[tex]\begin{array}{l}x^2+t^2=10 \\ \frac{1}{x}+\frac{1}{t}=4 / 3\end{array}[/tex]
Answer (2)
Rewrite the second equation to express the relationship between x and t .
Introduce new variables u = x + t and v = x t to simplify the equations.
Form a quadratic equation in terms of u and solve for u .
Find the corresponding values of v and solve for x and t . The solutions are ( 1 , 3 ) , ( 3 , 1 ) .
Explanation
Analyze the problem We are given the following system of equations: x 2 + t 2 = 10 x 1 + t 1 = 3 4 Our goal is to find the values of x and t that satisfy both equations.
Rewrite the second equation First, let's rewrite the second equation: x 1 + t 1 = x t x + t = 3 4 This implies: 3 ( x + t ) = 4 x t
Introduce new variables Now, let u = x + t and v = x t . We can rewrite the equation above as: 3 u = 4 v v = 4 3 u
Use the first equation We also know that: ( x + t ) 2 = x 2 + 2 x t + t 2 u 2 = x 2 + t 2 + 2 x t Since x 2 + t 2 = 10 , we have: u 2 = 10 + 2 v
Form a quadratic equation Substitute v = 4 3 u into the equation u 2 = 10 + 2 v :
u 2 = 10 + 2 ( 4 3 u ) u 2 = 10 + 2 3 u Rearrange the equation to get a quadratic equation in u :
u 2 − 2 3 u − 10 = 0
Solve for u Solve the quadratic equation for u using the quadratic formula: u = 2 a − b ± b 2 − 4 a c where a = 1 , b = − 2 3 , and c = − 10 .
u = 2 ( 1 ) 2 3 ± ( − 2 3 ) 2 − 4 ( 1 ) ( − 10 ) u = 2 2 3 ± 4 9 + 40 u = 2 2 3 ± 4 9 + 160 u = 2 2 3 ± 4 169 u = 2 2 3 ± 2 13 So, we have two possible values for u :
u 1 = 2 2 3 − 2 13 = 2 − 2 10 = 2 − 5 = − 2.5 u 2 = 2 2 3 + 2 13 = 2 2 16 = 2 8 = 4
Solve for v Now we find the corresponding values for v using v = 4 3 u :
For u 1 = − 2.5 :
v 1 = 4 3 ( − 2.5 ) = − 4 7.5 = − 1.875 For u 2 = 4 :
v 2 = 4 3 ( 4 ) = 3
Form quadratic equations for z Now we have x + t = u and x t = v . We can express x and t as the roots of a quadratic equation z 2 − u z + v = 0 .
For u 1 = − 2.5 and v 1 = − 1.875 :
z 2 − ( − 2.5 ) z + ( − 1.875 ) = 0 z 2 + 2.5 z − 1.875 = 0 For u 2 = 4 and v 2 = 3 :
z 2 − 4 z + 3 = 0
Solve for z Solving z 2 + 2.5 z − 1.875 = 0 , we find the roots to be approximately z 1 ≈ − 3.104 and z 2 ≈ 0.604 .
Solving z 2 − 4 z + 3 = 0 , we can factor it as ( z − 1 ) ( z − 3 ) = 0 , so the roots are z 3 = 1 and z 4 = 3 .
Find the solutions Therefore, the solutions for ( x , t ) are approximately ( − 3.104 , 0.604 ) and ( 1 , 3 ) or ( 3 , 1 ) .
Let's verify the solutions: For ( 1 , 3 ) :
1 2 + 3 2 = 1 + 9 = 10 1 1 + 3 1 = 3 4 For ( 3 , 1 ) :
3 2 + 1 2 = 9 + 1 = 10 3 1 + 1 1 = 3 4 For ( − 3.104 , 0.604 ) :
( − 3.104 ) 2 + ( 0.604 ) 2 ≈ 9.635 + 0.365 ≈ 10 − 3.104 1 + 0.604 1 ≈ − 0.322 + 1.656 ≈ 1.334 ≈ 3 4 Thus, the solutions are ( 1 , 3 ) and ( 3 , 1 ) .
State the final answer The solutions to the system of equations are ( x , t ) = ( 1 , 3 ) and ( x , t ) = ( 3 , 1 ) .
Examples
Systems of equations are used in various fields, such as physics, engineering, and economics, to model and solve problems involving multiple variables and constraints. For example, in circuit analysis, systems of equations are used to determine the currents and voltages in different parts of a circuit. In economics, they can model supply and demand curves to find equilibrium prices and quantities. Understanding how to solve systems of equations is crucial for analyzing and predicting the behavior of these systems. For instance, consider a simple economic model where the quantity demanded ( Q d ) and the quantity supplied ( Q s ) are given by: Q d = 10 − 2 P Q s = 3 P where P is the price. To find the equilibrium price and quantity, we set Q d = Q s and solve the resulting system of equations.
By rewriting the equations and introducing new variables, we transformed the system into manageable quadratic equations. The solutions for the values of (x, t) are (1, 3) and (3, 1), both satisfying the given equations. Therefore, the pairs of (x, t) are valid solutions to the system.
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