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need to find out why a+b\b = c+d\d using properties found in
- Thread starter jay34989
- Start date
Re: I need help on how this works.
(a+b)\b= (c+d)\d
Loren said:a+b\b = c+d\d means \(\displaystyle a+\frac{b}{b}=c+\frac{d}{d}\). I don't think that's what you mean. Please clarify by using grouping symbols.
(a+b)\b= (c+d)\d
Re: I need help on how this works.
If \(\displaystyle \frac{a}{b}=\frac{c}{d}\)
\(\displaystyle ad = bc\) <<< The product of the means is equal to the product of the extremes.
\(\displaystyle ad + bd = bc + bd\) <<< Add bd to both sides of the equation.
\(\displaystyle d(a+b) = b(c+d)\) <<< factor.
\(\displaystyle \frac{d(a+b)}{bd}=\frac{d(c+d)}{bd}\) <<< Divide both sides by bd.
\(\displaystyle \frac{a+b}{b}=\frac{c+d}{d}\) <<<Reduce fractions.
If \(\displaystyle \frac{a}{b}=\frac{c}{d}\)
\(\displaystyle ad = bc\) <<< The product of the means is equal to the product of the extremes.
\(\displaystyle ad + bd = bc + bd\) <<< Add bd to both sides of the equation.
\(\displaystyle d(a+b) = b(c+d)\) <<< factor.
\(\displaystyle \frac{d(a+b)}{bd}=\frac{d(c+d)}{bd}\) <<< Divide both sides by bd.
\(\displaystyle \frac{a+b}{b}=\frac{c+d}{d}\) <<<Reduce fractions.
Re: I need help on how this works.
Hello, jay34989!
Your statement needs some explanation.
But I recognize it from some studies on ratios and proportions,
. . so I think I know what you're asking.
There is a theorem that says:
. . \(\displaystyle \text{If }\,\frac{a}{b} \:=\:\frac{c}{d},\:\text{ then: }\:\frac{a+b}{b} \:=\:\frac{c+d}{d}\)
The proof is quite simple . . .
\(\displaystyle \text{We have: }\:\frac{a}{b} \:=\:\frac{c}{d}\)
\(\displaystyle \text{Add 1 to both sides: }\:\frac{a}{b} + 1 \:=\:\frac{c}{d}+1\)
\(\displaystyle \text{Then we have: }\:\frac{a+b}{b} \:=\:\frac{c+d}{d}\)
Hello, jay34989!
Your statement needs some explanation.
But I recognize it from some studies on ratios and proportions,
. . so I think I know what you're asking.
There is a theorem that says:
. . \(\displaystyle \text{If }\,\frac{a}{b} \:=\:\frac{c}{d},\:\text{ then: }\:\frac{a+b}{b} \:=\:\frac{c+d}{d}\)
The proof is quite simple . . .
\(\displaystyle \text{We have: }\:\frac{a}{b} \:=\:\frac{c}{d}\)
\(\displaystyle \text{Add 1 to both sides: }\:\frac{a}{b} + 1 \:=\:\frac{c}{d}+1\)
\(\displaystyle \text{Then we have: }\:\frac{a+b}{b} \:=\:\frac{c+d}{d}\)
D
Deleted member 4993
Guest
Re: I need help on how this works.
An interesting extension is:
If
\(\displaystyle \frac{a}{b} \, = \, \frac{c}{d}\)
then
\(\displaystyle \frac{a+b}{a-b} \, = \, \frac{c+d}{c-d}\)
An interesting extension is:
If
\(\displaystyle \frac{a}{b} \, = \, \frac{c}{d}\)
then
\(\displaystyle \frac{a+b}{a-b} \, = \, \frac{c+d}{c-d}\)