$\begin{array}{l}x^2+t^2=10 \\ \frac{1}{x}+\frac{1}{t}=\frac{4}{3}\end{array}$
Answer (2)
The equations lead to solutions: (x, t) = (1, 3), (3, 1), (\frac{-5 + \sqrt{55}}{4}, \frac{-5 - \sqrt{55}}{4}), and (\frac{-5 - \sqrt{55}}{4}, \frac{-5 + \sqrt{55}}{4}). The method involves eliminating fractions and using substitutions to create a quadratic equation in two variables. Finally, applying the quadratic formula yields the solutions for x and t.
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Rewrite the second equation to eliminate fractions and express it in terms of x + t and x t .
Substitute u = x + t and v = x t to simplify the equations.
Solve the resulting quadratic equation for u , and then find the corresponding values for v .
Solve for x and t using the quadratic formula or factoring, based on the values of u and v , resulting in the solutions: ( x , t ) = ( 1 , 3 ) , ( 3 , 1 ) , ( 4 − 5 + 55 , 4 − 5 − 55 ) , ( 4 − 5 − 55 , 4 − 5 + 55 ) .
Explanation
Problem Analysis We are given a system of two equations with two variables, x and t :
x 2 + t 2 = 10 x 1 + t 1 = 3 4
Our goal is to solve for x and t .
Rewriting the Equations First, let's rewrite the second equation to eliminate the fractions:
x 1 + t 1 = 3 4 ⇒ x t x + t = 3 4 ⇒ 3 ( x + t ) = 4 x t
Now we have two equations:
x 2 + t 2 = 10 3 ( x + t ) = 4 x t
Let u = x + t and v = x t . Then our equations become:
x 2 + t 2 = 10 3 u = 4 v
We also know that ( x + t ) 2 = x 2 + 2 x t + t 2 , so u 2 = x 2 + t 2 + 2 x t = 10 + 2 v .
Creating a Quadratic Equation Now we have a system of equations in terms of u and v :
u 2 = 10 + 2 v 3 u = 4 v
We can solve for v in the second equation: v = 4 3 u . Substituting this into the first equation, we get:
u 2 = 10 + 2 ( 4 3 u ) ⇒ u 2 = 10 + 2 3 u ⇒ u 2 − 2 3 u − 10 = 0
This is a quadratic equation in u .
Solving for u To solve the quadratic equation u 2 − 2 3 u − 10 = 0 , we can use the quadratic formula:
u = 2 a − b ± b 2 − 4 a c
where a = 1 , b = − 2 3 , and c = − 10 . Plugging in these values, we get:
u = 2 ( 1 ) 2 3 ± ( − 2 3 ) 2 − 4 ( 1 ) ( − 10 ) = 2 2 3 ± 4 9 + 40 = 2 2 3 ± 4 169 = 2 2 3 ± 2 13
So, the two possible values for u are:
u 1 = 2 2 3 + 2 13 = 2 2 16 = 2 8 = 4
u 2 = 2 2 3 − 2 13 = 2 2 − 10 = 2 − 5 = − 2.5
Thus, u = 4 or u = − 2.5 .
Solving for v Now we can find the corresponding values for v using v = 4 3 u :
If u = 4 , then v = 4 3 ( 4 ) = 3 .
If u = − 2.5 , then v = 4 3 ( − 2.5 ) = − 1.875 .
So we have two pairs of values for u and v : ( 4 , 3 ) and ( − 2.5 , − 1.875 ) .
Solving for x and t Now we need to find x and t . We know that x + t = u and x t = v . This means that x and t are the roots of the quadratic equation z 2 − u z + v = 0 . Let's solve for x and t for each pair of ( u , v ) .
Case 1: u = 4 and v = 3 . The quadratic equation is z 2 − 4 z + 3 = 0 . Factoring this equation, we get ( z − 1 ) ( z − 3 ) = 0 . So the roots are z = 1 and z = 3 . Thus, x = 1 and t = 3 or x = 3 and t = 1 .
Case 2: u = − 2.5 and v = − 1.875 . The quadratic equation is z 2 + 2.5 z − 1.875 = 0 . Using the quadratic formula:
z = 2 − 2.5 ± ( 2.5 ) 2 − 4 ( 1 ) ( − 1.875 ) = 2 − 2.5 ± 6.25 + 7.5 = 2 − 2.5 ± 13.75 = 2 − 2.5 ± 4 55 = 2 − 2.5 ± 2 55 = 4 − 5 ± 55
So the roots are z = 4 − 5 + 55 and z = 4 − 5 − 55 . Thus, x = 4 − 5 + 55 and t = 4 − 5 − 55 or x = 4 − 5 − 55 and t = 4 − 5 + 55 .
Final Answer Therefore, the solutions are:
( x , t ) = ( 1 , 3 ) , ( 3 , 1 ) , ( 4 − 5 + 55 , 4 − 5 − 55 ) , ( 4 − 5 − 55 , 4 − 5 + 55 )
Examples
Systems of equations are used extensively in engineering to model and analyze circuits, structures, and control systems. For instance, when designing a bridge, engineers use systems of equations to calculate the forces and stresses acting on different parts of the structure, ensuring it can withstand various loads and environmental conditions. Similarly, in electrical engineering, systems of equations are used to analyze complex circuits and determine the flow of current and voltage at different points, which is crucial for designing efficient and reliable electronic devices.
