Problem 40 - Algebra Form and Function

Question

Solve the systems of equations. $$ \left\\{\begin{array}{l} 3(e+f)=5 e+f+2 \\ 4(f-e)=e+2 f-4 \end{array}\right. $$

Step-by-Step Solution

Answer

Answer: e = 2 and f = 3

Step 1: Write the system of equations

Here are the given equations: \(1. \quad 3(e+f)=5e+f+2\) \(2. \quad 4(f-e)=e+2f-4\)

Step 2: Simplify the equations

We will now simplify both equations by distributing and combining like terms: For equation \(1\): \(3e + 3f = 5e + f + 2\) For equation \(2\): \(4f - 4e = e + 2f - 4\) Now our simplified system of equations looks like this: \(1. \quad 3e + 3f = 5e + f + 2\) \(2. \quad 4f - 4e = e + 2f - 4\)

Step 3: Use the elimination method to eliminate one variable

Let's eliminate the variable \(e\) from the equations above by adding the two equations. First, multiply equation \(2\) by \(3\) to make the coefficient of e in equation 2 equal to \(-3\): \(2. \quad 3 (4f - 4e) = 3 (e + 2f - 4)\) \(2. \quad 12f - 12e = 3e + 6f - 12\) Now, add the equations and eliminate \(e\): \((3e + 3f) + (-12e + 12f) = (5e + f + 2) + (3e + 6f - 12)\) \(9f = 8e + 7f - 10\)

Step 4: Solve for one variable

Now we can solve for f: \(9f - 7f = 8e - 10\) \(2f = 8e - 10\) Divide by 2: \(f= 4e - 5\)

Step 5: Substitute the value back into one of the original equations

Now, substitute the value of \(f\) back into the equation \(1\): \(3e + 3(4e - 5) = 5e + (4e - 5) + 2\)

Step 6: Solve for the other variable

Now solve for \(e\): \(3e + 12e - 15 = 5e + 4e - 5 + 2\) \(15e - 15 = 9e - 3\) Subtract \(9e\): \(6e - 15 = -3\) Add \(15\): \(6e = 12\) Divide by \(6\): \(e = 2\)

Step 7: Substitute this value back into the expression for f

Now, substitute the value of \(e\) back into the expression for f: \(f = 4(2) - 5\) \(f = 8 - 5\) \(f = 3\) The solution of the given system of equations is \(e=2\) and \(f=3\).

Key Concepts

Elimination Method

The elimination method is a popular technique to solve systems of linear equations by removing one variable, allowing us to solve for the other variable easily. This method involves two main steps:

• Equalizing coefficients: To eliminate a variable, first make their coefficients in both equations equal, or one positive and the other negative. This is usually done by multiplying one or both equations by appropriate numbers.
• Adding or Subtracting Equations: Then, add or subtract the equations to eliminate one variable, resulting in an equation with only one variable.

For example, in the given problem, multiplying the second equation by 3 made the coefficient of variable \(e\) in Equation 2 equal to \(-3\). This allowed us to add the equations together and eliminate \(e\), simplifying the solution process to focus solely on \(f\). The beauty of the elimination method lies in its ability to simplify complex systems into manageable single variable equations that can be easily solved, making it an efficient tool in algebra.

Substitution Method

The substitution method is another effective approach for solving systems of linear equations. It works by isolating one variable in one of the equations and then substituting this expression into the other equation. Here are the basic steps:

• Isolate a variable: Choose one equation and solve for one of the variables in terms of the other.
• Substitute: Substitute this expression into the other equation. This transforms the system into a single equation with one variable.
• Solve: Solve the resulting equation for the one remaining variable.

In the problem, after using elimination to find \(f\) in terms of \(e\), \(f = 4e - 5\), we substituted this expression back into one of the original equations to find \(e=2\). The substitution method is particularly handy when one equation is easily solvable for one variable, making it a flexible choice for various types of linear systems.

Linear Equations

Linear equations are equations of the first degree, meaning they graph as straight lines when plotted on a coordinate plane. Each equation in a system of linear equations represents a line, and the solution to the system is the point where the lines intersect. Here are some essential features of linear equations:

• Standard Form: The general form for a linear equation with two variables is \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants, and \(x\) and \(y\) are variables.
• Graphing: When graphed, a single linear equation will form a straight line, and multiple linear equations will form a system of lines that may intersect, be parallel, or coincide.
• Solving Systems: Solutions are sought by finding the intersecting point of the lines represented by the equations, using methods such as elimination or substitution.

In the given exercise, both equations were manipulated and solved using such methods, showcasing the critical role that understanding and manipulating linear equations plays in finding solutions to these systems. Knowing how to handle linear equations is key to graphing and understanding the relationships between lines in a coordinate plane.