Problem 24 - Intermediate Algebra

Question

For Problems 19-48, solve each system by using either the substitution or the elimination-by-addition method, whichever seems more appropriate. (Objective 2) $$ \left(\begin{array}{l} y=3 x+34 \\ y=-8 x-54 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \( x = -8 \) and \( y = 10 \).

Step 1: Identify the method

We have two linear equations in the form \( y = 3x + 34 \) and \( y = -8x - 54 \). Since both equations are already solved for \( y \), we can use the **substitution method** by setting the right-hand sides of the equations equal to each other.

Step 2: Set the equations equal

Since \( y = 3x + 34 \) and \( y = -8x - 54 \), we can set the equations equal to each other:\[3x + 34 = -8x - 54.\] This step eliminates \( y \) and allows us to solve for \( x \).

Step 3: Simplify the equation

Combine like terms by adding \( 8x \) to both sides:\[3x + 8x + 34 = -54.\]This simplifies to:\[11x + 34 = -54.\]

Step 4: Solve for x

Subtract 34 from both sides to isolate the term with \( x \):\[11x = -54 - 34.\]Simplify the right-hand side:\[11x = -88.\]Divide by 11:\[x = -8.\]

Step 5: Substitute back to find y

Substitute \( x = -8 \) back into the first equation \( y = 3x + 34 \):\[y = 3(-8) + 34.\]Simplify:\[y = -24 + 34 = 10.\]

Step 6: Verify the solution

Verify \( (x, y) = (-8, 10) \) satisfies the second equation \( y = -8x - 54 \):\[10 = -8(-8) - 54.\]Simplify:\[10 = 64 - 54 = 10.\]Both equations are satisfied, confirming the solution is correct.

Key Concepts

Substitution Method

The substitution method is a technique used to solve a system of linear equations. It's particularly handy when one of the equations is already solved for a single variable, as is the case here. We have:

• Equation 1: \( y = 3x + 34 \)
• Equation 2: \( y = -8x - 54 \)

Since both equations define \( y \), the substitution method allows us to eliminate \( y \) by equating the two expressions for it. This turns the problem into a simpler equation with just one unknown, which you solve to find \( x \). You then substitute back to find \( y \). This method works efficiently when equations are designed for direct substitution.

Elimination Method

The elimination method is another effective technique for solving systems of linear equations. In this approach, the goal is to eliminate one of the variables by adding or subtracting the equations. This can be done when both equations have the same variable coefficient, allowing straightforward elimination.

While in this particular exercise we didn't use it (due to the form of the equations favoring substitution), the elimination method is crucial for systems not easily formatted for substitution. You adjust the equations to align coefficients, ensuring that when you add or subtract, one variable cancels out. This leads to an equation with just one variable to solve, similar to substitution.

Linear Equations

Linear equations are algebraic expressions of the form \( ax + by = c \), representing straight lines when graphed on a coordinate plane. In this exercise, our system consisted of two linear equations in the slope-intercept form:

• Equation 1: \( y = 3x + 34 \)
• Equation 2: \( y = -8x - 54 \)

These equations represent lines with different slopes and y-intercepts. The solution to the system of equations is the point where the two lines intersect. For this case, the solution \( (x, y) = (-8, 10) \) is the intersection point.

Algebraic Solutions

Finding algebraic solutions for systems of linear equations involves using methods like substitution and elimination to find values of variables that satisfy all given equations simultaneously. Here, we identified a variable from one equation (\( y = 3x + 34 \)) and substituted it into the other (\( y = -8x - 54 \)) to solve for \( x \).

This step-by-step approach helps break down complex algebraic problems into manageable pieces, enabling us to find precise solutions. By substituting back, we verify that our solution meets both original equations. This ensures accuracy and confirms that the solution represents the intersection of the lines on a graph, solidifying the understanding of solving linear systems algebraically.