Problem 88 - Algebra for College Students

Question

Solve each system. \(\left(\begin{array}{l}\frac{4}{x}+\frac{1}{y}=11 \\\ \frac{3}{x}-\frac{5}{y}=-9\end{array}\right)\)

Step-by-Step Solution

Answer

The solution is \( x = \frac{1}{2} \), \( y = \frac{1}{3} \).

Step 1: Identify the System of Equations

We are given the system of equations: \( \frac{4}{x} + \frac{1}{y} = 11 \) and \( \frac{3}{x} - \frac{5}{y} = -9 \). We aim to solve these equations to find the values of \( x \) and \( y \).

Step 2: Substitute Variables

Let's substitute \( \frac{1}{x} = a \) and \( \frac{1}{y} = b \). This transforms our system into: \( 4a + b = 11 \) and \( 3a - 5b = -9 \).

Step 3: Solve the New System

We now have two equations: 1. \( 4a + b = 11 \)2. \( 3a - 5b = -9 \)From equation 1, express \( b \) in terms of \( a \): \( b = 11 - 4a \). Substitute this expression for \( b \) into equation 2.

Step 4: Substitute and Simplify

Substituting \( b = 11 - 4a \) into \( 3a - 5b = -9 \), we have: \( 3a - 5(11 - 4a) = -9 \). Simplify this equation: \( 3a - 55 + 20a = -9 \). Combine terms: \( 23a - 55 = -9 \).

Step 5: Solve for \( a \)

Add 55 to both sides: \( 23a = 46 \). Then divide by 23: \( a = 2 \).

Step 6: Solve for \( b \)

Substitute \( a = 2 \) back into \( b = 11 - 4a \): \( b = 11 - 4(2) \).Calculate: \( b = 11 - 8 = 3 \).

Step 7: Convert Back to Variables \( x \) and \( y \)

Since \( a = \frac{1}{x} \) and \( b = \frac{1}{y} \), we have:\( \frac{1}{x} = 2 \ Rightarrow x = \frac{1}{2} \).\( \frac{1}{y} = 3 \ Rightarrow y = \frac{1}{3} \).

Step 8: Verify Solution

Substitute \( x = \frac{1}{2} \) and \( y = \frac{1}{3} \) back into the original equations to verify correctness:For equation 1: \( \frac{4}{\frac{1}{2}} + \frac{1}{\frac{1}{3}} = 8 + 3 = 11 \). For equation 2: \( \frac{3}{\frac{1}{2}} - \frac{5}{\frac{1}{3}} = 6 - 15 = -9 \). Both equations are satisfied with \( x = \frac{1}{2} \) and \( y = \frac{1}{3} \).

Key Concepts

Substitution Method

The substitution method is a powerful technique used to solve systems of equations. It involves expressing one variable in terms of the other and then using this expression to replace the variable in the second equation.

This approach simplifies the problem, transforming a system of two equations into a single equation with one variable. In our exercise, we started by substituting:

• \( \frac{1}{x} = a \)
• \( \frac{1}{y} = b \)

This transformation transformed our original system into linear equations that were easier to handle:

• \( 4a + b = 11 \)
• \( 3a - 5b = -9 \)

By smartly using substitution, problem-solving becomes more manageable and helps systematically solve equations that might be complex to handle otherwise.

Solving Linear Equations

Solving linear equations is all about finding the value of the unknown variables that satisfy the equations. Linear equations form straight lines when graphed and usually take the form of \( ax + by = c \).

In our exercise, after substituting, the equations were already set up for easy solving. By rearranging terms and isolating a single variable, you can find specific values for each variable. We expressed \( b \) from the first equation as \( b = 11 - 4a \) then used that insight to replace \( b \) in the second equation.

This step-by-step approach ensures accuracy and is foundational in working through equations effectively. The objective is always to simplify until you have your variables neatly isolated and their values clearly determined.

Variable Elimination

Variable elimination is a strategic method used to remove a variable from a system of equations, allowing us to focus on a single variable to solve for first. It pairs nicely with the substitution method when you transform and simplify equations.

In our example, elimination was stealthily performed through substitution when \( b = 11 - 4a \) was plugged into the other equation. This automatically eliminated \( b \) from the consideration, reducing it to just an equation involving \( a \):

• \( 3a - 5(11 - 4a) = -9 \)
• Simplifying to \( 23a = 46 \)

Then it was straightforward to solve for \( a \). After solving for one variable, substitute back to find others. This tactical reduction simplifies balancing equations, especially when handling complex systems, demonstrating an essential skill in algebra.