Problem 12 - Algebra and Trigonometry

Question

Solve each system by the substitution method. \(\left\\{\begin{array}{l}{4 x+3 y=0} \\ {2 x-y=0}\end{array}\right.\)

Step-by-Step Solution

Answer

The solutions for the system of equations are (3,-5) and (-1,3).

Step 1: Solve Equation in Terms of One Variable

Choose the simpler equation to solve in terms of one variable. In this case, we're choosing the second equation \(2x+y=1\) and solve for y which becomes \(y = 1-2x\)

Step 2: Substitute the Value of the Variable into the Other Equation

Substitute \(y = 1-2x\) from the second equation into the first equation which becomes \(x^2 + (1 - 2x) = 4\). Simplify the equation to get \(x^2 - 2x + 1 - 4 = 0\), which gives \(x^2 - 2x - 3 = 0\).

Step 3: Solve for the Variable

Now factor the quadratic equation to get \((x - 3)(x + 1) = 0\). Setting each factor equal to zero gives the roots \(x = 3, -1\).

Key Concepts

Solving Systems of Equations

Systems of equations involve finding the value of variables that satisfy multiple equations simultaneously. A system can consist of linear equations, quadratic equations, or even more complex function types. In our example, we have one quadratic equation and one linear equation, namely:

• \(x^2 + y = 4\)
• \(2x + y = 1\)

The goal is to find values of \(x\) and \(y\) that satisfy both equations. We can solve systems by multiple methods such as graphing, elimination, or substitution. Choosing the right method depends on the complexity and type of equations involved. Always aim to simplify the solving process by picking the equation that appears easiest to solve first.

Quadratic Equations

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\). They are characterized by the squared variable \(x^2\), which means they graph as parabolas on a coordinate plane. In the exercise, the quadratic equation is derived after substituting \(y\) in \(x^2 + y = 4\), leading to:

• \(x^2 - 2x - 3 = 0\)

Quadratics can be solved using several methods including:

• Factoring: Rewriting the equation as a product of its factors.
• Quadratic Formula: Using the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the roots.
• Completing the Square: Rewriting the quadratic as a perfect square trinomial.

In this context, factoring was used, which yielded the solutions \((x - 3)(x + 1) = 0\), giving roots \(x = 3\) and \(x = -1\). Being familiar with these methods will help you tackle similar problems with confidence.

Algebraic Substitution

Algebraic substitution is a handy method for solving systems of equations. It involves replacing one variable with an expression derived from another equation. In our exercise, we strategically used the linear equation \(2x + y = 1\) to express \(y\) in terms of \(x\). We solved for \(y\) to get:

• \(y = 1 - 2x\)

This expression was then substituted into the quadratic equation \(x^2 + y = 4\), resulting in a single-variable quadratic equation. Substitution simplifies the system by reducing the number of variables, making it easier to isolate and solve. It's often effective when one equation is simpler, offering a neat pathway to reach variable expressions that can simplify the others. Understand the problem, identify substitutions, and simplify the system step-by-step.