We are tasked with finding a pair of integers \(\(x\)\) and \(\(y\)\) such that their product is 10 and their difference is 3. Let's denote these conditions mathematically:
1. \(\( x \cdot y = 10 \)\)
2. \(\( x - y = 3 \)\)
To solve this, we can use a system of equations approach.
### Step-by-Step Solution
1. Represent one variable in terms of the other using the second equation.
From the second equation:
\(\[
x - y = 3 \implies x = y + 3
\]\)
Now, we have \(\( x \)\) expressed in terms of \(\( y \)\).
2. Substitute \(\( x \)\) in the first equation.
Substitute \(\( x = y + 3 \)\) into \(\( x \cdot y = 10 \)\):
\(\[
(y + 3) \cdot y = 10
\]\)
Simplify to form a quadratic equation:
\(\[
y^2 + 3y = 10
\]\)
\(\[
y^2 + 3y - 10 = 0
\]\)
3. Solve the quadratic equation.
The standard form of a quadratic equation is \(\( ay^2 + by + c = 0 \)\), where \(\( a = 1 \)\), \(\( b = 3 \)\), and \(\( c = -10 \)\). The discriminant (\(\( \Delta \)\)) is given by:
\(\[
\Delta = b^2 - 4ac
\]\)
Plugging in the values:
\(\[
\Delta = 3^2 - 4 \cdot 1 \cdot (-10) = 9 + 40 = 49
\]\)
4. Find the roots of the quadratic equation using the quadratic formula.
The quadratic formula is:
\(\[
y = \frac{-b \pm \sqrt{\Delta}}{2a}
\]\)
Substitute \(\( a = 1 \)\), \(\( b = 3 \)\), and \(\( \Delta = 49 \)\):
\(\[
y = \frac{-3 \pm \sqrt{49}}{2 \cdot 1} = \frac{-3 \pm 7}{2}
\]\)
This gives us two solutions:
\(\[
y_1 = \frac{-3 + 7}{2} = \frac{4}{2} = 2
\]\)
\(\[
y_2 = \frac{-3 - 7}{2} = \frac{-10}{2} = -5
\]\)
5. Find the corresponding \(\( x \)\) values.
Using \(\( x = y + 3 \)\):
- When \(\( y = 2 \)\):
\(\[
x = 2 + 3 = 5
\]\)
- When \(\( y = -5 \)\):
\(\[
x = -5 + 3 = -2
\]\)
### Solutions
The pairs of integers that satisfy both conditions are:
1. \(\( (x, y) = (5, 2) \)\)
2. \(\( (x, y) = (-2, -5) \)\)
Thus, the pairs of integers whose product is 10 and whose difference is 3 are:
\(\[ (5, 2) \text{ and } (-2, -5) \]\)
