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