Problem 28 - Intermediate Algebra

Question

Solve each system of equations by substitution for real values of \(x\) and \(y.\) See Examples 2 and 3. $$ \left\\{\begin{array}{l} x^{2}+y^{2}=10 \\ y=3 x^{2} \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions are \(\left( \frac{\sqrt{5}}{3}, \frac{5}{3} \right)\) and \(\left( -\frac{\sqrt{5}}{3}, \frac{5}{3} \right)\).

Step 1: 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).

Step 2: Substitute and Simplify

Substitute \(y = 3x^2\) into the first equation:\[x^2 + (3x^2)^2 = 10\]Now simplify the equation:\[x^2 + 9x^4 = 10\]

Step 3: Rearrange to Form a Polynomial Equation

Rearrange the equation obtained in Step 2:\[9x^4 + x^2 - 10 = 0\]

Step 4: 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:\[(9x^4 + x^2 - 10) = (x^2 + 2)(9x^2 - 5) = 0\]

Step 5: 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 \(9x^2 - 5 = 0\):**\[9x^2 = 5\]\[x^2 = \frac{5}{9}\]\[x = \pm \frac{\sqrt{5}}{3}\]

Step 6: 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 \(x = -\frac{\sqrt{5}}{3}\):\[y = 3 \left( -\frac{\sqrt{5}}{3} \right)^2 = 3 \cdot \frac{5}{9} = \frac{5}{3}\]

Step 7: Write the Solution Set

Combine the \(x\) and \(y\) values from Step 6 to obtain the final solution set of the system of equations:\[\left( \frac{\sqrt{5}}{3}, \frac{5}{3} \right) \quad \text{and} \quad \left( -\frac{\sqrt{5}}{3}, \frac{5}{3} \right) \]

Key Concepts

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:

• 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.

Using this method efficiently turns a two-variable problem into a one-variable equation, making it much easier to solve.

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.

These equations often need special techniques for solving, and recognizing them is crucial for unravelling more complex algebra problems.

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 \( x = \pm \frac{\sqrt{5}}{3} \).

Real solutions are crucial for many practical problems, as they represent feasible answers within the constraints of reality.

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.

Factoring can be tricky, and not every polynomial can be factored easily, but it's a powerful tool for untangling complex equations.