+239 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 $d?$
+983 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:
\(b(c+1)+c=146.\)
Solving for b, we have:
\(b=\frac{146-c}{1+c}.\)
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:
\((524-a)+c(524-a)=146(1+a)-(524-a).\)
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:
\(c=\frac{7a-18}{25}.\)
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! :)
Gavin.
