The discussion focuses on solving the equations involving positive variables \(a\), \(b\), and \(c\) defined by the equations \(ab + ac = 518\), \(bc + ab = 468\), and \(ac + bc = 650\). Participants confirm the validity of the solution approach, emphasizing the importance of maintaining the condition \(a > 0\), \(b > 0\), and \(c > 0\). The goal is to determine the product \(a \times b \times c\) based on these equations.
PREREQUISITESMathematics students, educators, and anyone interested in solving algebraic equations involving multiple variables.
Albert said:$a>0,b>0,c>0$ and :
$ ab+ac=518-----(1)$
$ bc+ab=468-----(2)$
$ ac+bc=650-----(3)$
find :
$a\times b\times c=?$
very good!anemone said:My solution:
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 $abc=4200$.
Albert said:$a>0,b>0,c>0$ and :
$ ab+ac=518-----(1)$
$ bc+ab=468-----(2)$
$ ac+bc=650-----(3)$
find :
$a\times b\times c=?$
nice solution!greg1313 said:$$(3)-(2)\Rightarrow ac-ab=182\,(4)$$
$$(4)+(1)\Rightarrow2ac=700\Rightarrow ac=350$$
$$\Rightarrow bc=300$$
$$\Rightarrow ab=168$$
$$a^2b^2c^2=350\cdot300\cdot168=2\cdot5^2\cdot7\cdot2^2\cdot3\cdot5^2\cdot2^3\cdot3\cdot7$$
$$abc=2^3\cdot3\cdot5^2\cdot7=4200$$
nice job!kaliprasad said: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 $abc= 50 * 42 *2 = 4200$