Problem 7 - Algebra and Trigonometry with Analytic Geometry

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

\( x = \frac{212}{265} \), \( y = \frac{28}{53} \).

Step 1: Identify the System of Equations

The given system of linear equations consists of two separate equations: \( 5x - 6y = 4 \) and \( 3x + 7y = 8 \). Our goal is to find the values of \( x \) and \( y \) that satisfy both equations simultaneously.

Step 2: Use the Elimination Method

To eliminate one of the variables, we'll multiply the first equation by \( 3 \) and the second equation by \( 5 \) so that the coefficients of \( x \) are the same:\[3(5x - 6y) = 3(4) \implies 15x - 18y = 12 \]\[5(3x + 7y) = 5(8) \implies 15x + 35y = 40 \]

Step 3: Subtract the Equations

Subtract the first modified equation from the second modified equation to eliminate \( x \): \[(15x + 35y) - (15x - 18y) = 40 - 12 \]Simplify to get:\[ 53y = 28 \]

Step 4: Solve for \( y \)

To find \( y \), divide both sides of the equation by \( 53 \): \[ y = \frac{28}{53} \]

Step 5: Substitute back to Find \( x \)

Use the value of \( y \) in one of the original equations to find \( x \). Substitute \( y = \frac{28}{53} \) into the first equation \( 5x - 6y = 4 \): \[ 5x - 6\left(\frac{28}{53}\right) = 4 \] Simplify and solve for \( x \): \[ 5x = 4 + \frac{168}{53} \] \[ 5x = \frac{212}{53} \] \[ x = \frac{212}{53 \times 5} \] \[ x = \frac{212}{265} \]

Step 6: Simplify \( x \)

Simplify \( x \): \[ x = \frac{212}{265} \]. Check if it can be reduced; since the greatest common divisor is 1, it is already in simplest form.

Key Concepts

Elimination Method

When solving linear systems, the elimination method offers an efficient way to find variable values. The key idea is to eliminate one variable by adding or subtracting the equations, making it simpler to solve for the remaining variable. This often involves aligning the coefficients of one variable between the equations.

First, you might need to adjust the equations so their coefficients (the numbers in front of the variables) are the same. In our problem, we want to eliminate the variable \(x\) first. By multiplying the first equation by 3 and the second by 5, we achieve equal coefficients of 15 for \(x\) in both equations.

With matching coefficients, we subtract the two new equations:

• \(15x - 18y = 12\)
• \(15x + 35y = 40\)

Through subtraction, the \(x\) terms cancel each other out, leaving only the \(y\) term. This step forms a single linear equation, enabling us to solve for \(y\) easily.

Solving Equations

Solving equations is the process of finding unknown variable values that make an equation true. Each step carefully simplifies the equation, following algebraic rules until we isolate and determine the variables.

After eliminating \(x\), we have a simplified single equation in \(y\):

• \(53y = 28\)

To find \(y\), divide both sides by 53, resulting in \(y = \frac{28}{53}\). This fraction is the simplest form of \(y\), meaning it can't be reduced further as the greatest common divisor is 1.

Once we find \(y\), we substitute it back into one of the original equations to solve for \(x\). Choosing the first equation, we replace \(y\) with \(\frac{28}{53}\) and solve:

• \(5x - 6\left(\frac{28}{53}\right) = 4\)

Through simplification and arithmetic, we determine \(x\) as \(\frac{212}{265}\). Checking for simplification confirms it is in the simplest form.

Simultaneous Equations

Simultaneous equations, or systems of equations, consist of multiple equations that share a set of variables. The goal is to find a solution where the same values satisfy all equations at the same time.

In our example, the system is defined as:

• \(5x - 6y = 4\)
• \(3x + 7y = 8\)

Solving simultaneous equations often involves methods like substitution and elimination. Here, we used elimination to systematically reduce the problem to a single solvable equation.

Once we determine values for \(x\) and \(y\), they together form the solution to the system: \(x = \frac{212}{265}\), \(y = \frac{28}{53}\). Checking these values in the original equations ensures they satisfy both, confirming the solution's correctness.