Problem 175 - Elementary Algebra

Question

In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination. (a) \(\left\\{\begin{array}{l}8 x-15 y=-32 \\ 6 x+3 y=-5\end{array}\right.\) (B) \(\left\\{\begin{array}{l}x=4 y-3 \\ 4 x-2 y=-6\end{array}\right.\)

Step-by-Step Solution

Answer

System (a) is more convenient to solve by elimination. System (b) is more convenient to solve by substitution.

Step 1: Identify the given systems of equations

The systems of equations provided are: (a) \(\begin{cases}8x - 15y = -32 \ 6x + 3y = -5 \end{cases}\) and (b) \(\begin{cases}x = 4y - 3 \ 4x - 2y = -6 \end{cases}\).

Step 2: Check for Substitution (System (a))

For substitution, one equation should be easily solved for one variable. Let's check if either equation in System (a) is easy to solve for one variable. Neither 8x - 15y = -32 nor 6x + 3y = -5 can be easily solved for one variable initially. Therefore, substitution might not be convenient.

Step 3: Check for Elimination (System (a))

For elimination, we can multiply equations by constants to create coefficients that cancel each other out. Multiply the second equation by 5 to align the coefficients of y: \(5(6x + 3y = -5)\) gives us \(30x + 15y = -25\). Now we add this to the first equation to eliminate y: \(8x - 15y + 30x + 15y = -32 - 25\) which simplifies to \(38x = -57\). Therefore, elimination is more convenient for System (a).

Step 4: Check for Substitution (System (b))

For substitution, check if an equation can be easily solved for one variable. In System (b), the first equation is already solved for x: \(x = 4y - 3\). Substitution is very convenient here.

Key Concepts

Substitution Method

The substitution method is one of the primary ways to solve a system of linear equations. It involves solving one of the equations for one variable and then substituting that expression into the other equation.

Here's the step-by-step approach:

• First, solve one of the equations for one variable.
• Next, substitute that expression into the other equation. This will give you an equation with only one variable.
• Solve the single-variable equation.
• Finally, substitute the value back into the first equation to find the value of the other variable.

The substitution method works best when one of the equations is already solved for one variable or can be easily rearranged to solve for one variable. In the example, System (b) is ideal for substitution because the first equation is already solved for x: \( x = 4y - 3 \). By substituting \( 4y - 3 \) for \( x \) in the second equation, we simplify our work significantly.

Elimination Method

The elimination method is used to eliminate one variable by adding or subtracting the equations. This method is particularly useful when the coefficients of one of the variables are opposites or can be made into opposites easily through multiplication.

Here’s how you do it:

• Adjust the equations so that the coefficients of one variable are opposites. You may need to multiply one or both equations by a constant.
• Add or subtract the equations to eliminate one of the variables. You'll get an equation with only one variable.
• Solve the single-variable equation.
• Then substitute the solved value back into one of the original equations to find the second variable.

In the example, System (a) is more convenient for the elimination method. By multiplying the second equation by 5, we align the coefficients of y: \( 5(6x + 3y) = 30x + 15y \). When we add this to the first equation \( 8x - 15y \), the \( y \) terms cancel out, making it easier to solve for \( x \).

Linear Equations

Linear equations are equations of the first degree, meaning they involve only the first power of the variables. The general form of a linear equation in two variables is \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants. Linear equations graph as straight lines on the coordinate plane.

When solving systems of linear equations, you need to find the values of the variables that satisfy all equations in the system simultaneously. There are three possible outcomes when solving these systems:

• One unique solution: The lines intersect at exactly one point.
• No solution: The lines are parallel and never intersect.
• Infinitely many solutions: The lines coincide, meaning they are the same line.

Understanding the nature of linear equations helps in deciding the appropriate method (substitution or elimination) for solving them. The goal is to simplify the system to make one of these methods work efficiently.