Startseite » Forum » what is a/b equal to..

# what is a/b equal to..

## Tags: Mathematik

20:27 Uhr, 23.11.2022

Tag,
need help with this

I started with the original equation and took the reciprocal of both sides to solve for $b\phantom{\rule{0.12em}{0ex}}$.

$\frac{a\phantom{\rule{0.12em}{0ex}}}{b\phantom{\rule{0.12em}{0ex}}}=\frac{3}{5}$

$\frac{b\phantom{\rule{0.12em}{0ex}}}{a\phantom{\rule{0.12em}{0ex}}}=\frac{5}{3}$

$b\phantom{\rule{0.12em}{0ex}}=\frac{5a\phantom{\rule{0.12em}{0ex}}}{3}$

now I have the value of $b\phantom{\rule{0.12em}{0ex}},$ what can I do now,

Vielen Dank!

Für alle, die mir helfen möchten (automatisch von OnlineMathe generiert):
"Ich möchte die Lösung in Zusammenarbeit mit anderen erstellen."

21:00 Uhr, 23.11.2022

Hello,

now you can insert $\frac{5a\phantom{\rule{0.12em}{0ex}}}{3}$ into $\frac{b\phantom{\rule{0.12em}{0ex}}}{c\phantom{\rule{0.12em}{0ex}}}=\frac{7}{9}$:

$\frac{\frac{5a\phantom{\rule{0.12em}{0ex}}}{3}}{c\phantom{\rule{0.12em}{0ex}}}=\frac{7}{9}$

$\frac{5a\phantom{\rule{0.12em}{0ex}}}{3c\phantom{\rule{0.12em}{0ex}}}=\frac{7}{9}$

...

Solve the equation for $\frac{a\phantom{\rule{0.12em}{0ex}}}{c\phantom{\rule{0.12em}{0ex}}}$

Alternatively you can calculate $\frac{a\phantom{\rule{0.12em}{0ex}}}{b\phantom{\rule{0.12em}{0ex}}}\cdot \frac{b\phantom{\rule{0.12em}{0ex}}}{c\phantom{\rule{0.12em}{0ex}}}$, with the given values.

greetings,
pivot

21:29 Uhr, 23.11.2022

okay, let me try that, thanks.
$\frac{5a\phantom{\rule{0.12em}{0ex}}}{3c\phantom{\rule{0.12em}{0ex}}}=\frac{7}{9}$

now $i\phantom{\rule{0.12em}{0ex}}$ will multiply both sides by $3c\phantom{\rule{0.12em}{0ex}}$ to clear the a variable on the left side.

$3c\phantom{\rule{0.12em}{0ex}}\left(\frac{5a\phantom{\rule{0.12em}{0ex}}}{3c\phantom{\rule{0.12em}{0ex}}}\right)=3c\phantom{\rule{0.12em}{0ex}}\left(\frac{7}{9}\right)$

$5a\phantom{\rule{0.12em}{0ex}}=\frac{7c\phantom{\rule{0.12em}{0ex}}}{3}$

now $i\phantom{\rule{0.12em}{0ex}}$ will divide both sides by 5 to clear variable $a\phantom{\rule{0.12em}{0ex}},$

$\frac{\frac{5}{a\phantom{\rule{0.12em}{0ex}}}}{5}=\frac{\frac{7c\phantom{\rule{0.12em}{0ex}}}{3}}{5}$

$a\phantom{\rule{0.12em}{0ex}}=\frac{7c\phantom{\rule{0.12em}{0ex}}}{3}\cdot \frac{1}{5}$

$a\phantom{\rule{0.12em}{0ex}}=\frac{7c\phantom{\rule{0.12em}{0ex}}}{15}$

now that I have the value of a and $b\phantom{\rule{0.12em}{0ex}},$ what can I do to find the equivalent fraction to a over $b\phantom{\rule{0.12em}{0ex}}$
this is where I am really lost.

21:33 Uhr, 23.11.2022

Your transformation is right, but you have gone a little bit too far. Remember that is aksed for the value of $\frac{a\phantom{\rule{0.12em}{0ex}}}{c\phantom{\rule{0.12em}{0ex}}}$

Dividing your result by $c\phantom{\rule{0.12em}{0ex}}$ you get $\frac{a\phantom{\rule{0.12em}{0ex}}}{c\phantom{\rule{0.12em}{0ex}}}=\frac{7}{15}$

21:41 Uhr, 23.11.2022

Don't understand well,
do you mean this is the operation that $i\phantom{\rule{0.12em}{0ex}}$ need to do ?.
$\frac{a\phantom{\rule{0.12em}{0ex}}}{b\phantom{\rule{0.12em}{0ex}}}\cdot \frac{b\phantom{\rule{0.12em}{0ex}}}{c\phantom{\rule{0.12em}{0ex}}}$

$\frac{3}{5}=\frac{\frac{7c\phantom{\rule{0.12em}{0ex}}}{15}}{c\phantom{\rule{0.12em}{0ex}}}$

21:44 Uhr, 23.11.2022

or this one?

$\frac{3}{5}\cdot \frac{7}{9}=\frac{21}{45}$

21:46 Uhr, 23.11.2022

but they are asking what the value of $\frac{a\phantom{\rule{0.12em}{0ex}}}{c\phantom{\rule{0.12em}{0ex}}}$ is.

22:01 Uhr, 23.11.2022

>>Don't understand well, do you mean this is the operation that i need to do ? ...<<

This is one (second) approach. It is a little bit easier than your approach.
Have you notice that b is cancelling out?

>>or this one?<<

That is right. Finally you can cancel the fraction by $3$.

>>but they are asking what the value of a/c is<<
I know.

22:03 Uhr, 23.11.2022

Ohh, $i\phantom{\rule{0.12em}{0ex}}$ see it now,
you mean having the value of a then $i\phantom{\rule{0.12em}{0ex}}$ can substitute $a\phantom{\rule{0.12em}{0ex}}\in \frac{a\phantom{\rule{0.12em}{0ex}}}{c\phantom{\rule{0.12em}{0ex}}}$
and just simplifiy it to get $\frac{7}{15}$

$\frac{a\phantom{\rule{0.12em}{0ex}}}{c\phantom{\rule{0.12em}{0ex}}}=7\frac{c\phantom{\rule{0.12em}{0ex}}}{15}/c\phantom{\rule{0.12em}{0ex}}=\frac{7}{15}$

so $\frac{a\phantom{\rule{0.12em}{0ex}}}{c\phantom{\rule{0.12em}{0ex}}}=\frac{7}{15}$

is it correct?

thanks

22:38 Uhr, 23.11.2022

That's right.

22:46 Uhr, 23.11.2022

Thanks.

05:16 Uhr, 24.11.2022

my approach:

$\frac{a\phantom{\rule{0.12em}{0ex}}}{b\phantom{\rule{0.12em}{0ex}}}=\frac{3}{5}$
$a\phantom{\rule{0.12em}{0ex}}=\frac{3}{5}b\phantom{\rule{0.12em}{0ex}}$

$\frac{b\phantom{\rule{0.12em}{0ex}}}{c\phantom{\rule{0.12em}{0ex}}}=\frac{7}{9}$
$c\phantom{\rule{0.12em}{0ex}}=\frac{9}{7}b\phantom{\rule{0.12em}{0ex}}$
$\frac{\frac{3}{5}b\phantom{\rule{0.12em}{0ex}}}{\frac{7}{9}b\phantom{\rule{0.12em}{0ex}}}=\frac{3}{5}\cdot \frac{9}{7}=\frac{27}{35}$

05:36 Uhr, 24.11.2022

There are so many ways to skin a cat and as we seem to start to collect different ways, here are my two cents:

$1\right)$

(the point is to generate the same number at "equal" positions $\to$ least common multiple)
So we see $a\phantom{\rule{0.12em}{0ex}}:b\phantom{\rule{0.12em}{0ex}}:c\phantom{\rule{0.12em}{0ex}}=21:35:45$ and therefore $a\phantom{\rule{0.12em}{0ex}}:c\phantom{\rule{0.12em}{0ex}}=21:45=7:15$

$2\right)$
$\frac{a\phantom{\rule{0.12em}{0ex}}}{c\phantom{\rule{0.12em}{0ex}}}=\frac{a\phantom{\rule{0.12em}{0ex}}}{b\phantom{\rule{0.12em}{0ex}}}\cdot \frac{b\phantom{\rule{0.12em}{0ex}}}{c\phantom{\rule{0.12em}{0ex}}}=\frac{3}{5}\cdot \frac{7}{9}=\frac{1}{5}\cdot \frac{7}{3}=\frac{7}{15}$