Problem 20 - College Algebra

Question

Solve each system by the addition method. \(\left\\{\begin{array}{l}x+y-6 \\ x-y--2\end{array}\right.\)

Step-by-Step Solution

Answer

The solutions are (1,0) and (-1,0)

Step 1: Identify the System of Equations

The system of equations is given by: \[\begin{cases}4 x^{2}-y^{2}=4, \4 x^{2}+y^{2}=4. \\end{cases}\]

Step 2: Add the Equations

Add the two equations to eliminate \(y^{2}\). \[4x^{2} - y^{2} + 4x^{2} + y^{2} = 4 + 4\]which simplifies to:\[8x^{2} = 8\]

Step 3: Simplify the Resulting Equation

Divide both sides of the equation by 8:\[x^{2} = 1\]This gives two solutions for \(x\):\[x = \pm 1\]

Step 4: Solve for y

Substitute \(x\) = 1 & -1 into one of the original equations (let's use the second one: \(4 x^{2} + y^{2} = 4\)):For \(x = 1\):\[4(1)^{2} + y^{2} = 4 \rightarrow y^{2} = 0\]So \(y = 0\)For \(x = -1\):\[4(-1)^{2} + y^{2} = 4 \rightarrow y^{2} = 0 \]So \(y = 0\) again

Key Concepts

System of Equations

A system of equations consists of multiple equations that share common variables. The goal is to find values for these variables that make all the equations true at the same time. In many real-life situations, systems of equations allow us to model and solve complex problems with multiple constraints.

In our example, the system of equations given contains quadratic terms, specifically:

• \(4x^2 - y^2 = 4\)
• \(4x^2 + y^2 = 4\)

The key to solving this system is finding a method that effectively deals with all the equations simultaneously. Each equation represents a different constraint on the values of \(x\) and \(y\). In this case, we can use the addition or elimination method to solve the system and find the pairs of \((x, y)\) that satisfy both equations.

Quadratic Equations

Quadratic equations are polynomials of degree 2, often written in the form \(ax^2 + bx + c = 0\). Solving quadratic equations is fundamental in algebra and can be done using various methods such as factoring, completing the square, or using the quadratic formula.

In our system, each equation involves a quadratic term in \(x\), that is \(4x^2\). Interestingly, when combined, these equations also feature the interaction of a quadratic \((-y^2)\) and a linear \((+y^2)\) term. When we add these two equations, carefully orchestrating this addition cancels out the linear terms because they have equal coefficients with opposite signs. This simplification reduces the system to a single variable quadratic equation, which is more straightforward to solve.

For example, after eliminating \(y^2\), we were left with \(8x^2 = 8\), which easily simplifies to \(x^2 = 1\), giving us two possible solutions for \(x\), namely \(x = 1\) and \(x = -1\). This highlights the elegance of the elimination method when dealing with quadratic systems.

Elimination Method

The elimination method, also known as the addition method, is used to solve systems of equations by eliminating one variable at a time. This is done by adding or subtracting the equations in such a way that one variable cancels out. This is particularly effective when the coefficients of one variable are equal and opposite.

In the provided example, the equations were added as follows:

• \(4x^2 - y^2\) and \(4x^2 + y^2\) were added
• The \(y^2\) terms cancelled each other out, leaving \(8x^2 = 8\).

From there, it was straightforward to solve for \(x\) because the equation was reduced to one variable. The elimination method is powerful because it simplifies the system, making it easier to find solutions. By focusing on simplifying the system's complexity through elimination, we uncovered a clear path to the solution, which are the values for \(x\) and \(y\) that satisfy both equations in the system. By understanding the elimination process, solving complex systems becomes manageable and opens doors to understanding more intricate algebraic concepts.