Problem 11 - Elementary Algebra Within Reach

Question

In Exercises 7-12, solve the system by the method of elimination. $$ \left\\{\begin{array}{l} 2 x-5 y=-1 \\ 2 x-y=1 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \( x = 0.75 \) and \( y = 0.5 \).

Step 1: Subtract Second Equation from the First

By subtracting the second equation from the first, we can eliminate the variable \(x\). We get \(2x - 2x - 5y - (-y) = -1 - 1\), which simplifies to \(-4y = -2\).

Step 2: Solve for the Eliminated Variable

Now, we divide by -4 on both sides of the equation to solve for \(y\). This gives us \(y = \frac{-2}{-4} = 0.5\).

Step 3: Substitute and Solve for the other Variable

Now that we have \(y = 0.5\), we can substitute \(y\) into either of the original equations. Here we substitute it into the second equation \(2x - y = 1\), which gives us \(2x - 0.5 = 1\), from here we solve for \(x\) to find that \(x = \frac{1+0.5}{2} = 0.75\).

Key Concepts

Solving Systems of Equations

Solving systems of equations involves finding the values of variables that satisfy all equations in the system. When we have two or more equations working together, these are known as a system of equations. Systems can be solved by several methods, including graphing, substitution, and elimination.

In this particular example, the elimination method is used, where the goal is to eliminate one variable to easily solve for another.

• Start by arranging both equations in a consistent form: Typically, we write them with variables on one side and constants on the other.
• Choose which variable to eliminate: Often, it's simpler to eliminate the variable that can be easily canceled out by addition or subtraction.
• Perform the elimination: Manipulate the equations if necessary, such as by multiplying, to cancel one variable when equations are added or subtracted.
• Solve for the remaining variable: Once one variable is eliminated, determine the value of the other variable. Afterward, substitute back to find the eliminated variable.

By systematically eliminating a variable, solving the system becomes more straightforward and reduces potential errors.

Algebraic Methods

Algebraic methods, like elimination, substitution, and others, allow us to handle variables symbolically and solve equations efficiently. They're vital tools for deciphering unknown quantities in mathematical problems.

• Elimination Method: Here, by adding or subtracting equations from each other, one variable is "eliminated," simplifying the system into a single equation. This method requires careful arithmetic but often requires less rewriting of equations compared to substitution.
• Substitution Method: This involves solving one equation for a particular variable and substituting this expression into another equation. While intuitive, it may involve more manipulations to reach the solution.

Given the benefits of each, deciding which technique to use often depends on the specific equations at hand. Techniques that simplify variables, such as elimination, are always valuable when striving for a quicker solution.

Linear Equations

Linear equations are the backbone of algebra and consist of variables that are not raised to any power other than one. This structure ensures their graphs are straight lines, and systems of these equations represent points where these lines intersect.

• Standard Form: A linear equation in the standard form is expressed as \(Ax + By = C\), where \(A\), \(B\), and \(C\) are constants.
• Solution of Linear Equations: Solving a single linear equation is simple; solving a system means finding the intersection point (or points) of the lines represented by the equations.

For the pair of equations given in the exercise, we find the solution where both lines intersect each other. Such a solution is the set of values for \(x\) and \(y\) that satisfy both equations simultaneously. This forms the theoretical basis upon which much of linear algebra is built.