Problem 41 - Elementary Algebra Within Reach

Question

In Exercises 35-46, solve the system by the method of substitution. $$ \left\\{\begin{array}{l} \frac{x}{3}-\frac{y}{4}=2 \\ \frac{x}{2}+\frac{y}{6}=3 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution of the system of equations is \(x = 5.25\) and \(y = -1\).

Step 1: Express one of the variables

To solve the system using substitution, express the variable \(x\) from the first equation in terms of \(y\): \[x = 3 (2 + \frac{y}{4}) = 6 + \frac{3y}{4}\]

Step 2: Substitute \(x\) into the second equation

Now substitute \(x\) in terms of \(y\) into the second equation. So the second equation becomes: \[ \frac{6+\frac{3y}{4}}{2} + \frac{y}{6} = 3 \] It's better to get rid of fractions, to that multiply the equation by 12 (common multiple of 2, 4, and 6): \[6(6+ \frac{3y}{4}) + 2y = 36 \] which simplifies to \(45 + \frac{9y}{2} + 2y = 36\). By furhter simplification we get to \( \frac{18y}{2} = -9 \) which results in \( y = -1 \).

Step 3: Solve for \(x\)

Finally, substitute \(y = -1\) back into the equation for \(x\) from Step 1: \[ x = 6 + \frac{3(-1)}{4} = \frac{21}{4} = 5.25 \]

Key Concepts

Understanding the Method of Substitution

The method of substitution is a strategic approach used to solve systems of equations, where we express one variable in terms of the other. This method is particularly useful when at least one equation is easy to solve for one of the variables. Our main goal is to simplify the system by reducing the number of variables involved in each equation.

Here's how it works:

• Choose one equation and solve it for one variable, in terms of the other variable.
• Substitute this expression into the other equation, effectively eliminating one variable.
• Solve the resulting equation for the remaining variable.
• Use this solution to find the value of the first variable.

Throughout this process, we are essentially turning the problem back into a simpler single-variable equation. This substitution clears up the way, allowing us to focus on one equation and one unknown at a time.

Handling Fraction Equations

Dealing with fraction equations can often seem intimidating due to the complexity of fractional expressions. However, careful steps can simplify the task.

In the example we solved, each equation involved fractions. The secret to handling these effectively is to eliminate them by finding a common denominator.

• In our case, the first step was to express \(x\) in terms of \(y\) in the first equation. This manipulation helped us to prepare for substitution by isolating \(x\).
• When performing the substitution in the second equation, we noticed that fractions were still present. We tackled this by multiplying through by the least common multiple of the denominators. Here, multiplying by 12 helped eliminate the fractions, translating the equation into a form that is easier to manage.

Removing fractions simplifies equations and aids in easier arithmetic operations and clearer logical progression. This is a crucial step when you encounter fraction equations in a system.

Step-by-Step Solutions for Clarity

Following a step-by-step approach is vital for solving systems of equations logically and comprehensibly. This method breaks down complex problems into manageable parts, promoting organized problem-solving techniques.

Here is a generalized step-by-step approach:

• Analyze the problem and identify which equation can easily be rearranged to express one variable in terms of the other(s).
• Perform the required algebraic manipulations to isolate one variable.
• Substitute this expression into the other equation and solve for the remaining variable. Careful simplification is key here.
• Back substitute the found value to solve for the first variable and complete the solution.

The step-by-step approach ensures that each operation is accounted for without skipping any critical arithmetic steps. This clear, methodical approach to solving problems not only aids in comprehension but also reduces errors, boosting confidence in working through even the most complex equation systems.