Problem 48 - 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) $$ \left(\begin{array}{l} \frac{x-y}{4}-\frac{2 x-y}{3}=-\frac{1}{4} \\ \frac{2 x+y}{3}+\frac{x+y}{2}=\frac{17}{6} \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \( x = 1 \) and \( y = 2 \).

Step 1: Simplify the First Equation

Begin by simplifying the first equation \( \frac{x-y}{4}-\frac{2x-y}{3}=-\frac{1}{4} \). To do this, find a common denominator for the fractions, which in this case is 12. Rewrite the equation:\[ \frac{3(x-y) - 4(2x-y)}{12} = -\frac{3}{12} \]This simplifies to :\[ 3(x-y) - 8x + 4y = -3 \]Simplify further by performing the multiplication:\[ 3x - 3y - 8x + 4y = -3 \]Combine like terms:\[ -5x + y = -3 \].

Step 2: Simplify the Second Equation

Next, simplify the second equation \( \frac{2x+y}{3} + \frac{x+y}{2} = \frac{17}{6} \). Find a common denominator, which is 6:\[ \frac{2(2x+y) + 3(x+y)}{6} = \frac{17}{6} \]This leads to :\[ 4x + 2y + 3x + 3y = 17 \]Combine like terms:\[ 7x + 5y = 17 \].

Step 3: Express y from the First Equation

From the simplified first equation \( -5x + y = -3 \), express \( y \) in terms of \( x \): \[ y = 5x - 3 \].

Step 4: Substitute y in the Second Equation

Substitute \( y = 5x - 3 \) into the second equation \( 7x + 5y = 17 \):\[ 7x + 5(5x - 3) = 17 \]Solve for \( x \):\[ 7x + 25x - 15 = 17 \]\[ 32x - 15 = 17 \]\[ 32x = 32 \]\[ x = 1 \].

Step 5: Find y Using x = 1

Use \( x = 1 \) to find \( y \) using the expression derived in Step 3:\[ y = 5(1) - 3 \]\[ y = 5 - 3 \]\[ y = 2 \].

Step 6: Conclusion: Verify the Solutions

We have \( x = 1 \) and \( y = 2 \). Substitute back into the original equations to verify:- First equation: \( \frac{1-2}{4} - \frac{2(1)-2}{3} = -\frac{1}{4} \) checks out.- Second equation: \( \frac{2(1)+2}{3} + \frac{1+2}{2} = \frac{17}{6} \) checks out.Both equations are satisfied, verifying the solution is correct.

Key Concepts

Substitution Method

The substitution method is a straightforward way to solve systems of linear equations. It involves replacing one variable in an equation with an equivalent expression obtained from another equation. This transforms the system into a single equation with one variable. Here’s how it works:

• First, solve one of the equations for one variable in terms of the other variable.
• Substitute this expression into the other equation. This step eliminates one of the variables, simplifying the problem to a single-variable equation.
• Solve this new equation to find the value of one variable.
• Finally, substitute back to find the value of the second variable.

In our original step-by-step solution, we saw this method in action. After simplifying the first equation, we solved for \( y \) in terms of \( x \), and then we used this expression to substitute \( y \) in the second equation, making it simpler to solve for \( x \). Once we had \( x \), we substituted back to find \( y \). This method is particularly helpful when one equation has a clear isolated variable, making substitution easy and efficient.

Elimination Method

The elimination method, also known as the addition method, is used to solve systems of linear equations by removing one variable. This involves aligning equations so that adding or subtracting them will eliminate one variable.

• Adjust the coefficients through multiplication if necessary, so they match for easy elimination.
• Add or subtract the equations to eliminate one of the variables.
• Solve the resulting single-variable equation.
• Substitute back to find the other variable in one of the original equations.

This method is very effective, particularly when both equations are in a similar form and can be easily manipulated to cancel one of the variables. In our exercise, although the substitution method was used, elimination could have been another suitable approach to consider. By applying elimination, equations could be aligned such that the coefficients of either \( x \) or \( y \) cancel each other upon addition or subtraction.

Linear Equations

Linear equations are algebraic expressions where each term is either a constant or the product of a constant and a single variable. They form straight lines when graphed and are the simplest type of polynomial equations.

• They have the general form \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants.
• The solution to a linear equation is the set of all possible values that satisfy the equation simultaneously.
• When dealing with a system of linear equations, we are looking for common solutions that satisfy all equations in the system.

In our original problem, we were dealing with a set of two linear equations. Our task was to find the values of \( x \) and \( y \) that satisfy both equations at the same time. By manipulating these simple linear relationships—in our exercise through simplification—we could methodically solve the problem, demonstrating the foundational principles of linear equations and their solutions.