To find the values of [tex]\(x - y\)[/tex] given the equations [tex]\(x + y = 10\)[/tex] and [tex]\(\frac{1}{x} + \frac{1}{y} = \frac{5}{12}\)[/tex], follow these steps:
1. Set Up the System of Equations:
[tex]\[
x + y = 10 \quad \text{(Equation 1)}
\][/tex]
[tex]\[
\frac{1}{x} + \frac{1}{y} = \frac{5}{12} \quad \text{(Equation 2)}
\][/tex]
2. Express [tex]\(\frac{1}{x} + \frac{1}{y}\)[/tex] in Terms of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[
\frac{1}{x} + \frac{1}{y} = \frac{x + y}{xy}
\][/tex]
Substituting the value from Equation 1:
[tex]\[
\frac{x + y}{xy} = \frac{10}{xy}
\][/tex]
Setting this equal to the given value from Equation 2:
[tex]\[
\frac{10}{xy} = \frac{5}{12}
\][/tex]
Solve for [tex]\(xy\)[/tex]:
[tex]\[
10 \cdot 12 = 5 \cdot xy
\][/tex]
[tex]\[
120 = 5xy
\][/tex]
[tex]\[
xy = \frac{120}{5} = 24
\][/tex]
3. Form a Quadratic Equation Using [tex]\(x + y\)[/tex] and [tex]\(xy\)[/tex]:
[tex]\[
t^2 - (x + y)t + xy = 0
\][/tex]
Substituting [tex]\(x + y = 10\)[/tex] and [tex]\(xy = 24\)[/tex]:
[tex]\[
t^2 - 10t + 24 = 0
\][/tex]
4. Solve the Quadratic Equation:
Use the quadratic formula:
[tex]\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, [tex]\(a = 1\)[/tex], [tex]\(b = -10\)[/tex], and [tex]\(c = 24\)[/tex]:
[tex]\[
t = \frac{10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot 24}}{2 \cdot 1}
\][/tex]
[tex]\[
t = \frac{10 \pm \sqrt{100 - 96}}{2}
\][/tex]
[tex]\[
t = \frac{10 \pm \sqrt{4}}{2}
\][/tex]
[tex]\[
t = \frac{10 \pm 2}{2}
\][/tex]
Therefore, the solutions are:
[tex]\[
t = \frac{10 + 2}{2} = 6
\][/tex]
[tex]\[
t = \frac{10 - 2}{2} = 4
\][/tex]
Thus, [tex]\(x = 6\)[/tex] and [tex]\(y = 4\)[/tex], or [tex]\(x = 4\)[/tex] and [tex]\(y = 6\)[/tex].
5. Find [tex]\(x - y\)[/tex] for Each Pair:
For [tex]\(x = 6\)[/tex] and [tex]\(y = 4\)[/tex]:
[tex]\[
x - y = 6 - 4 = 2
\][/tex]
For [tex]\(x = 4\)[/tex] and [tex]\(y = 6\)[/tex]:
[tex]\[
x - y = 4 - 6 = -2
\][/tex]
6. Conclusion:
The possible values of [tex]\(x - y\)[/tex] are [tex]\(\boxed{2 \ \text{and} \ -2}\)[/tex].
