Problem 49 - Intermediate Algebra

Question

Solve each system of equations for real values of \(x\) and \(y.\) $$ \left\\{\begin{array}{l} 3 x-y=-3 \\ 25 y^{2}-9 x^{2}=225 \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions are \((x, y) = (0, 3)\) and \((-\frac{25}{12}, -\frac{13}{4})\).

Step 1: Solve the First Equation for y

Start by solving the first equation for \( y \) in terms of \( x \). Given \( 3x - y = -3 \), we can rearrange it as \( y = 3x + 3 \).

Step 2: Substitute y in the Second Equation

Substitute \( y = 3x + 3 \) into the second equation \( 25y^2 - 9x^2 = 225 \). This leads to \( 25(3x + 3)^2 - 9x^2 = 225 \).

Step 3: Expand and Simplify

Expand \( (3x + 3)^2 \) to get \( 9x^2 + 18x + 9 \). The equation becomes \( 25(9x^2 + 18x + 9) - 9x^2 = 225 \). Simplify this to get \( 225x^2 + 450x + 225 - 9x^2 = 225 \).

Step 4: Combine Like Terms

Combine like terms to simplify the equation: \( (225x^2 - 9x^2) + 450x + 225 = 225 \) simplifies to \( 216x^2 + 450x + 225 = 225 \).

Step 5: Solve for x

Subtract 225 from both sides to obtain \( 216x^2 + 450x = 0 \). Factor out \( x \) to get \( x(216x + 450) = 0 \). This results in \( x = 0 \) or \( 216x + 450 = 0 \).

Step 6: Solve 216x + 450 = 0 for x

Solve \( 216x + 450 = 0 \) by subtracting 450 from both sides to get \( 216x = -450 \). Divide both sides by 216 to find \( x = -\frac{450}{216} = -\frac{75}{36} \), which simplifies to \( x = -\frac{25}{12} \).

Step 7: Solve for y Corresponding to Each x Value

For \( x = 0 \): substitute into \( y = 3x + 3 \) to get \( y = 3(0) + 3 = 3 \). For \( x = -\frac{25}{12} \): substitute into \( y = 3x + 3 \) to get \( y = 3(-\frac{25}{12}) + 3 = -\frac{75}{12} + 3 \). Convert 3 to \( \frac{36}{12} \): \( y = -\frac{75}{12} + \frac{36}{12} = -\frac{39}{12} \), which reduces to \( y = -\frac{13}{4} \).

Step 8: Identify the Solutions

The solution set for the system of equations is \( (x, y) = (0, 3) \) and \( (x, y) = (-\frac{25}{12}, -\frac{13}{4}) \).

Key Concepts

Substitution Method

The substitution method is a common technique for solving systems of equations. It involves solving one of the equations for one variable, and then substituting that expression into the other equation. By doing this, you can reduce two equations to a single equation with one variable.

In our system, we start by solving the first equation, \(3x - y = -3\), for \(y\). Rearranging the terms, we find \(y = 3x + 3\). This forms our substitution expression.

Next, we substitute this \(y\) expression into the second equation, \(25y^2 - 9x^2 = 225\). This changes the equation to involve only \(x\), allowing us to focus on one unknown. This strategy simplifies the problem, making it easier to solve.

Quadratic Equations

A quadratic equation is any equation that can be written in the form \(ax^2 + bx + c = 0\). It is called 'quadratic' because 'quad' means square, and the variable \(x\) is squared.

In our problem, once we substitute \(y = 3x + 3\) into the second original equation, we get a single quadratic equation in terms of \(x\): \(225x^2 + 450x + 225 = 225\). When simplified, this becomes \(216x^2 + 450x = 0\).

Solving this involves factoring or using the quadratic formula. In our solution, we factored it: \(x(216x + 450) = 0\). The solutions for \(x\) are found by setting each factor equal to zero, generating two potential values for \(x\).

Solution Sets

The solution set for a system of equations is the set of all values that satisfy every equation in the system.

In this exercise, upon solving our quadratic equation, we found two potential \(x\) values: \(x = 0\) and \(x = -\frac{25}{12}\). Each \(x\) value was then substituted back into our expression for \(y\), \(y = 3x + 3\).

This gave us two corresponding \(y\) values, forming the ordered pairs that constitute the solution set:

• For \(x = 0\), \(y = 3\). The ordered pair is \((0, 3)\).
• For \(x = -\frac{25}{12}\), \(y = -\frac{13}{4}\). The ordered pair is \((-\frac{25}{12}, -\frac{13}{4})\).

The solution set contains these two pairs.

Real Numbers

Real numbers include all the numbers on the number line. This set includes both positive and negative numbers, zero, and all fractions and decimals. They are the numbers we typically use in algebra and everyday life to denote measurements and quantities.

In this exercise, we are interested in finding solutions that are real numbers. The original problem explicitly states to solve for real values of \(x\) and \(y\).

Both solutions found, \((x, y) = (0, 3)\) and \((x, y) = (-\frac{25}{12}, -\frac{13}{4})\), are indeed real numbers. This ensures they are valid solutions within the context of the problem, satisfying the original equations while existing on the number line.