Problem 27 - College Algebra and Calculus: An Applied Approach

Question

Solve the system by the method of substitution. $$\left\\{\begin{array}{l}6 x+5 y=-3 \\ -x-\frac{5}{6} y=-7\end{array}\right.$$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = -1.0\) and \(y = -3.6\).

Step 1: Isolate Variable in One Equation

Let's isolate \(x\) in the second equation: \(-x = 7+ \frac{5}{6}y\), hence \(x= -7 - \frac{5}{6}y \)

Step 2: Substitute Variable in the Other Equation

Substitute \(x\) in the first equation: \( 6(-7 - \frac{5}{6}y) +5y = -3 \) which simplifies to \( -42 -5y +5y = -3 \). Solving for \(y\) yields \(y = -3.6 \)

Step 3: Find Other Variable

Substitute \(y = -3.6\) back into the isolated \(x\) equation: \(x = -7 - \frac{5}{6}(-3.6) \), which simplifies to \(x = -1.0\)

Key Concepts

Substitution Method

The substitution method is a powerful technique to solve systems of equations, particularly useful when you have two equations with two variables. This method involves isolating one variable in one of the equations and then substituting this expression into the other equation. This effectively reduces the system to a single equation with one variable.

Here are the general steps in a more digestible format:

• First, pick one of the equations and solve for one variable. It can be either the one with the simplest coefficient or whichever is easiest to isolate.
• Second, substitute this expression into the other equation and solve for the remaining variable.
• Finally, substitute back to find the first variable using the value you just found.

The great thing about the substitution method is its efficiency with smaller systems and its clarity, especially when equations become complex or equations include fractions.

Algebraic Solution

An algebraic solution is reached when we solve systems of equations using algebraic manipulations rather than graphical or numerical methods. With algebra, we rely entirely on the manipulation of symbols and numbers.

This process ensures that we can pinpoint exact solutions. For instance, in our exercise, through carefully guided steps, we isolated variables and substituted them in equations, leading to a precise solution of \(x = -1.0\) and \(y = -3.6\).

Key advantages of algebraic methods include:

• Exact solutions: Unlike approximations or visual interpretations, algebraic solutions provide precise answers.
• Clear logical steps: Each manipulation can be traced back, making it easy to understand or verify each step.

Understanding algebraic solutions requires a focus on methodical steps and practicing correctly applying these algebraic rules.

Linear Equations

Linear equations are equations of the first order, meaning they involve constants and variables with no exponents other than one. They form straight lines when graphed. In systems of equations, a linear system represents two or more of these equations.

Our system from the exercise, \(6x + 5y = -3\) and \(-x - \frac{5}{6}y = -7\), is a perfect example of a linear system. Each component involves linear terms of \(x\) and \(y\).

These are pivotal in both real-world applications and mathematics education because:

• They represent various relationships: in physics, economics, and more, defining how different variables interrelate.
• Simplicity and power: while they are simple, their solutions can model complex systems when combined with multiple equations.

By getting a firm grasp of linear equations, you lay the groundwork for understanding more complex algebraic concepts and real-life problem-solving scenarios.