Problem 19 - Algebra Form and Function

Question

Solve the systems of equations. $$ \left\\{\begin{array}{l} 3 e+2 f=4 \\ 4 e+5 f=-11 \end{array}\right. $$

Step-by-Step Solution

Answer

Solution: The solution for the system of equations is e = -217/26 and f = 2/13.

Step 1: Write down the system of equations

We are given the following system of linear equations: $$ \left\{\begin{array}{l} 3e + 2f = 4 \\ 4e + 5f = -11 \end{array}\right. $$

Step 2: Rearrange equations to make it suitable for elimination method

We want to create a common coefficient for either e or f in both equations to eliminate one variable. In this case, we can multiply the first equation by 2 and the second equation by 3 to create a common coefficient for f. New system of equations will look like: $$ \left\{\begin{array}{l} 6e + 4f = 8 \\ 12e + 15f = -33 \end{array}\right. $$

Step 3: Eliminate one variable

To eliminate f from the system, we can subtract the first equation from the second equation: $$ (12e + 15f) - (6e + 4f) = -33 - 8 \\ 6e + 11f = -41 $$

Step 4: Solve for e

We got a simplified equation, now divide the equation by the coefficient of e, which is 6: $$ e + \frac{11}{6}f = -\frac{41}{6} $$ Now we can substitute this expression for e back into the first equation of the original system of equations: $$ 3(e + \frac{11}{6}f) + 2f = 4 $$

Step 5: Solve for f

Now we simplify the equation and solve for f: $$ 3e + \frac{33}{2}f + 2f = 4 \\ 3e + \frac{37}{2}f = 4 $$ Substitute the expression for e back into this new equation: $$ 3(-\frac{41}{6} - \frac{11}{6}f) + \frac{37}{2}f = 4 \\ - \frac{123}{6} - \frac{11}{2}f + \frac{37}{2}f = 4 \\ \frac{26}{2}f = 4 + \frac{123}{6} \\ 13f = 2 \\ f = \frac{2}{13} $$

Step 6: Solve for e

Plug the value of f into the expression for e: $$ e + \frac{11}{6}\left(\frac{2}{13}\right) = -\frac{41}{6} \\ e + \frac{22}{78} = -\frac{41}{6} \\ e = -\frac{41}{6} - \frac{22}{78} \\ e = -\frac{217}{26} $$

Step 7: Write the solution

The solution for the system of equations is: $$ e = -\frac{217}{26}, \quad f = \frac{2}{13} $$

Key Concepts

Elimination Method

The elimination method is a popular algebraic technique used for solving systems of linear equations. The basic idea is to eliminate one of the variables by adding or subtracting equations, resulting in a single equation with one variable. This makes it easier to solve the system step by step. Here's a simple breakdown on how the elimination method works:

• Align the equations: Begin by writing the system of equations in standard form, ensuring that similar variables are aligned vertically.
• Create equal coefficients: Adjust the equations, often by multiplying, so that the coefficients of one variable are the same in both equations.
• Eliminate a variable: Add or subtract the equations to eliminate one variable, simplifying to solve the other.
• Substitute back: Use the found value and substitute it back into one of the original equations to find the remaining variable.

The elimination method is generally straightforward, but it requires careful handling of signs and arithmetic to ensure that variables are correctly eliminated.

Linear Equations

Linear equations are equations of the first order. This means they involve only the first power of the variables with no exponents, products of variables, or division by variables. The general form of a linear equation in two variables looks like this:\[ ax + by = c \]Where:

• a, b, and c are constants.
• x and y are variables.

Linear equations can describe various lines on a plane, and when you have two such equations, their graphs will intersect at a point. This intersection represents the solution to the system of equations. The solution, when present, is always a single point that satisfies both equations simultaneously.These equations form the backbone of linear algebra, a fundamental part of high school math and are used to model and solve real-world problems involving constant rates of change.

Solution of Equations

The solution of a system of equations refers to the values of the variables that satisfy all equations in the system simultaneously. There are several scenarios when solving such systems:

• Unique solution: This occurs when the system of equations intersects at exactly one point. This is typically the case in the given example where both lines intersect, yielding a single point \( (e, f) \).
• No solution: This happens if the lines are parallel and never intersect, indicating that there's no common solution that satisfies all equations.
• Infinite solutions: This rare case is when the lines coincide entirely, meaning they overlap and have all points in common.

In our specific example, the solution \( e = -\frac{217}{26}, f = \frac{2}{13} \) is obtained after eliminating one variable and then simplifying the resulting equations. These values make both the original equations true when substituted back. Understanding the solution and the graph of these equations helps students visualize algebraically how solving systems of equations works in practical scenarios.