$\frac{x}{t}+\frac{t}{x}=\frac{17}{4}, x^2-4 x t+t^2=1$
Answer (1)
Substitute u = t x into the first equation and solve for u , obtaining u = 4 and u = 4 1 .
For u = 4 , substitute x = 4 t into the second equation to find t = ± 1 , giving solutions ( 4 , 1 ) and ( − 4 , − 1 ) .
For u = 4 1 , substitute x = 4 1 t into the second equation to find t = ± 4 , giving solutions ( 1 , 4 ) and ( − 1 , − 4 ) .
The solutions to the system of equations are ( 4 , 1 ) , ( − 4 , − 1 ) , ( 1 , 4 ) , ( − 1 , − 4 ) .
Explanation
Problem Analysis We are given the equations t x + x t = 4 17 x 2 − 4 x t + t 2 = 1 Our objective is to solve this system of equations for x and t .
Substitution Let's introduce a new variable u = t x . Then the first equation can be rewritten as u + u 1 = 4 17 Multiplying both sides by 4 u , we get 4 u 2 + 4 = 17 u 4 u 2 − 17 u + 4 = 0 We can solve this quadratic equation for u .
Solving for u The solutions for the quadratic equation 4 u 2 − 17 u + 4 = 0 are u = 4 and u = 4 1 .
Case 1: x/t = 4 Case 1: u = t x = 4 , so x = 4 t . Substituting x = 4 t into the second equation, we have ( 4 t ) 2 − 4 ( 4 t ) t + t 2 = 1 16 t 2 − 16 t 2 + t 2 = 1 t 2 = 1 Thus, t = ± 1 . If t = 1 , then x = 4 ( 1 ) = 4 . If t = − 1 , then x = 4 ( − 1 ) = − 4 .
Case 2: x/t = 1/4 Case 2: u = t x = 4 1 , so x = 4 1 t . Substituting x = 4 1 t into the second equation, we have ( 4 1 t ) 2 − 4 ( 4 1 t ) t + t 2 = 1 16 1 t 2 − t 2 + t 2 = 1 16 1 t 2 = 1 Thus, t 2 = 16 , so t = ± 4 . If t = 4 , then x = 4 1 ( 4 ) = 1 . If t = − 4 , then x = 4 1 ( − 4 ) = − 1 .
Final Solutions Therefore, the solutions are ( x , t ) = ( 4 , 1 ) , ( − 4 , − 1 ) , ( 1 , 4 ) , ( − 1 , − 4 ) .
Examples
This system of equations can be used to model relationships between two variables where their ratio and a quadratic relationship involving them are known. For example, in electrical circuits, x and t could represent current and voltage, where the first equation describes their proportional relationship and the second equation represents a specific energy constraint. Solving such systems helps engineers determine specific operating points that satisfy both the proportional requirement and the energy constraint.
