\begin{array}{l}x^2+t^2=5 \\ \frac{1}{x^2}+\frac{1}{t^2}=\frac{5}{4}\end{array}
Answer (1)
Substitute u = x 2 and v = t 2 to simplify the equations.
Rewrite the second equation and substitute u + v = 5 to get uv = 4 .
Solve the system u + v = 5 and uv = 4 to find u = 1 , v = 4 or u = 4 , v = 1 .
Substitute back to find the solutions for x and t : ( 1 , 2 ) , ( 1 , − 2 ) , ( − 1 , 2 ) , ( − 1 , − 2 ) , ( 2 , 1 ) , ( 2 , − 1 ) , ( − 2 , 1 ) , ( − 2 , − 1 ) . The final answer is ( 1 , 2 ) , ( 1 , − 2 ) , ( − 1 , 2 ) , ( − 1 , − 2 ) , ( 2 , 1 ) , ( 2 , − 1 ) , ( − 2 , 1 ) , ( − 2 , − 1 ) .
Explanation
Analyze the problem We are given a system of two equations with two variables, x and t . Our goal is to find all pairs ( x , t ) that satisfy both equations simultaneously. The equations are:
x 2 + t 2 = 5 x 2 1 + t 2 1 = 4 5
We will use algebraic manipulation and substitution to solve this system.
Introduce new variables Let's introduce new variables to simplify the equations. Let u = x 2 and v = t 2 . Then the system becomes:
u + v = 5 u 1 + v 1 = 4 5
This substitution makes the equations easier to work with.
Rewrite the second equation We can rewrite the second equation as:
uv u + v = 4 5
Now, substitute the first equation, u + v = 5 , into this equation:
uv 5 = 4 5
This simplifies to:
uv = 4
Solve for u and v Now we have a new system of equations:
u + v = 5 uv = 4
We can solve for v in the first equation: v = 5 − u . Substitute this into the second equation:
u ( 5 − u ) = 4
Expanding and rearranging, we get a quadratic equation:
5 u − u 2 = 4 u 2 − 5 u + 4 = 0
Find the solutions for u We can factor the quadratic equation:
( u − 1 ) ( u − 4 ) = 0
So the solutions for u are u = 1 and u = 4 .
If u = 1 , then v = 5 − 1 = 4 .
If u = 4 , then v = 5 − 4 = 1 .
Substitute back and find x and t Now we substitute back x 2 for u and t 2 for v . We have two cases:
Case 1: x 2 = 1 and t 2 = 4 . This means x = ± 1 and t = ± 2 .
Case 2: x 2 = 4 and t 2 = 1 . This means x = ± 2 and t = ± 1 .
Therefore, the solutions are ( x , t ) = ( 1 , 2 ) , ( 1 , − 2 ) , ( − 1 , 2 ) , ( − 1 , − 2 ) , ( 2 , 1 ) , ( 2 , − 1 ) , ( − 2 , 1 ) , ( − 2 , − 1 ) .
State the final answer The solutions to the system of equations are:
( 1 , 2 ) , ( 1 , − 2 ) , ( − 1 , 2 ) , ( − 1 , − 2 ) , ( 2 , 1 ) , ( 2 , − 1 ) , ( − 2 , 1 ) , ( − 2 , − 1 )
These are all the pairs of ( x , t ) that satisfy both equations.
Examples
Systems of equations like this appear in various fields, such as physics, engineering, and economics. For instance, in circuit analysis, you might have two equations describing the relationships between voltage and current in different parts of a circuit. Solving the system helps you determine the specific values of voltage and current that satisfy all conditions. Similarly, in economics, you might have equations representing supply and demand, and solving the system gives you the equilibrium price and quantity. Understanding how to solve these systems is crucial for making predictions and designing systems effectively.
