Problem 4 - Elementary Algebra Within Reach

Question

In Exercises \(1-6\), solve the system by the method of elimination. $$ \left\\{\begin{array}{l} x+2 y=14 \\ x-2 y=10 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 12, y = 1\).

Step 1: Addition of Equations

Add the two equations in the system to eliminate \(y\). \[\begin{align*} x+2 y + (x - 2y) & = 14 + 10 \cr2x & = 24 \end{align*}\]

Step 2: Solve for \(x\)

Divide the resulting equation by 2 to solve for \(x\) \[ x = \frac{24}{2} = 12 \]

Step 3: Substitution to find \(y\)

Substitute \(x = 12\) into the first original equation to find \(y\):\[\begin{align*}12 + 2y & = 14 \cr2y & = 14 - 12 = 2 \end{align*}\]Divide the result by 2 to solve for \(y\):\[y = \frac{2}{2} = 1\]

Key Concepts

Elimination Method

The elimination method is a technique used in algebra to solve systems of linear equations. It involves manipulating the equations to eliminate one variable, making it possible to solve for the other variable. This method is particularly useful when equations are set up in a way that makes addition or subtraction a straightforward approach to eliminating one of the variables.

For example, consider a system with the equations:

• \(x + 2y = 14\)
• \(x - 2y = 10\)

By adding these two equations together, we can eliminate the \(y\) variable because \(+2y\) and \(-2y\) are additive inverses and sum to zero. This simplification leads to an equation in one variable, \(2x = 24\), which can be easily solved. The key to the elimination method is to set up the equations suitably or to multiply them by necessary factors to cancel out one of the variables upon addition or subtraction.

Linear Equations

Linear equations are fundamental in algebra and represent relationships between variables where each term is either a constant or the product of a constant and a single variable. The general form of a linear equation in two variables, \(x\) and \(y\), is given by \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants.

In the context of systems of equations, two linear equations can graphically represent two lines on a coordinate plane. The solution to the system is the point where the two lines intersect. If the lines are parallel and distinct, they do not intersect, and no solution exists. If the lines coincide, they have an infinite number of solutions. In our problem, we have a pair of linear equations:

• \(x + 2y = 14\)
• \(x - 2y = 10\)

These equate to two lines that intersect at a single point, which is the solution to the system.

Algebraic Solutions

Algebraic solutions refer to the values of variables that satisfy all equations within a system. Finding these solutions involves using algebraic methods such as substitution, elimination, or graphing. In the case of our problem, the algebraic solution comes through the elimination method.

After determining the value of \(x\) to be 12, we substitute it back into one of the original equations to solve for \(y\). This process turns the system of equations into a problem with only one variable—something that is much simpler to solve. Once we find that \(y = 1\), we have the solution to our system: \((x, y) = (12, 1)\). It is crucial in algebra to carry out each step correctly, as a mistake at any point can lead to the wrong solution. Moreover, checking the solution by substituting the values back into both original equations can confirm the correctness of the obtained algebraic solution.