Problem 42 - Algebra and Trigonometry

Question

Solve each system by the method of your choice. $$ \left\\{\begin{array}{l} {x-3 y=-5} \\ {x^{2}+y^{2}-25=0} \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions are \((-5,0)\) and \((4,3)\)

Step 1: Isolate variable x in the first equation

Rearrange the first equation to solve for x by adding \(3y\) to both sides, which gives: \(x = 3y - 5\).

Step 2: Substitute x in the second equation

Substitute \(x = 3y - 5\) into the second equation: \((3y - 5)^{2} + y^{2} = 25\). This leads to a quadratic equation: \(9y^{2} - 30y + 25 + y^{2} = 25\), which simplifies to \(10y^{2} - 30y = 0\).

Step 3: Solve for y

Solve the quadratic equation: \(10y(y - 3) = 0\). This gives two values for y: \(y = 0\) and \(y = 3\).

Step 4: Solve for x

Substitute the obtained values for y into the equation obtained in step 1: For \(y = 0\), we get \(x = -5\). For \(y = 3\), we get \(x = 4\).

Step 5: Solutions of the system

Hence, the solutions to the system of equations are \((-5,0)\) and \((4,3)\)

Key Concepts

Quadratic Equation

A quadratic equation is essentially an algebraic equation of degree 2, known for its characteristic parabolic graph when plotted in Cartesian coordinates. The general form of a quadratic equation is \[ ax^2 + bx + c = 0 \]where:

• \(a\) is the coefficient of \(x^2\) (quadratic term)
• \(b\) is the coefficient of \(x\) (linear term)
• \(c\) is the constant.

To solve a quadratic equation, one can employ various methods such as factoring, using the quadratic formula, or graphing. In our problem, the form \[ 10y^2 - 30y = 0 \] is a classic quadratic equation. This particular equation can be factored as it has common factors, resulting in a simpler format: \[ 10y(y - 3) = 0 \].This gives us the solutions directly by setting each factor equal to zero. Quadratic equations frequently appear in problems involving physics, geometry, and everyday situations where relationships between varying quantities are squared.

Substitution Method

The substitution method is a technique used to solve a system of equations whereby one equation is solved for one variable, which is then substituted into the other equation. This method can effectively simplify a system, especially when dealing with equations of different types like a linear and a quadratic equation.Let's delve into how it worked for this problem:

• We began with two equations:\[\begin{align*}x - 3y &= -5 \x^2 + y^2 &= 25\end{align*}\]
• Rewriting the first equation to express \(x\) in terms of \(y\): \[x = 3y - 5\]
• This expression for \(x\) was substituted into the second equation, transforming it from a mixed form into a pure quadratic form in terms of \(y\).

The substitution method is particularly useful and ensures that complex relationships involving multiple variables can be untangled, step by step, until we arrive at the solution.

Solving Algebraic Equations

Solving algebraic equations involves finding the values of variables that make the equation true. It’s a fundamental skill in algebra, critical to a wide range of applications. Each equation type and form can require different strategies for finding solutions.In this problem, we demonstrated this skill through the following steps:

• **Rewriting for simplicity:**In the given system, the first linear equation was manipulated to solve for one variable, making substitution possible.
• **Quadratic resolution:**The resulting quadratic equation was solved by factoring, identifying the values for \(y\). The key was recognizing and isolating terms to factor them more easily.
• **Variable substitution:**Once \(y\) values were found, substituting back into the linear equation helped find the respective \(x\) values. This back-and-forth process between equations is typical when solving systems.

This systematic approach not only solves the given equations but also illuminates concepts and methods used in broader algebraic problem-solving scenarios. By practicing these techniques, students can enhance their problem-solving skills in algebra.