x+y = 5, x-y = 1 find out (x^2+y^2) / xy
You asked:
Given the following equations: \[\begin{aligned}x + y &= 5\\ x - y &= 1\end{aligned}\] Evaluate \(\frac{{x}^{2} + {y}^{2}}{x y}\).
MathBot Answer:
The system of linear equations has exactly one solution.
The substitutions are \( x = 3 \), \( y = 2 \).
\[\frac{{x}^{2} + {y}^{2}}{x y} = \frac{13}{6}\]
Solve \(x + y = 5\) for \(x\). \[x = 5 - y\]
Substitute \(5 - y\) for \(x\) in \(x - y = 1\) and simplify. $$\begin{aligned}x - y &= 1 \\ \left(5 - y\right) - y &= 1 \\ y &= 2 \end{aligned}$$
Use substitution of the numerical value of \(y\) to get the values of \(x\). $$\begin{aligned}x &= 5 - y \\ x &= 5 - 2 \\ x &= 3\end{aligned}$$
By substituting these values into the expression, we find that:$$\begin{aligned} \frac{{x}^{2} + {y}^{2}}{x y} &= \frac{2^{2} + 3^{2}}{2 \cdot 3}\\ &=\frac{13}{6}\end{aligned}$$