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Set up an algebraic equation and then solve the following problem.

1. An integer is 3 less than 5 times another. If the product of the two integers is 36, find the integers.


Sagot :

Sure, let's break this problem down step by step. We're given that one integer is 3 less than 5 times another integer, and we're also given that the product of these two integers is 36. We need to find these integers.

### Step 1: Define the Variables
Let's denote the two integers as [tex]\( x \)[/tex] and [tex]\( y \)[/tex].

### Step 2: Set Up the Equations
According to the problem statement:
1. One integer (let's call it [tex]\( y \)[/tex]) is 3 less than 5 times another integer (let's call it [tex]\( x \)[/tex]).
[tex]\[ y = 5x - 3 \][/tex]

2. The product of the two integers is 36.
[tex]\[ x \cdot y = 36 \][/tex]

### Step 3: Substitute the First Equation into the Second Equation
We substitute [tex]\( y \)[/tex] from the first equation into the second equation:
[tex]\[ x \cdot (5x - 3) = 36 \][/tex]

### Step 4: Formulate the Quadratic Equation
Expand the equation:
[tex]\[ 5x^2 - 3x = 36 \][/tex]

Bring all terms to one side to set the equation to zero:
[tex]\[ 5x^2 - 3x - 36 = 0 \][/tex]

### Step 5: Solve the Quadratic Equation
We solve the quadratic equation [tex]\( 5x^2 - 3x - 36 = 0 \)[/tex]. The solutions to this equation are found to be:
[tex]\[ x = -\frac{12}{5} \][/tex]
[tex]\[ x = 3 \][/tex]

### Step 6: Find the Corresponding [tex]\( y \)[/tex] Values
For each [tex]\( x \)[/tex], we find the corresponding [tex]\( y \)[/tex]:
1. If [tex]\( x = -\frac{12}{5} \)[/tex]:
[tex]\[ y = 5 \left(-\frac{12}{5}\right) - 3 = -12 - 3 = -15 \][/tex]

2. If [tex]\( x = 3 \)[/tex]:
[tex]\[ y = 5(3) - 3 = 15 - 3 = 12 \][/tex]

### Conclusion
So, the pairs of integers that satisfy the given conditions are:
1. [tex]\( x = -\frac{12}{5} \)[/tex] and [tex]\( y = -15 \)[/tex] (although [tex]\(-\frac{12}{5}\)[/tex] is not an integer, this would be the result of solving the equation).
2. [tex]\( x = 3 \)[/tex] and [tex]\( y = 12 \)[/tex]
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