Problem 44 - Elementary Algebra

Question

Solve each system by using either the substitution or the elimination-by- addition method, whichever seems more appropriate. $$\left(\begin{array}{l}\frac{x-y}{4}-\frac{2 x-y}{3}=-\frac{1}{4} \\ \frac{2 x+y}{3}+\frac{x+y}{2}=\frac{17}{6}\end{array}\right)$$

Step-by-Step Solution

Answer

x = 1, y = 2.

Step 1: Simplify the First Equation

The first equation is \( \frac{x-y}{4} - \frac{2x-y}{3} = -\frac{1}{4} \). Find a common denominator for the fractions, which is 12. Rewrite each term with this denominator:\( \frac{3(x-y)}{12} - \frac{4(2x-y)}{12} = -\frac{3}{12} \).This gives: \( \frac{3x - 3y - 8x + 4y}{12} = -\frac{3}{12} \).Simplify to get: \( \frac{-5x + y}{12} = -\frac{3}{12} \).Multiply every term by 12 to eliminate the fraction: \( -5x + y = -3 \).

Step 2: Express y in terms of x from the Simplified First Equation

From the equation \( -5x + y = -3 \), add \( 5x \) to both sides to isolate \( y \):\( y = 5x - 3 \).

Step 3: Substitute y in the Second Equation

The second equation is \( \frac{2x+y}{3} + \frac{x+y}{2} = \frac{17}{6} \).Substitute \( y = 5x - 3 \) into the equation:\( \frac{2x + (5x - 3)}{3} + \frac{x + (5x - 3)}{2} = \frac{17}{6} \).Simplify to \( \frac{7x - 3}{3} + \frac{6x - 3}{2} = \frac{17}{6} \).

Step 4: Solve for x

First find a common denominator for the two fractions on the left side, which is 6, and rewrite:\( \frac{2(7x - 3)}{6} + \frac{3(6x - 3)}{6} = \frac{17}{6} \).This gives the equation:\( \frac{14x - 6 + 18x - 9}{6} = \frac{17}{6} \).Simplify the numerator: \( \frac{32x - 15}{6} = \frac{17}{6} \).Since the denominators are equal, equate the numerators:32x - 15 = 17.Add 15 to both sides: 32x = 32.Divide by 32: x = 1.

Step 5: Find y Using the Expression from Step 2

Substitute \( x = 1 \) back into the expression for \( y \):\( y = 5(1) - 3 \).This simplifies to \( y = 2 \).

Key Concepts

Substitution Method

The substitution method is a great way to solve systems of equations, especially when one equation is already simplified to express one variable in terms of another. In the given problem, after simplifying the first equation, we found it convenient to express \( y \) in terms of \( x \), resulting in \( y = 5x - 3 \).

• Start by isolating one variable in one of the equations if it's possible.
• Next, substitute this expression into the other equation. This step is crucial as it allows you to work with a single equation involving only one variable.
• Solve this new equation for the remaining variable, which in our problem was \( x = 1 \).

After obtaining the value of \( x \), you substitute it back into the expression for \( y \), providing a complete solution for both variables. This method is a straightforward way to tackle systems of equations when substitution seems easiest.

Elimination Method

The elimination method, also known as the addition method, is another powerful technique for solving systems of linear equations. It involves adding or subtracting equations to eliminate one of the variables. In scenarios where equations are already somewhat aligned or coefficients are suitable for elimination, this method shines.

• First, adjust the equations by multiplying them by constants so that the coefficients of one of the variables are opposites or equal.
• When you add or subtract these equations, your aim is to cancel out one of the variables.
• This will leave you with a single equation with one variable, making it easy to solve for that variable.

In our original problem, simplification led us to use substitution, but if the coefficients were more aligned, we could have used the elimination method instead. It's important to be flexible and choose the most convenient approach.

Fractions in Equations

Dealing with fractions in equations can be tricky but with some straightforward steps, it becomes manageable. Finding a common denominator is often the first step, as seen in the original problem. For example, both equations involved fractions with denominators of 4, 3, and 2.

• Convert each term to have a common denominator. This simplifies the transition to working with whole numbers, which are easier to handle.
• Once you have a common denominator, multiply the entire equation by this denominator to eliminate the fractions entirely.
• This process makes the system of equations clearer and more straightforward to solve.

By transforming each term and eliminating fractions, the equations become more manageable and reduce the likelihood of errors during the solution process.