Problem 10 - Intermediate Algebra

Question

For Problems \(1-18\), use the elimination-by-addition method to solve each system. (Objective 1 ) $$ \left(\begin{array}{l} 4 x+3 y=-4 \\ 3 x-7 y=34 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution to the system is \(x = 2\) and \(y = -4\).

Step 1: Choose a variable to eliminate

We'll start by choosing one of the variables, either \(x\) or \(y\), to eliminate. \(y\) looks like a suitable candidate because the coefficients of \(y\) in the two equations can easily be manipulated to cancel each other out.

Step 2: Make the coefficients of \(y\) equal in magnitude

The coefficient of \(y\) in the first equation is 3, and in the second it's -7. To eliminate \(y\), we'll make the coefficients equal in magnitude by finding the least common multiple. Here, it is 21. Multiply the entire first equation by 7 and the second equation by 3.

Step 3: Multiply the equations

Multiply the first equation by 7: \(7(4x + 3y) = 7(-4)\) yields \(28x + 21y = -28\). Multiply the second equation by 3: \(3(3x - 7y) = 3(34)\) yields \(9x - 21y = 102\).

Step 4: Add the equations

Now we add the two resulting equations to eliminate \(y\). \((28x + 21y) + (9x - 21y) = -28 + 102\). This simplifies to \(37x = 74\).

Step 5: Solve for \(x\)

Solve the equation \(37x = 74\) for \(x\) by dividing both sides by 37. This yields \(x = 2\).

Step 6: Substitute \(x = 2\) back into one of the original equations

To find \(y\), substitute \(x = 2\) back into the first original equation: \(4(2) + 3y = -4\), which simplifies to \(8 + 3y = -4\).

Step 7: Solve for \(y\)

Subtract 8 from both sides to isolate \(3y\): \(3y = -12\). Divide both sides by 3 to solve for \(y\), obtaining \(y = -4\).

Step 8: Check the solution

Substitute \((x, y) = (2, -4)\) into both original equations to ensure correctness. For the first equation: \(4(2) + 3(-4) = 8 - 12 = -4\). For the second equation: \(3(2) - 7(-4) = 6 + 28 = 34\). Both equations hold true, confirming our solution is correct.

Key Concepts

Solving Systems of Equations

When we refer to solving systems of equations, we are essentially looking for a solution that satisfies all given equations simultaneously. In the context of linear equations, these solutions typically represent points of intersection on a graph.
The elimination method, sometimes known as the elimination-by-addition method, is a straightforward approach to solving these systems. By systematically adding or subtracting equations, we can eliminate one of the variables, making it easier to solve for the remaining variable.
Key steps in this process include:

• Selecting a variable to eliminate based on convenience, such as having coefficients that can easily be manipulated.
• Adjusting the coefficients by multiplying the equations to make one variable's coefficients opposites.
• Adding the equations to cancel out one variable, thus reducing the system from two equations to one.

Once you solve for one variable, substitute it back to find the other variable and verify by substituting both values into the original equations.

Linear Equations

Linear equations are equations that make a straight line when graphed. They have the general form: \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants.
This simplicity allows for several straightforward methods to find where lines intersect, like the elimination method used above.
When dealing with linear equations in a system, it’s important to:

• Keep track of coefficients as they directly affect the manipulation needed during elimination.
• Recognize that the solution represents the x and y coordinates of the intersection point of the lines.

Understanding the nature of linear equations is crucial for solving them effectively, as it allows for predictions about the nature of solutions, such as whether they are unique, infinite, or nonexistent.

Intermediate Algebra

Intermediate algebra serves as a bridge between basic algebraic concepts and advanced topics. It involves a deeper engagement with equations and manipulations, as seen in solving systems of equations.
The concepts covered in intermediate algebra are foundational for techniques such as the elimination method. Skills developed in this stage include:

• Proficiency with the manipulation of equations, such as scaling and combining them.
• Enhanced problem-solving strategies that apply algebraic manipulations to arrive at solutions efficiently.
• Understanding the importance of verification to ensure solutions are correct.

A good grasp of intermediate algebra not only prepares students for more complex mathematical topics but also enhances analytical skills, beneficial in many disciplines.