Problem 35 - College Algebra

Question

Solve each system by any method. $$ \begin{array}{l} 7 x-4 y=\frac{7}{6} \\ 2 x+4 y=\frac{1}{3} \end{array} $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{1}{6}\) and \(y = 0\).

Step 1: Adding Equations

Start by adding the two equations together. The second equation is: \[2x + 4y = \frac{1}{3}\]Adding the first equation:\[7x - 4y + 2x + 4y = \frac{7}{6} + \frac{1}{3}\] The \(-4y\) and \(4y\) cancel each other out.

Step 2: Simplify and Solve for x

After cancelling \(y\):\[9x = \frac{7}{6} + \frac{1}{3}\]Convert \(\frac{1}{3}\) to have a common denominator:\[\frac{1}{3} = \frac{2}{6}\]So:\[\frac{7}{6} + \frac{2}{6} = \frac{9}{6} = \frac{3}{2}\]Now solve for \(x\) by dividing both sides by 9:\[x = \frac{3}{2} \times \frac{1}{9} = \frac{3}{18} = \frac{1}{6}\]

Step 3: Substitute x into Second Equation

Substitute \(x = \frac{1}{6}\) back into the second equation:\[2x + 4y = \frac{1}{3}\]\[2\left(\frac{1}{6}\right) + 4y = \frac{1}{3}\]Simplify:\[\frac{2}{6} + 4y = \frac{1}{3}\]\[\frac{1}{3} + 4y = \frac{1}{3}\]

Step 4: Solve for y

Simplify further for \(y\):Subtract \(\frac{1}{3}\) from both sides:\[4y = \frac{1}{3} - \frac{1}{3} = 0\]Divide by 4:\[y = 0\]

Key Concepts

Linear Equations

Linear equations are fundamental in algebra. They are equations of the first degree, which means the highest exponent of the variable(s) involved is 1.
They typically take the form of \(ax + by = c\), where \(a, b,\) and \(c\) are constants, and \(x\) and \(y\) are variables. In the exercise, our system of equations consists of two linear equations:

• \(7x - 4y = \frac{7}{6}\)
• \(2x + 4y = \frac{1}{3}\)

The key aspect of linear equations is that they describe straight lines on a coordinate plane.
When solving systems of linear equations, such as in this exercise, we are essentially finding the point where these lines intersect. It represents the solution of the system, the values of \(x\) and \(y\) that satisfy both equations.Understanding linear equations sets the foundation for tackling more complex algebraic problems.
They allow us to model and solve real-world situations, involving intersecting quantities and constant rates of change.

Substitution Method

The substitution method is a strategic way to solve a system of equations. It involves solving one of the equations for one variable and then substituting this solution into the other equation.
This reduces the system to a single equation with one variable.In practice, it often looks like this:

• Select one of the equations and solve for either \(x\) or \(y\).
• Substitute this expression into the other equation.
• Solve the resulting single-variable equation.

In the step-by-step solution provided, the substitution method is employed after finding \(x = \frac{1}{6}\).
By substituting \(x\) into the second equation \(2x + 4y = \frac{1}{3}\), we directly find the value of \(y\).Substitution is particularly useful when one equation is easy to solve for a variable or when one of the coefficients is 1 or -1.
It's a clear and logical method that simplifies the problem to a solvable equation.

Elimination Method

The elimination method, sometimes called the addition or subtraction method, is another powerful technique for solving systems of equations.
The goal is to eliminate one of the variables by adding or subtracting the equations from each other.Here's a breakdown of how it works:

• Arrange the equations so that adding or subtracting them cancels out one variable.
• Solve the resulting single-variable equation.
• Substitute back into one of the original equations to find the other variable.

In our exercise, the elimination method is applied initially. By adding the given equations:
\(7x - 4y = \frac{7}{6}\)
\(2x + 4y = \frac{1}{3}\)
The \(-4y\) and \(4y\) are eliminated, simplifying the problem to an equation in \(x\) alone.
This step reduces the system to an equation we can straightforwardly solve for \(x\), making the rest of the process manageable.The elimination method is effective when coefficients are easily matched or when they are opposites, as in this exercise.
It provides a clear path to finding the solution by reducing complexity in the equations.