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Magazine Problem 98
Because variables appear in denominators, the system $$\begin{aligned}&\frac{5}{x}+\frac{15}{y}=16\\\&\frac{5}{x}+\frac{4}{y}=5\end{aligned}$$ is not a linear system. However, we can solve it in a manner similar to the method for solving a linear system by using a substitution-of-variable technique. Let \(t=\frac{1}{x}\) and let \(u=\frac{1}{y} \). Solve the system in Exercise 97 for \(t\) and \(u\).
Problem 99
Because variables appear in denominators, the system $$\begin{aligned}&\frac{5}{x}+\frac{15}{y}=16\\\&\frac{5}{x}+\frac{4}{y}=5\end{aligned}$$ is not a linear system. However, we can solve it in a manner similar to the method for solving a linear system by using a substitution-of-variable technique. Let \(t=\frac{1}{x}\) and let \(u=\frac{1}{y} \). Solve the given system for \(x\) and \(y\) by using the equations relating \(t\) to \(x\) and \(u\) to \(y\).
Problem 100
Because variables appear in denominators, the system $$\begin{aligned}&\frac{5}{x}+\frac{15}{y}=16\\\&\frac{5}{x}+\frac{4}{y}=5\end{aligned}$$ is not a linear system. However, we can solve it in a manner similar to the method for solving a linear system by using a substitution-of-variable technique. Let \(t=\frac{1}{x}\) and let \(u=\frac{1}{y} \). Refer to the first equation in the given system, and solve for \(y\) in terms of \(x\) to obtain a rational function.
Problem 101
Because variables appear in denominators, the system $$\begin{aligned}&\frac{5}{x}+\frac{15}{y}=16\\\&\frac{5}{x}+\frac{4}{y}=5\end{aligned}$$ is not a linear system. However, we can solve it in a manner similar to the method for solving a linear system by using a substitution-of-variable technique. Let \(t=\frac{1}{x}\) and let \(u=\frac{1}{y} \). Repeat Exercise 100 for the second equation in the given system.
