Problem:
A positive integer is 39 more than 18 times another. Their product is 13182. Find the two integers.
Solution Steps:
1. Define the variables:
Let \(x\) be the first integer and \(y\) be the second integer.
2. Set up the equations:
According to the problem, we have the following relationships:
\(x=18y+39\)
and
\(x.y=13182\)
3. Substitute the first equation into the second:
Substitute \(x=18y+39\) into the product equation:
\((18y+39).y=13182\)
4. Form a quadratic equation:
Expand and rearrange the equation to form a quadratic equation. To
make the calculations easier, divide the entire equation by 3:
\(18y^2+39y=13182\\\)
\(\text{or, }18y^2+39y-13182=0\)
\(\text{or, }6y^2+13y-4394=0\)
5. Solve the quadratic equation:
To solve the quadratic equation \(6y^2+13y-4394=0\), we can use the
quadratic formula:
\(y=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\)
where \(a=6,\ b=13\text{ and }c=-4394\)
Calculate the discriminant (Δ):
\(\Delta=b^2-4ac\\\Delta=13^2-4(6)(-4394)\\\Delta=169+105456\\\Delta=105625\)
Find the square root of the discriminant:
\(\sqrt{\Delta}=\sqrt{105625}=325\)
Now, apply the quadratic formula:
\(y=\dfrac{-13\pm325}{12}\)
This gives us two possible solutions for \(y\):
\(y=\dfrac{-13+325}{12}=\dfrac{312}{12}=26\)
\(y=\dfrac{-13-325}{12}=-\dfrac{338}{12}=-\dfrac{169}{6}\ \ \ \text{(Not an integer.)}\)
Since \(y\) is the positive integer, we take \(y=26\).
6. Find the corresponding value of \(x:\)
Using \(y=26\), we find \(x\) using the first equation:
\(x=18y+39\\x=18.26+39\\x=507\)
7. Verify the solution:
Verify the product:
\(x.y=507\times26=13182\)
Conclusion:
The two integers are \(x=507\) and \(y=26\)
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Final Answer:
The two positive integers are 507 and 26.
