Problem 7 - Precalculus: Functions and Graphs

Question

Solve the system. $$\left\\{\begin{array}{l} 5 x-6 y=4 \\ 3 x+7 y=8 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \( x = \frac{76}{53} \) and \( y = \frac{28}{53} \).

Step 1: Write Down the Equations

The given system of equations is: 1. \( 5x - 6y = 4 \) 2. \( 3x + 7y = 8 \)

Step 2: Solve One Equation for One Variable

We'll solve the first equation for \( x \). From \( 5x - 6y = 4 \), solve for \( x \):\[ 5x = 6y + 4 \]\[ x = \frac{6y + 4}{5} \]

Step 3: Substitute in the Second Equation

Substitute \( x = \frac{6y + 4}{5} \) into the second equation \( 3x + 7y = 8 \):\[ 3\left(\frac{6y + 4}{5}\right) + 7y = 8 \]Simplifying, we get:\[ \frac{18y + 12}{5} + 7y = 8 \]

Step 4: Eliminate the Fraction

Multiply every term by 5 to eliminate the fraction:\[ 18y + 12 + 35y = 40 \]

Step 5: Combine Like Terms

Combine the \( y \) terms:\[ 53y + 12 = 40 \]

Step 6: Solve for \( y \)

Isolate \( y \) by subtracting 12 from both sides:\[ 53y = 28 \]Then, divide by 53:\[ y = \frac{28}{53} \]

Step 7: Solve for \( x \)

Substitute \( y = \frac{28}{53} \) back into the equation for \( x \):\[ x = \frac{6\left(\frac{28}{53}\right) + 4}{5} \]Simplifying the expression:\[ x = \frac{\frac{168}{53} + \frac{212}{53}}{5} \]\[ x = \frac{380}{53 \times 5} \]\[ x = \frac{380}{265} \]The fraction can be simplified:\[ x = \frac{76}{53} \]

Step 8: Conclude the Solution

The solution to the system is \( x = \frac{76}{53} \) and \( y = \frac{28}{53} \).

Key Concepts

Substitution Method

The substitution method is a powerful technique for solving systems of linear equations. It's especially useful when equations include variables that can easily be expressed in terms of one another. In essence, you solve one of the equations for one variable and substitute this expression into the other equation.Here’s how it typically works:

• Identify one of the equations in the system that can be easily manipulated to express one variable in terms of the other(s). This equation is referred to as the "solved" equation.
• Rewrite this equation, aiming to isolate one variable on one side of the equation.
• Replace the isolated variable in the other equation with the expression derived from the solved equation.
• Simplify and solve the resulting equation, which now contains only one variable.

In the given exercise, we solved for \( x \) in terms of \( y \) from the first equation and then substituted \( x \)'s expression into the second equation. This approach reduces the system from two equations in two variables to just one equation in one variable.

Solving Equations

Solving equations is a fundamental process in algebra, focusing on finding the value of variables that satisfy a given equation. Whether an equation is simple or complex, the goal is always to solve for the unknown.For linear equations, such as those found in systems, the process involves:

• Isolating the variable to make it the subject of the equation. This could involve operations such as addition, subtraction, multiplication, division, or a combination of these.
• Applying inverse operations — if a term is added, subtract it from both sides; if a term is multiplied, divide both sides by that term, and vice versa.

In our exercise, solving for \( y \) in the substituted equation involved combining like terms and isolating \( y \) by subtracting and dividing using inverse operations. These steps led us to determine \( y = \frac{28}{53} \).Finally, having determined the value of one variable, it’s essential to substitute back into one of the original or derived equations to find the other variable.

Fraction Simplification

Fractions often emerge when dealing with linear equations, especially when solving by substitution or elimination. Simplifying fractions is important for ensuring that the solution is presented in its simplest form, making it easier to interpret.Fraction simplification involves:

• Finding common factors of the numerator and the denominator.
• Dividing both the numerator and the denominator by their greatest common factor (GCF).
• Ensuring the fraction is in its lowest terms.

In the exercise, fraction simplification occurs when we determine \( x = \frac{380}{265} \). By finding the GCF of 380 and 265 — which is 5 — and dividing both by this number, we simplify the fraction to \( x = \frac{76}{53} \).Fraction simplification is more than a mathematical necessity; it's an expressive way to ensure results are clear and accurate when presenting mathematical solutions.