Problem 25 - College Algebra Essentials

Question

In Exercises \(19-30,\) solve each system by the addition method. $$ \left\\{\begin{array}{l} 4 x+3 y=15 \\ 2 x-5 y=1 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is x = 3 and y = 1.

Step 1: Plan the elimination

First, plan to eliminate one of the variables when the two equations are added together. If 3y from Equation 1 and -5y from Equation 2 could be made equal but opposite, they would cancel out. This can be achieved by multiplying Equation 1 by 5 and Equation 2 by 3.

Step 2: Multiply the equations

Next, multiply Equation 1 by 5 and Equation 2 by 3 to make the coefficients in front of y equal and opposite. This gives: 20x + 15y = 75 and 6x - 15y = 3.

Step 3: Add the two equations

Following on from Step 2, add the two changed equations together. This means 20x and 6x are added together and +15y and -15y are also added together, achieving an equation of: 26x = 78.

Step 4: Solve for x

For solving the value of x, divide both sides of the equation 26x = 78 by 26, leading to: x = 3.

Step 5: Substitute x = 3 into Equation 1

Now that the value of x is known (x = 3), substitute this into the first original equation so that the value can be found for y. Substituting x = 3 into Equation 1 results in: 4*3 + 3y = 15, which simplifies to: 3y = 15 - 12, or: 3y = 3.

Step 6: Solve for y

After substituting x into the first equation, solve the equation for y. This involves dividing both sides by 3 to get: y = 1.

Key Concepts

Addition Method

The addition method helps you solve systems of linear equations effectively. The idea is to organize and manipulate the equations so that one variable gets eliminated when you add them.
For the given system:

• Start by aligning the equations.
• Notice that if you add them as they are, no variable will cancel out.
• So, adjust the equations by multiplication to set them up for elimination.

This adjustment makes the addition method so powerful. It prepares the equations for seamless elimination of one unknown, making it simpler to find the solution.

Elimination Technique

The elimination technique in algebra involves removing one variable from a system of equations. This is done by adding or subtracting equations after aligning their terms properly.
Here's how it works:

• You need to make the coefficients of one variable opposite in both equations.
• In this case, multiply Equation 1 by 5 and Equation 2 by 3 to balance the y terms to +15y and -15y.
• Add these adjusted equations; the y terms will cancel each other, leaving a single-variable equation to solve.

This method highlights the core advantage of elimination: reducing the system to a simpler, single-variable equation.

Solving Linear Equations

Solving linear equations is a fundamental skill in algebra. Once you eliminate one variable, you're left with a linear equation in the form ax = b.
Here's the approach:

• Simplify the equation if necessary, ensuring only one variable is present.
• Solve for that variable by isolating it on one side of the equation. For instance, divide both sides by the coefficient of the variable.

In our example, after eliminating y, you're left with the equation 26x = 78. Then simply divide by 26 to get x = 3. Remembering these steps helps with clarity and efficiency in finding the solutions.

Algebraic Methods

Algebraic methods provide various techniques for tackling equations, and systems of equations, including the addition method and elimination technique.
They are vital because they offer:

• The capacity to handle complex systems in a step-by-step manner.
• Flexibility, as you can choose the most convenient approach for the problem at hand.
• A solid foundation for more advanced mathematics.

Understanding these methods builds confidence and competence, enabling easy navigation through algebraic challenges from solving basic linear equations to more intricate system resolutions.