Problem 17 - Algebra 2

Question

Find the exact solution(s) of each system of equations. $$ \begin{array}{l}{y^{2}+x^{2}=25} \\ {y^{2}+9 x^{2}=25}\end{array} $$

Step-by-Step Solution

Answer

The solutions are \((0, 5)\) and \((0, -5)\).

Step 1: Identify the System of Equations

We have two equations in the system: \( y^2 + x^2 = 25 \) and \( y^2 + 9x^2 = 25 \). Both are quadratic equations.

Step 2: Eliminate y^2 by Subtraction

Subtract the second equation from the first. \[ (y^2 + x^2) - (y^2 + 9x^2) = 25 - 25 \] This simplifies to: \[ -8x^2 = 0 \] which gives \( x^2 = 0 \). Hence, \( x = 0 \).

Step 3: Substitute x into the First Equation

Substitute \( x = 0 \) into the first equation: \[ y^2 + 0^2 = 25 \] Simplifying gives \( y^2 = 25 \).

Step 4: Solve for y

Take the square root of both sides to solve for \( y \): \( y = \pm 5 \). So the solutions for \( y \) are \( y = 5 \) and \( y = -5 \).

Step 5: Verify the Solutions

According to the solutions \( x = 0, y = 5 \) and \( x = 0, y = -5 \), substituting back into the second equation: \[ 5^2 + 9(0)^2 = 25 \] which holds true, confirming they are correct solutions. Similarly, \[ (-5)^2 + 9(0)^2 = 25 \] is valid as well.

Key Concepts

Quadratic Equations

A quadratic equation is an equation that can be rearranged in the form of \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( x \) represents an unknown variable. Quadratic equations tell us about parabolic curves on a graph. In the exercise's equations, \( y^2 + x^2 = 25 \) and \( y^2 + 9x^2 = 25 \), both equations are quadratic as they contain squared variable terms.

Quadratic equations often show up in systems of equations, where two or more equations are analyzed together. These require us to find solutions that satisfy all equations at once.
Understanding their characteristic curves and intersection points helps us find these solutions effectively.

Systems involving quadratic equations can have multiple solutions, determined by the intersection points of the parabolas represented by each equation.

• When graphed, parabola curves can intersect each other in up to two points.
• This means a system of quadratic equations can have zero, one, or two solutions depending on how these curves intersect each other.

Substitution Method

The substitution method is a technique for solving systems of equations. It involves solving one equation for one variable and then substituting that expression into the other equation. This effectively reduces the system to a single equation in one variable.

In this exercise, after identifying the equations \( y^2 + x^2 = 25 \) and \( y^2 + 9x^2 = 25 \), we use subtraction to eliminate \( y^2 \). This manipulation simplifies one of the equations, leading to \( x^2 = 0 \).

The substitution method is useful because:

• It reduces the complexity by focusing on one equation at a time.
• It allows step-by-step isolation of each variable individually, making the system more manageable.
• It provides an analytical approach to gain insights into relationships between equations.

By substituting \( x = 0 \) into the first equation, \( y^2 = 25 \), we effectively simplify the problem into finding the square roots of 25. This results in the possible values for \( y \), which can be \( y = 5 \) or \( y = -5 \).

Solving Polynomial Equations

Solving polynomial equations involves finding the values of variables that make the equation true. A polynomial equation can include a variety of powers of variables, but following standard methods can simplify the process.

In dealing with our system of equations derived from the exercise, the simplified form \( x^2 = 0 \) and its substitution into the other equation helps in reducing the equations into solvable polynomials.

To solve these:

• Focus on one variable at a time, simplifying whenever possible.
• For \( x^2 = 0 \), we know that \( x \) must equal zero, as any number squared resulting in zero must be zero itself.
• The equation \( y^2 = 25 \) leads to identifying \( y \), where the square root provides potential solutions \( y = 5 \) and \( y = -5 \).

In polynomial solving, verification is crucial by back-substituting the results into the original equations—in this case, confirming both solutions satisfy the system. This ensures correctness and completeness of the solution.