Problem 21 - Elementary Algebra

Question

Solve each system by using either the substitution or the elimination-by- addition method, whichever seems more appropriate. $$\left(\begin{array}{l}x=-6 y+79 \\ x=4 y-41\end{array}\right)$$

Step-by-Step Solution

Answer

The solution is \(x = 7\) and \(y = 12\).

Step 1: Set up equations

The given system of equations is \(x = -6y + 79\) and \(x = 4y - 41\). Both equations express \(x\) in terms of \(y\).

Step 2: Equate expressions for x

Since both equations equal \(x\), we set the expressions equal to each other: \(-6y + 79 = 4y - 41\).

Step 3: Simplify the equation

Rearrange the terms to isolate \(y\) on one side: \(-6y - 4y = -41 - 79\). This simplifies to \(-10y = -120\).

Step 4: Solve for y

Divide both sides by \(-10\) to solve for \(y\): \(y = 12\).

Step 5: Substitute y back to find x

Substitute \(y = 12\) back into one of the original equations to find \(x\). Using \(x = 4y - 41\), we get \(x = 4(12) - 41 = 48 - 41 = 7\).

Step 6: Verify the solution

Substitute both \(x = 7\) and \(y = 12\) back into the original equations to ensure they satisfy both: - For \(x = -6y + 79\): \(7 = -6(12) + 79 = -72 + 79 = 7\).- For \(x = 4y - 41\): \(7 = 4(12) - 41 = 48 - 41 = 7\).Both are correct.

Key Concepts

Substitution Method

The substitution method is a technique for solving systems of equations, where you solve one equation for one variable and then substitute this expression into another equation. This can transform the system into one equation with one unknown, making it possible to solve.

For instance, consider two equations:

• Equation 1: \( x = -6y + 79 \)
• Equation 2: \( x = 4y - 41 \)

Both equations are already solved for \(x\), which makes substitution straightforward.

Since both expressions equal \(x\), set them equal to each other: - \(-6y + 79 = 4y - 41\).

This results in one equation, which can be simplified to find \(y\), and consequently used to find \(x\). Substituting and simplifying helps eliminate one variable early on and is extremely helpful when equations are in this format.

Elimination Method

The elimination method involves adding or subtracting equations to cancel out one of the variables. This strategy is particularly effective when systems of equations are aligned and ready to eliminate a variable by addition or subtraction.

This method works best when two equations are structured in a similar way, with the coefficients of one variable allowing them to cancel each other out when equations are added or subtracted.

Let’s consider an example using these simplified general forms:

• Equation 1: \(a_1x + b_1y = c_1\)
• Equation 2: \(a_2x + b_2y = c_2\)

To eliminate \(x\), multiply the equations so the coefficients of \(x\) are opposites. Then add the equations together, cancel \(x\), and solve for \(y\). Once \(y\) is found, substitute back into either original equation to solve for \(x\).

This method can be less intuitive than substitution but is incredibly powerful for larger systems.

Linear Equations

Linear equations are a fundamental part of algebra, representing relationships between variables with constant rates of change. They are typically written in the form \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants.

When you graph linear equations on a coordinate plane, they yield straight lines, hence the name 'linear'. The intersection point of two lines represents the solution to a system of equations.

In systems of linear equations, solutions can be found where the lines intersect:

• No intersection: This means the lines are parallel, implying no solutions.
• Exactly one intersection: The lines intersect at one point, indicating a unique solution.
• Infinite intersections: The lines coincide and overlap entirely, suggesting infinitely many solutions.

Understanding linear equations is crucial for solving more complex systems, as they form the building blocks of many algebraic concepts.

They offer a visual way to interpret solutions, thus deepening comprehension and intuition around solving equations.