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Magazine Chapter 10: Problem 28
Solve each system of equations by substitution for real values of \(x\) and \(y.\) See Examples 2 and 3. $$ \left\\{\begin{array}{l} x^{2}+y^{2}=10 \\ y=3 x^{2} \end{array}\right. $$
Short Answer
Expert verified
The solutions are \(\left( \frac{\sqrt{5}}{3}, \frac{5}{3} \right)\) and \(\left( -\frac{\sqrt{5}}{3}, \frac{5}{3} \right)\).
Step by step solution
01
Identify What to Substitute
Look at the system of equations:1. \(x^2 + y^2 = 10\)2. \(y = 3x^2\)In this case, equation (2) already isolates \(y\) as \(3x^2\). This means we can substitute \(3x^2\) for \(y\) in equation (1).
02
Substitute and Simplify
Substitute \(y = 3x^2\) into the first equation:\[x^2 + (3x^2)^2 = 10\]Now simplify the equation:\[x^2 + 9x^4 = 10\]
03
Rearrange to Form a Polynomial Equation
Rearrange the equation obtained in Step 2:\[9x^4 + x^2 - 10 = 0\]
04
Factor the Polynomial
Factor the polynomial equation if possible. This step might require some trial and error or use of factoring techniques. Here, we try:\[(9x^4 + x^2 - 10) = (x^2 + 2)(9x^2 - 5) = 0\]
05
Solve the Factored Equations
From the factorization, we have two simpler equations:\[x^2 + 2 = 0\] and \[9x^2 - 5 = 0\].**1. Solve \(x^2 + 2 = 0\):**This equation has no real solutions because \(x^2 = -2\) which is not possible with real numbers.**2. Solve \(9x^2 - 5 = 0\):**\[9x^2 = 5\]\[x^2 = \frac{5}{9}\]\[x = \pm \frac{\sqrt{5}}{3}\]
06
Solve for y using x-values
Use the equation \(y = 3x^2\) to find \(y\) for each value of \(x\):For \(x = \frac{\sqrt{5}}{3}\):\[y = 3 \left( \frac{\sqrt{5}}{3} \right)^2 = 3 \cdot \frac{5}{9} = \frac{5}{3}\]For \(x = -\frac{\sqrt{5}}{3}\):\[y = 3 \left( -\frac{\sqrt{5}}{3} \right)^2 = 3 \cdot \frac{5}{9} = \frac{5}{3}\]
07
Write the Solution Set
Combine the \(x\) and \(y\) values from Step 6 to obtain the final solution set of the system of equations:\[\left( \frac{\sqrt{5}}{3}, \frac{5}{3} \right) \quad \text{and} \quad \left( -\frac{\sqrt{5}}{3}, \frac{5}{3} \right) \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a powerful technique used to solve systems of equations by replacing one variable with an expression in terms of another variable.
Let's dive into this concept:
Let's dive into this concept:
- Identify which variable to isolate. Preferably, find one of the equations where a single variable is already isolated or can be easily isolated.
- In our problem, we have the equations: \[ \begin{align*} x^2 + y^2 & = 10, \ y &= 3x^2. \end{align*} \]
- Notice that the second equation already expresses \( y \) in terms of \( x \), specifically \( y = 3x^2 \).
- This allows us to substitute \( 3x^2 \) for \( y \) in the first equation.
- This step significantly simplifies our system, allowing us to work with just one variable in the equation.
Polynomial Equations
A polynomial equation involves a sum of powers of variables, with each term multiplied by coefficients. They can appear intimidating, but let's break it down.
- After substituting \( y = 3x^2 \) into \( x^2 + y^2 = 10 \), our equation becomes: \[ x^2 + (3x^2)^2 = 10. \]
- Simplifying gives us: \[ x^2 + 9x^4 = 10. \]
- We rearrange it to form a polynomial: \[ 9x^4 + x^2 - 10 = 0. \]
- This is a fourth-degree polynomial because the highest power of \( x \) is 4.
Real Solutions
Real solutions of an equation are the values that satisfy the equation within the real number system. When solving polynomial equations, identifying real solutions is key.
- Our polynomial \( 9x^4 + x^2 - 10 = 0 \) produces two equations when factored: \[\begin{align*} x^2 + 2 &= 0, \ 9x^2 - 5 &= 0. \end{align*}\]
- Solve each separately:
- For \( x^2 + 2 = 0 \), no real solutions exist, since a square can't be negative in real numbers.
- For \( 9x^2 - 5 = 0 \), we find real solutions. Solving gives \( x = \pm \frac{\sqrt{5}}{3} \).
Factoring Techniques
Factoring is a process of breaking down equations into simpler products to solve them more easily. It's essential for solving complex polynomial equations.
- To factor \( 9x^4 + x^2 - 10 = 0 \), look for patterns or use trial and error to break it down.
- After attempting various techniques, we find: \[ (x^2 + 2)(9x^2 - 5) = 0. \]
- This factorization breaks the initial polynomial into two more manageable equations, each more straightforward to solve.
- Factoring simplifies solving by reducing complex problems.
