Solving the Simultaneous Equations: A Comprehensive Guide

Step-by-step resolution of y = x² – 7x – 7 and y = 3x + 5

Key Insights

• Substitution Method: Setting the two expressions for y equal allows us to derive a quadratic equation.
• Quadratic Resolution: Solving the quadratic equation using the quadratic formula yields the precise x-values.
• Back Substitution: Substituting the x-values into the simpler linear equation gives the corresponding y-values.

---

Understanding the Problem

In our study of simultaneous equations, we are given a quadratic equation and a linear equation:

Equations Provided

1. The quadratic equation: y = x² – 7x – 7.
2. The linear equation: y = 3x + 5.

The goal is to find the values of x and y that satisfy both equations simultaneously. Typically, when two equations are set equal to each other, the values of x found will be those at which the graphs of these equations intersect.

---

Detailed Explanation of the Approach

Step 1: Setting the Equations Equal to Each Other

Given that both equations are equal to y, we begin by setting the right-hand sides equal:

x² – 7x – 7 = 3x + 5

This step is pivotal because it transforms the problem from one involving two separate equations into a single equation in one variable.

Step 2: Rearranging into a Standard Quadratic Form

Next, we rearrange the equation by moving all terms to one side to set it equal to zero:

x² – 7x – 7 – (3x + 5) = 0

Simplify by combining like terms:

x² – 10x – 12 = 0

This is now in the standard quadratic form, where a quadratic equation is generally written as:

ax² + bx + c = 0

In this case, a = 1, b = -10, and c = -12.

Step 3: Solving the Quadratic Equation

Using the Quadratic Formula

The quadratic formula is a versatile method used when the quadratic does not factor easily. The formula is given by:

\( \displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Substituting our coefficients:

\( \displaystyle x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} \)

Simplify within the formula step by step:

\( \displaystyle x = \frac{10 \pm \sqrt{100 + 48}}{2} \)

\( \displaystyle x = \frac{10 \pm \sqrt{148}}{2} \)

Notice that 148 factors as 4 × 37, so:

\( \displaystyle \sqrt{148} = \sqrt{4 \cdot 37} = 2\sqrt{37} \)

Thus we obtain:

\( \displaystyle x = \frac{10 \pm 2\sqrt{37}}{2} \)

Simplifying further by dividing numerator and denominator by 2:

\( \displaystyle x = 5 \pm \sqrt{37} \)

This solution provides us with two possible x-values:

• First solution: \( x_1 = 5 + \sqrt{37} \)
• Second solution: \( x_2 = 5 - \sqrt{37} \)

Step 4: Substituting Back to Find Corresponding y-values

With the solutions for x in hand, the next task is to substitute these values into one of the original equations to solve for y. The linear equation, y = 3x + 5, is a natural choice as it is straightforward to compute.

Substitution for \( x_1 = 5 + \sqrt{37} \)

Substitute into y = 3x + 5:

\( \displaystyle y_1 = 3(5 + \sqrt{37}) + 5 \)

Perform the multiplication:

\( \displaystyle y_1 = 15 + 3\sqrt{37} + 5 \)

Combine constant terms:

\( \displaystyle y_1 = 20 + 3\sqrt{37} \)

Substitution for \( x_2 = 5 - \sqrt{37} \)

Similarly, substitute into y = 3x + 5:

\( \displaystyle y_2 = 3(5 - \sqrt{37}) + 5 \)

Distribute the multiplication:

\( \displaystyle y_2 = 15 - 3\sqrt{37} + 5 \)

Combine the constants:

\( \displaystyle y_2 = 20 - 3\sqrt{37} \)

---

Summary Table of Steps and Solutions

Step	Description	Result / Expression
1	Set the equations equal, since y equals both expressions	\( x^2 - 7x - 7 = 3x + 5 \)
2	Rearrange the equation into standard quadratic form	\( x^2 - 10x - 12 = 0 \)
3	Use the quadratic formula	\( x = \frac{10 \pm \sqrt{148}}{2} = 5 \pm \sqrt{37} \)
4	Substitute \( x_1 \) into y = 3x + 5	\( y_1 = 20 + 3\sqrt{37} \)
5	Substitute \( x_2 \) into y = 3x + 5	\( y_2 = 20 - 3\sqrt{37} \)

---

In-Depth Explanation and Verification

Understanding the Derivation Process

The problem involves finding the intersection of a parabola and a straight line. In analytical geometry, the point(s) of intersection satisfy both equations. By equating the two expressions for y, it confirms that at the intersection, the outputs of both functions are the same.

The key challenge here is to correctly set up and solve the resulting quadratic equation. The quadratic equation derived is:

\( x^2 - 10x - 12 = 0 \)

This equation is not immediately factorable into simple integers, and the quadratic formula becomes the most reliable method to find the exact roots. The computation of the discriminant \( \Delta = b^2 - 4ac \) reveals important characteristics:

• The discriminant is \( \Delta = 100 + 48 = 148 \).
• Since \( \Delta > 0 \), there are two distinct real solutions.
• The square root of the discriminant simplifies to \( 2\sqrt{37} \), leading to the solutions \( x = 5 \pm \sqrt{37} \).

Accuracy of the Back Substitution

Substituting the x-values back into the linear equation ensures the y-values are derived with minimal complication. The choice of y = 3x + 5 is intentional due to its simplicity compared to substituting back into a quadratic relation. This approach solidifies the final answers:

• For \( x = 5 + \sqrt{37} \): \( y = 20 + 3\sqrt{37} \).
• For \( x = 5 - \sqrt{37} \): \( y = 20 - 3\sqrt{37} \).

These solutions precisely define the intersection points of the two curves, establishing the analytical validity of the approach.

---

Alternative Perspectives and Verification

Graphical Interpretation

Visually, these equations represent a parabola and a line. The intersection points calculated provide the exact coordinates where the graph of the parabola (representing \( y = x^2 - 7x - 7 \)) meets the graph of the line (representing \( y = 3x + 5 \)). One would observe that the quadratic function opens upward and the linear function, being a straight line with a positive slope, cuts through the parabola at precisely two distinct points when plotted on a standard coordinate plane. This affirms the presence of two intersection points.

Verification Using the Quadratic Formula

Reiterating the quadratic formula calculation:

\( \displaystyle x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4\cdot1\cdot(-12)}}{2\cdot1} = \frac{10 \pm \sqrt{148}}{2} = 5 \pm \sqrt{37} \)

This method not only provides confidence in the computed results but also demonstrates the robustness of the quadratic formula in handling such equations reliably.

---

Final Results

After performing the substitution and algebraic manipulations, the solutions to the system of simultaneous equations are:

• For \( x = 5 + \sqrt{37} \), we find \( y = 20 + 3\sqrt{37} \).
• For \( x = 5 - \sqrt{37} \), we find \( y = 20 - 3\sqrt{37} \).

These results are exact, written in terms of radicals, and represent the precise intersection points between the two functions. Their derivation is both systematic and verified by substituting back into the original equations.

---

Comparison of Methods

Methodological Overview

The resolution of these equations can be approached primarily via:

• Substitution: Equate the two expressions for y, then solve the resulting quadratic equation.
• Graphical Analysis: Plot both equations and identify the intersection points.
• Analytical Verification: Use the quadratic formula as a robust method when factoring is not straightforward.

Among these, the substitution method coupled with the quadratic formula provides the most direct and algebraically rigorous pathway to the solution.

---

Final Answers in Mathematical Notation

The simultaneous equations:

\( \displaystyle y = x^2 - 7x - 7 \)
\( \displaystyle y = 3x + 5 \)

have the solutions:

1. \( \displaystyle x = 5 + \sqrt{37}, \quad y = 20 + 3\sqrt{37} \)
2. \( \displaystyle x = 5 - \sqrt{37}, \quad y = 20 - 3\sqrt{37} \)

---

References

• System Calculator - MathPapa
• System Equation Calculator - Wolfram|Alpha
• Simultaneous Equations Solver - Symbolab
• Simultaneous Equations Solver - eMathHelp
• Advanced Equation Solver - QuickMath

Recommended Related Queries