Problem 9 - Precalculus Mathematics for Calculus

Question

Use the elimination method to find all solutions of the system of equations. $$\left\\{\begin{array}{l} x^{2}-2 y=1 \\ x^{2}+5 y=29 \end{array}\right.$$

Step-by-Step Solution

Answer

The solutions are \( (3, 4) \) and \( (-3, 4) \).

Step 1: Set Up the System of Equations

We are given the system of equations:\[ \begin{align*} \text{(i) } & x^2 - 2y = 1, \ \text{(ii) } & x^2 + 5y = 29. \end{align*} \]The goal is to eliminate one of the variables, usually through subtraction of equations, but first we'll manipulate these so that subtraction of terms can eliminate a variable.

Step 2: Subtract the Equations

To eliminate the variable \(x^2\), subtract equation (i) from equation (ii): \[ (x^2 + 5y) - (x^2 - 2y) = 29 - 1. \] This simplifies to \[ (x^2 + 5y - x^2 + 2y) = 28. \] Which further simplifies to \[ 7y = 28. \]

Step 3: Solve for y

From the equation \( 7y = 28 \), divide both sides by 7: \[ y = \frac{28}{7} = 4. \] We find that \( y = 4 \).

Step 4: Substitute to Find x

Substitute \( y = 4 \) back into one of the original equations. We'll use equation (i):\[ x^2 - 2(4) = 1. \] Simplify to \[ x^2 - 8 = 1. \] Add 8 to both sides: \[ x^2 = 9. \] Take the square root of both sides to find \( x \): \[ x = 3 \text{ or } x = -3. \]

Step 5: Verify Solutions

The solutions are \( (3, 4) \) and \( (-3, 4) \). Verify by substituting back into both original equations. Both solutions satisfy:\[(3^2 - 8 = 1) \text{ and } ((-3)^2 - 8 = 1)\] and equally valid for the second equation\[(3^2 + 20 = 29) \text{ and } ((-3)^2 + 20 = 29).\] All checks are correct.

Key Concepts

System of Equations

A system of equations consists of two or more equations that are related to each other through common variables. In this exercise, we are working with the following system of equations:

• \(x^2 - 2y = 1\)
• \(x^2 + 5y = 29\)

The goal is to find values for \(x\) and \(y\) that satisfy both equations simultaneously. When we solve a system of equations, we look for an ordered pair \((x, y)\) that answers each equation in the system. This ordered pair is known as the solution of the system. To determine this, we use different methods such as substitution, graphical methods, and elimination.
It's important to ensure that both equations hold true for the values we find. A system of equations can have a single solution, no solution, or an infinite number of solutions.

Variable Elimination

Variable elimination is a technique used to solve a system of equations by removing one variable, allowing us to solve for the other. In this exercise, we applied the elimination method to remove \( x^2 \) from the system and find the value of \( y \).
To achieve this, subtract equation (i) \(x^2 - 2y = 1\) from equation (ii) \(x^2 + 5y = 29\). By subtracting the left side of both equations, we eliminate \( x^2 \):

• \((x^2 + 5y) - (x^2 - 2y) = 29 - 1\)
• Simplifying gives us \(7y = 28\)

Now, solving for \( y \), divide both sides by 7 to get \( y = 4 \).
Variable elimination makes the problem more manageable by reducing the number of variables. Once \( y \) has been determined, it is easier to substitute this value back into one of the original equations to find \( x \). This systematic approach avoids errors and ensures accuracy in solving the system.

Verify Solutions

After solving a system of equations, it is crucial to verify that the solutions are correct. Verifying ensures that the values found are true solutions to the original equations.
For the solutions \((3, 4)\) and \((-3, 4)\), we substitute back into both equations:

• Substituting \((3, 4)\) into \(x^2 - 2y = 1\):
  • \(3^2 - 2(4) = 9 - 8 = 1\)
• Substituting \((3, 4)\) into \(x^2 + 5y = 29\):
  • \(3^2 + 5(4) = 9 + 20 = 29\)

Both checks confirm that \((3, 4)\) satisfies the system of equations. The same can be verified for \((-3, 4)\) as well, showing they are both valid solutions. Additional attempts to substitute and confirm as done here bolster confidence in the obtained results, ensuring accuracy and precision in problem-solving.