Problem 48 - Algebra and Trigonometry

Question

Solve each system by the method of your choice. \(\left\\{\begin{array}{l}{\frac{x-y}{3}=\frac{x+y}{2}-\frac{1}{2}} \\\ {\frac{x+2}{2}-4=\frac{y+4}{3}}\end{array}\right.\)

Step-by-Step Solution

Answer

The solution to the system of equations is \(x=6, y=-1\)

Step 1: Simplify both equations

First, clear the fractions from both equations. Multiply the first equation by 6 and the second equation by 6: Equation 1: \(6*\frac{x-y}{3} = 6*\frac{x+y}{2}-\frac{1}{2}*6\) This simplifies to: \(2x-2y = 3x+3y - 3\), simplified further to \(-x-5y=-3\) or \(x + 5y = 3\)Equation 2: \(6*\frac{x+2}{2} -4*6 = 6*\frac{y+4}{3}\) This simplifies to: \(3x+12-24 = 2y + 8\), simplified further to \(3x-2y = 20\) or \(2y = 3x - 20\)

Step 2: Solve one equation for one variable

Isolate \(y\) in the second equation by dividing all terms by 2: \(y = \frac{3x}{2} - 10\)

Step 3: Substitute and solve for the other variable

Substitute \(y = \frac{3x}{2} - 10\) into the first equation \(x + 5y = 3\). This gives: \(x + 5(\frac{3x}{2} - 10) = 3\). Simplify to get: \(x + \frac{15x}{2} - 50 = 3\) or \( \frac{17x}{2} = 53\), \(x = \frac{53*2}{17} = 6\)

Step 4: Find the value of the other variable

Substitute \(x = 6\) into \(y = \frac{3x}{2} - 10\). This renders \(y = \frac{3*6}{2} - 10 = -1\)

Step 5: Confirm the solution

Check these values by substituting them into both original equations and verify that both sides of the equation are equal. \(x=6, y=-1\)

Key Concepts

Algebraic Methods

Algebraic methods are at the heart of solving mathematical problems, particularly equations. When faced with a system of equations, algebra offers various techniques to find the unknown variables that make the equations true. These methods involve manipulations using the fundamental operations of addition, subtraction, multiplication, and division, along with properties like the distributive, associative, and commutative properties.

For systems of equations, algebraic methods include the substitution method, the elimination method, and the graphical method, among others. The final goal is to deduce the precise solution or set of solutions that satisfy all the equations simultaneously. When applied meticulously, these approaches reveal the underlying relationships encapsulated within the mathematical expressions.

Substitution Method

The substitution method is a powerful algebraic approach designed to solve systems of equations by finding the value of one variable and then 'substituting' this value into another equation. This method is particularly useful when one of the equations can be easily solved for one variable either because of its simplicity or because of how the equation is structured.

To apply this method, follow these steps:

• First, identify which variable in one of the equations can be isolated with the least complexity.
• Express that variable in terms of the other(s) within the equation.
• Substitute this expression into the other equation(s), replacing the isolated variable.
• Solve the resulting equation for the other variable.
• Finally, substitute the value of the second variable back into the initially rearranged equation to find the value of the first variable.

This technique not only yields the solution to the system but also serves as a stepping stone to understanding more complex systems of equations.

Elimination Method

The elimination method, also known as the addition method, is another algebraic tool used to solve systems of equations. This method involves lining up two equations and strategically adding or subtracting them in order to eliminate one of the variables, making it possible to solve for the remaining variable.

Use the elimination method by following these steps:

• Rearrange the equations, if necessary, so that like terms align vertically.
• Decide which variable you want to eliminate first. Multiply one or both of the equations by necessary coefficients to make sure the coefficients of that variable are opposites.
• Add or subtract the equations, depending on which operation will eliminate the selected variable.
• Solve the resulting single-variable equation.
• Substitute the found value back into one of the original equations to find the other variable.

This method is especially useful when both equations are set up in a way that makes it easy to eliminate one variable without elaborate manipulation.

System of Linear Equations

A system of linear equations consists of two or more linear equations with the same variables aimed to be solved simultaneously. A solution to the system is an ordered pair (or set of pairs in higher dimensions) that satisfies all of the equations at once. Systems of linear equations can have one solution, no solution, or infinitely many solutions.

These equations represent lines in a Cartesian plane, and the solution to the system corresponds to the point where the lines intersect. For a two-variable system, like the one presented in this exercise, graphically they can be viewed as two lines which may:

• Intersect at one point, implying a unique solution for both variables.
• Be parallel and never intersect, implying that there is no solution (inconsistent system).
• Coincide with each other, implying infinitely many solutions (dependent system).

Understanding the concept of solving a system of linear equations is essential to many fields including mathematics, engineering, economics, and the natural sciences.