Certainly! Let's solve the problem step-by-step.
1. Identify the equations given:
We're given two pieces of information:
- Two numbers differ by 3.
- The sum of their reciprocals is \(\frac{7}{10}\).
2. Assign variables:
Let the two numbers be \( x \) and \( y \).
3. Set up the equations:
From the problem statements:
- The difference between the numbers: \( x - y = 3 \) (or equivalently, \( y - x = 3 \)).
- The sum of the reciprocals: \(\frac{1}{x} + \frac{1}{y} = \frac{7}{10}\).
4. Manipulate and solve the system of equations:
We will use the first equation \( x - y = 3 \).
The sum of the reciprocals equation is:
[tex]\[
\frac{1}{x} + \frac{1}{y} = \frac{7}{10}
\][/tex]
Let's express this using a common denominator:
[tex]\[
\frac{y + x}{xy} = \frac{7}{10}
\][/tex]
From \( x - y = 3 \), solve for \( x \) in terms of \( y \):
[tex]\[
x = y + 3
\][/tex]
Substitute \( x = y + 3 \) into the reciprocal equation:
[tex]\[
\frac{y + (y + 3)}{y(y + 3)} = \frac{7}{10}
\][/tex]
Simplify the numerator:
[tex]\[
\frac{2y + 3}{y(y + 3)} = \frac{7}{10}
\][/tex]
Cross multiply to clear the fraction:
[tex]\[
10(2y + 3) = 7y(y + 3)
\][/tex]
Distribute and simplify:
[tex]\[
20y + 30 = 7y^2 + 21y
\][/tex]
Move all terms to one side to form a quadratic equation:
[tex]\[
7y^2 + 21y - 20y - 30 = 0
\][/tex]
[tex]\[
7y^2 + y - 30 = 0
\][/tex]
5. Solve the quadratic equation:
Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 7 \), \( b = 1 \), and \( c = -30 \):
[tex]\[
y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 7 \cdot (-30)}}{2 \cdot 7}
\][/tex]
[tex]\[
y = \frac{-1 \pm \sqrt{1 + 840}}{14}
\][/tex]
[tex]\[
y = \frac{-1 \pm \sqrt{841}}{14}
\][/tex]
[tex]\[
y = \frac{-1 \pm 29}{14}
\][/tex]
This gives us two values for \( y \):
[tex]\[
y = \frac{28}{14} = 2 \quad \text{or} \quad y = \frac{-30}{14} = -2.142857
\][/tex]
6. Find corresponding \( x \) values:
Use \( x = y + 3 \) for each value of \( y \):
- For \( y = 2 \):
[tex]\[
x = 2 + 3 = 5
\][/tex]
- For \( y = -2.142857 \):
[tex]\[
x = -2.142857 + 3 = 0.857142857
\][/tex]
7. Conclusion:
The two number pairs that satisfy the conditions are approximately:
- \( x = 5 \) and \( y = 2 \)
- [tex]\( x = 0.857142857 \)[/tex] and [tex]\( y = -2.142857 \)[/tex]
