+0

help

+1
302
2

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

#1
+4

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.

Apr 20, 2018
#2
0

Nice, Gavin.....!!!   CPhill  Apr 21, 2018