 1:54
CuriousCurious
111
$endgroup$
$ab$ is:
A. $18$
B. $36$
C. $54$
D. $72$
E. $90$
Would like to know if there is a way other than testing out each option to get the answer.
algebra-precalculus
algebra-precalculus
edited May 6 '17 at 2:47
ΘΣΦGenSan
7,93992346
asked May 6 '17 at 1:54
CuriousCurious
111
edited May 6 '17 at 2:47
ΘΣΦGenSan
7,93992346
asked May 6 '17 at 1:54
CuriousCurious
111
edited May 6 '17 at 2:47
ΘΣΦGenSan
7,93992346
edited May 6 '17 at 2:47
ΘΣΦGenSan
7,93992346
edited May 6 '17 at 2:47
ΘΣΦGenSan
7,93992346
7,93992346
asked May 6 '17 at 1:54
CuriousCurious
111
asked May 6 '17 at 1:54
CuriousCurious
111
asked May 6 '17 at 1:54
CuriousCurious
111
111
closed as off-topic by user21820, Jyrki Lahtonen, Carl Schildkraut, metamorphy, Did Dec 20 '18 at 8:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Jyrki Lahtonen, Carl Schildkraut, metamorphy, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user21820, Jyrki Lahtonen, Carl Schildkraut, metamorphy, Did Dec 20 '18 at 8:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Jyrki Lahtonen, Carl Schildkraut, metamorphy, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
You already have 2 good answers (you should accept one). This is not an answer, just a comment - but this problem illustrates how, given exactly two of the three commonly used means of two numbers - i.e. the arithmetic (AM), geometric (GM) and harmonic mean (HM) - you can derive the third mean.
$endgroup$
– Deepak
May 6 '17 at 3:54
add a comment |
3
$begingroup$
You already have 2 good answers (you should accept one). This is not an answer, just a comment - but this problem illustrates how, given exactly two of the three commonly used means of two numbers - i.e. the arithmetic (AM), geometric (GM) and harmonic mean (HM) - you can derive the third mean.
$endgroup$
– Deepak
May 6 '17 at 3:54
3
3
$begingroup$
You already have 2 good answers (you should accept one). This is not an answer, just a comment - but this problem illustrates how, given exactly two of the three commonly used means of two numbers - i.e. the arithmetic (AM), geometric (GM) and harmonic mean (HM) - you can derive the third mean.
$endgroup$
– Deepak
May 6 '17 at 3:54
$begingroup$
You already have 2 good answers (you should accept one). This is not an answer, just a comment - but this problem illustrates how, given exactly two of the three commonly used means of two numbers - i.e. the arithmetic (AM), geometric (GM) and harmonic mean (HM) - you can derive the third mean.
$endgroup$
– Deepak
May 6 '17 at 3:54
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Well, $$frac{7}{18}=frac{1}{a}+frac{1}{b}=frac{b+a}{ab}=frac{21}{ab}$$ and so $$ab=frac{21(18)}{7}=54$$
answered May 6 '17 at 1:59
ΘΣΦGenSanΘΣΦGenSan
7,93992346
$endgroup$
add a comment |
$begingroup$
$$begin{array}{rcl}
dfrac1a + dfrac1b &=& dfrac7{18} \
dfrac b{ab} + dfrac a{ab} &=& dfrac7{18} \
dfrac{b+a}{ab} &=& dfrac7{18} \
dfrac{a+b}{ab} &=& dfrac7{18} \
dfrac{21}{ab} &=& dfrac7{18} \
dfrac{ab}{21} &=& dfrac{18}7 \
ab &=& 54
end{array}$$
answered May 6 '17 at 1:55
Kenny LauKenny Lau
19.9k2159
$endgroup$
add a comment |
$begingroup$
Just from a style point of view, I prefer to avoid fractions until the very end, so I'll just multiply both sides of the 2nd equation by $18ab$.
$$begin{align}
frac{1}{a}+frac{1}{b} &= frac{7}{18}\
18abbigg(frac{1}{a}+frac{1}{b}bigg) &= 18abbigg(frac{7}{18}bigg)\
frac{18ab}{a}+frac{18ab}{b} &= frac{7cdot18ab}{18}\
18b+18a &=7ab\
7ab &= 18(color{green}{a+b)}\
7ab &= 18cdot color{green}{21}\
ab & = frac{18cdot 21}{7} = frac{18cdot 3 cdot 7}{7} = 18cdot 3 = 54
end{align}$$
answered May 6 '17 at 10:59
John JoyJohn Joy
6,20611526
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Well, $$frac{7}{18}=frac{1}{a}+frac{1}{b}=frac{b+a}{ab}=frac{21}{ab}$$ and so $$ab=frac{21(18)}{7}=54$$
answered May 6 '17 at 1:59
ΘΣΦGenSanΘΣΦGenSan
7,93992346
$endgroup$
add a comment |
$begingroup$
Well, $$frac{7}{18}=frac{1}{a}+frac{1}{b}=frac{b+a}{ab}=frac{21}{ab}$$ and so $$ab=frac{21(18)}{7}=54$$
answered May 6 '17 at 1:59
ΘΣΦGenSanΘΣΦGenSan
7,93992346
$endgroup$
add a comment |
$begingroup$
Well, $$frac{7}{18}=frac{1}{a}+frac{1}{b}=frac{b+a}{ab}=frac{21}{ab}$$ and so $$ab=frac{21(18)}{7}=54$$
answered May 6 '17 at 1:59
ΘΣΦGenSanΘΣΦGenSan
7,93992346
$endgroup$
Well, $$frac{7}{18}=frac{1}{a}+frac{1}{b}=frac{b+a}{ab}=frac{21}{ab}$$ and so $$ab=frac{21(18)}{7}=54$$
answered May 6 '17 at 1:59
ΘΣΦGenSanΘΣΦGenSan
7,93992346
edited May 6 '17 at 2:23
answered May 6 '17 at 1:59
ΘΣΦGenSanΘΣΦGenSan
7,93992346
answered May 6 '17 at 1:59
ΘΣΦGenSanΘΣΦGenSan
7,93992346
answered May 6 '17 at 1:59
ΘΣΦGenSanΘΣΦGenSan
7,93992346
7,93992346
add a comment |
add a comment |
$begingroup$
$$begin{array}{rcl}
dfrac1a + dfrac1b &=& dfrac7{18} \
dfrac b{ab} + dfrac a{ab} &=& dfrac7{18} \
dfrac{b+a}{ab} &=& dfrac7{18} \
dfrac{a+b}{ab} &=& dfrac7{18} \
dfrac{21}{ab} &=& dfrac7{18} \
dfrac{ab}{21} &=& dfrac{18}7 \
ab &=& 54
end{array}$$
answered May 6 '17 at 1:55
Kenny LauKenny Lau
19.9k2159
$endgroup$
add a comment |
$begingroup$
$$begin{array}{rcl}
dfrac1a + dfrac1b &=& dfrac7{18} \
dfrac b{ab} + dfrac a{ab} &=& dfrac7{18} \
dfrac{b+a}{ab} &=& dfrac7{18} \
dfrac{a+b}{ab} &=& dfrac7{18} \
dfrac{21}{ab} &=& dfrac7{18} \
dfrac{ab}{21} &=& dfrac{18}7 \
ab &=& 54
end{array}$$
answered May 6 '17 at 1:55
Kenny LauKenny Lau
19.9k2159
$endgroup$
add a comment |
$begingroup$
$$begin{array}{rcl}
dfrac1a + dfrac1b &=& dfrac7{18} \
dfrac b{ab} + dfrac a{ab} &=& dfrac7{18} \
dfrac{b+a}{ab} &=& dfrac7{18} \
dfrac{a+b}{ab} &=& dfrac7{18} \
dfrac{21}{ab} &=& dfrac7{18} \
dfrac{ab}{21} &=& dfrac{18}7 \
ab &=& 54
end{array}$$
answered May 6 '17 at 1:55
Kenny LauKenny Lau
19.9k2159
$endgroup$
$$begin{array}{rcl}
dfrac1a + dfrac1b &=& dfrac7{18} \
dfrac b{ab} + dfrac a{ab} &=& dfrac7{18} \
dfrac{b+a}{ab} &=& dfrac7{18} \
dfrac{a+b}{ab} &=& dfrac7{18} \
dfrac{21}{ab} &=& dfrac7{18} \
dfrac{ab}{21} &=& dfrac{18}7 \
ab &=& 54
end{array}$$
answered May 6 '17 at 1:55
Kenny LauKenny Lau
19.9k2159
edited May 6 '17 at 2:03
answered May 6 '17 at 1:55
Kenny LauKenny Lau
19.9k2159
answered May 6 '17 at 1:55
Kenny LauKenny Lau
19.9k2159
answered May 6 '17 at 1:55
Kenny LauKenny Lau
19.9k2159
19.9k2159
add a comment |
add a comment |
$begingroup$
Just from a style point of view, I prefer to avoid fractions until the very end, so I'll just multiply both sides of the 2nd equation by $18ab$.
$$begin{align}
frac{1}{a}+frac{1}{b} &= frac{7}{18}\
18abbigg(frac{1}{a}+frac{1}{b}bigg) &= 18abbigg(frac{7}{18}bigg)\
frac{18ab}{a}+frac{18ab}{b} &= frac{7cdot18ab}{18}\
18b+18a &=7ab\
7ab &= 18(color{green}{a+b)}\
7ab &= 18cdot color{green}{21}\
ab & = frac{18cdot 21}{7} = frac{18cdot 3 cdot 7}{7} = 18cdot 3 = 54
end{align}$$
answered May 6 '17 at 10:59
John JoyJohn Joy
6,20611526
$endgroup$
add a comment |
$begingroup$
Just from a style point of view, I prefer to avoid fractions until the very end, so I'll just multiply both sides of the 2nd equation by $18ab$.
$$begin{align}
frac{1}{a}+frac{1}{b} &= frac{7}{18}\
18abbigg(frac{1}{a}+frac{1}{b}bigg) &= 18abbigg(frac{7}{18}bigg)\
frac{18ab}{a}+frac{18ab}{b} &= frac{7cdot18ab}{18}\
18b+18a &=7ab\
7ab &= 18(color{green}{a+b)}\
7ab &= 18cdot color{green}{21}\
ab & = frac{18cdot 21}{7} = frac{18cdot 3 cdot 7}{7} = 18cdot 3 = 54
end{align}$$
answered May 6 '17 at 10:59
John JoyJohn Joy
6,20611526
$endgroup$
add a comment |
$begingroup$
Just from a style point of view, I prefer to avoid fractions until the very end, so I'll just multiply both sides of the 2nd equation by $18ab$.
$$begin{align}
frac{1}{a}+frac{1}{b} &= frac{7}{18}\
18abbigg(frac{1}{a}+frac{1}{b}bigg) &= 18abbigg(frac{7}{18}bigg)\
frac{18ab}{a}+frac{18ab}{b} &= frac{7cdot18ab}{18}\
18b+18a &=7ab\
7ab &= 18(color{green}{a+b)}\
7ab &= 18cdot color{green}{21}\
ab & = frac{18cdot 21}{7} = frac{18cdot 3 cdot 7}{7} = 18cdot 3 = 54
end{align}$$
answered May 6 '17 at 10:59
John JoyJohn Joy
6,20611526
$endgroup$
Just from a style point of view, I prefer to avoid fractions until the very end, so I'll just multiply both sides of the 2nd equation by $18ab$.
$$begin{align}
frac{1}{a}+frac{1}{b} &= frac{7}{18}\
18abbigg(frac{1}{a}+frac{1}{b}bigg) &= 18abbigg(frac{7}{18}bigg)\
frac{18ab}{a}+frac{18ab}{b} &= frac{7cdot18ab}{18}\
18b+18a &=7ab\
7ab &= 18(color{green}{a+b)}\
7ab &= 18cdot color{green}{21}\
ab & = frac{18cdot 21}{7} = frac{18cdot 3 cdot 7}{7} = 18cdot 3 = 54
end{align}$$
answered May 6 '17 at 10:59
John JoyJohn Joy
6,20611526
answered May 6 '17 at 10:59
John JoyJohn Joy
6,20611526
answered May 6 '17 at 10:59
John JoyJohn Joy
6,20611526
answered May 6 '17 at 10:59
John JoyJohn Joy
6,20611526
6,20611526
add a comment |
add a comment |
3
$begingroup$
You already have 2 good answers (you should accept one). This is not an answer, just a comment - but this problem illustrates how, given exactly two of the three commonly used means of two numbers - i.e. the arithmetic (AM), geometric (GM) and harmonic mean (HM) - you can derive the third mean.
$endgroup$
– Deepak
May 6 '17 at 3:54