Solving for $a\times b \times c$: Equations (1), (2), and (3)

[Albert1]

$a>0,b>0,c>0$ and :
$ ab+ac=518-----(1)$
$ bc+ab=468-----(2)$
$ ac+bc=650-----(3)$
find :
$a\times b\times c=?$

 

[anemone]

$a>0,b>0,c>0$ and :
$ ab+ac=518-----(1)$
$ bc+ab=468-----(2)$
$ ac+bc=650-----(3)$
find :
$a\times b\times c=?$

My solution:

Spoiler

Note that $650=13(50)$ so we get $ac+bc=13(ac-bc)$ which simplifies to $c(14b-12a)=0$, since $c>0$, we get $\dfrac{a}{b}=\dfrac{14}{12}$.

Also note that $650-148=182$ and $\dfrac{182}{468}=\dfrac{7}{18}$. From this we get $\dfrac{ac-ab}{bc+ab}=\dfrac{7}{18}$, i.e. $\dfrac{a(c-b)}{b(c+a)}=\dfrac{7}{18}$ or $\dfrac{14}{12}\left(\dfrac{c-b}{c+\dfrac{14b}{12}}\right)=\dfrac{7}{18}$, which gives $\dfrac{b}{c}=\dfrac{12}{25}$.

So $a,\,b$ and $c$ are in the ratio $14:12:25$ and a check shows that $a=14$, $b=12$ and $c=25$ satisfy the given system and so $abc=4200$.

 

[Albert1]

My solution:

Spoiler

Note that $650=13(50)$ so we get $ac+bc=13(ac-bc)$ which simplifies to $c(14b-12a)=0$, since $c>0$, we get $\dfrac{a}{b}=\dfrac{14}{12}$.

Also note that $650-148=182$ and $\dfrac{182}{468}=\dfrac{7}{18}$. From this we get $\dfrac{ac-ab}{bc+ab}=\dfrac{7}{18}$, i.e. $\dfrac{a(c-b)}{b(c+a)}=\dfrac{7}{18}$ or $\dfrac{14}{12}\left(\dfrac{c-b}{c+\dfrac{14b}{12}}\right)=\dfrac{7}{18}$, which gives $\dfrac{b}{c}=\dfrac{12}{25}$.

So $a,\,b$ and $c$ are in the ratio $14:12:25$ and a check shows that $a=14$, $b=12$ and $c=25$ satisfy the given system and so $abc=4200$.

very good!

 

[Greg]

$a>0,b>0,c>0$ and :
$ ab+ac=518-----(1)$
$ bc+ab=468-----(2)$
$ ac+bc=650-----(3)$
find :
$a\times b\times c=?$

Spoiler

\(\displaystyle (3)-(2)\Rightarrow ac-ab=182\,(4)\)

\(\displaystyle (4)+(1)\Rightarrow2ac=700\Rightarrow ac=350\)

\(\displaystyle \Rightarrow bc=300\)

\(\displaystyle \Rightarrow ab=168\)

\(\displaystyle a^2b^2c^2=350\cdot300\cdot168=2\cdot5^2\cdot7\cdot2^2\cdot3\cdot5^2\cdot2^3\cdot3\cdot7\)

\(\displaystyle abc=2^3\cdot3\cdot5^2\cdot7=4200\)

 

[Albert1]

Spoiler

\(\displaystyle (3)-(2)\Rightarrow ac-ab=182\,(4)\)

\(\displaystyle (4)+(1)\Rightarrow2ac=700\Rightarrow ac=350\)

\(\displaystyle \Rightarrow bc=300\)

\(\displaystyle \Rightarrow ab=168\)

\(\displaystyle a^2b^2c^2=350\cdot300\cdot168=2\cdot5^2\cdot7\cdot2^2\cdot3\cdot5^2\cdot2^3\cdot3\cdot7\)

\(\displaystyle abc=2^3\cdot3\cdot5^2\cdot7=4200\)

nice solution!

 

[kaliprasad]

Spoiler

add all three to get
$2(ab+bc+ca) = 1636$
or $ab+bc+ca = 818$
subtracting the 3 given equations we get
$bc= 300,ca=350,ab= 168$
multiply above 3 to get
$(abc)^2= 300 * 350 * 168 = 50 * 6 * 50 * 7 * 42 * 4 = 50^2 * 42^2 * 2^2$
or $abc= 50 * 42 *2 = 4200$

 

[Albert1]

Spoiler

add all three to get
$2(ab+bc+ca) = 1636$
or $ab+bc+ca = 818$
subtracting the 3 given equations we get
$bc= 300,ca=350,ab= 168$
multiply above 3 to get
$(abc)^2= 300 * 350 * 168 = 50 * 6 * 50 * 7 * 42 * 4 = 50^2 * 42^2 * 2^2$
or $abc= 50 * 42 *2 = 4200$

nice job!