n Part a of this problem, you found the values of integers that satisfied the system of equations. In this part of the problem you will write up a full solution that describes how to solve this problem. Four positive integers $a$, $b$, $c$, and $d$ satisfy \begin{align*} ab + a + b &= 524, \\ bc + b + c &= 146, \\ cd + c + d &= 104, \\ \end{align*} and $abcd=8!$. What are $a$, $b$, $c$, and $d?$

 Apr 20, 2018

First of all, we have the equations:


\(\begin{align*} ab + a + b &= 524, \\ bc + b + c &= 146, \\ cd + c + d &= 104, \\ \end{align*}\)


and abcd=8!.


From the first equation, we have: 


\(ab + a + b = 524.\)


After factoring out the b, we have:


\(b(1 + a)  =  524 - a.\)


Solving for b, we have:


\(b =  \frac{524 -a}{1 + a}\)


From the second equation, we are going to try to write c in the form of a.


\(bc + b + c =  146.\)


Factoring out the b, we have:




Solving for b, we have:




Since b = b, we have:


\(\frac{146-c}{1+c}=\frac{524 -a}{1 + a}.\)After cross-multiplying, we end up with:


\( (1+c)(524-a)=(1+a)(146-c).\)


Using the distributive property, we have:




Isolating the expressions with c's and simplifying the left side, we have:


\(c(524-a+1+a)=146 + 146a-524+a.\)


Finalizing, we have:




Using the same method, we solve for d, and we end up with:


\(d =   \frac{374 - a}{1+a}.\)


Since abcd=8!, we have:


\(a\cdot\frac{524 - a}{1 + a}\cdot\frac{7a-18}{25}\cdot\frac{374 - a}{1 +a}  = 40320.\)


After simplifying and factoring, we have:


\((a - 24) (7a^3 -6136a^2 +232732a + 42000)   = 0.\)


The only way a can be an integer is when a=24.


From this, we can plug in a to find the other values.


\(\boxed{a=24}. \boxed{b=20}. \boxed{c=6}. \boxed{d=14}.\)


I hope this helped! :)



 Apr 20, 2018

Nice, Gavin.....!!!



cool cool cool

CPhill  Apr 21, 2018

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