Problem 46 - Algebra A Combined Function

Question

Solve. $$ 6(5+c)=5(c-4) $$

Step-by-Step Solution

Answer

The solution is \(c = -50\).

Step 1: Distribute the numbers in the parentheses

Start by applying the distributive property to both sides of the equation. For the left side, distribute the 6 across the terms in the parenthesis: \(6 \times 5 + 6 \times c = 30 + 6c\) For the right side, distribute the 5: \(5 \times c - 5 \times 4 = 5c - 20\). Now, the equation reads: \(30 + 6c = 5c - 20\).

Step 2: Move variables to one side

Subtract \(5c\) from both sides of the equation to bring all terms involving \(c\) to one side: \((30 + 6c) - 5c = (5c - 20) - 5c\) This simplifies to: \(30 + c = -20\).

Step 3: Isolate the variable

Subtract 30 from both sides to solve for \(c\): \(c = -20 - 30\) Simplifying the right side gives: \(c = -50\).

Step 4: Verify the solution

Substitute \(c = -50\) back into the original equation to ensure the solution is accurate:Left side: \(6(5 - 50) = 6 \times (-45) = -270\) Right side: \(5(-50 - 4) = 5 \times (-54) = -270\) Both sides are equal, confirming that \(c = -50\) is the correct solution.

Key Concepts

Linear Equations

Linear equations are fundamental in algebra, and they involve variables raised only to the first power. They typically take the form \(ax + b = c\), where \(a\), \(b\), and \(c\) are constants. In our exercise, we have the equation \(6(5+c) = 5(c-4)\), which is a linear equation involving the variable \(c\).

The primary characteristic of a linear equation is that when graphed, it forms a straight line. This straight-line relationship means that the rate of change is constant throughout the equation. Understanding how to manipulate these equations is essential as they form the basis for more complex algebraic problems. As you become more comfortable with solving linear equations, you'll find it easier to tackle different forms and systems of equations.

Solving Equations

Solving equations involves a series of logical steps aimed at finding the value of the unknown variable. The goal is to isolate the variable on one side of the equation and a number on the other side. In this process, we consider maintaining the balance of the equation by performing the same operations on both sides.

In our example:

• We begin by distributing the terms to eliminate the parentheses: \(6 \times 5 + 6 \times c\) on the left and \(5 \times c - 5 \times 4\) on the right.
• Next, we move all terms involving the variable \(c\) to one side and constants to the other. This often involves operations like addition or subtraction.
• Finally, the equation is simplified step-by-step to find the value of the variable (\(c = -50\) in this case).

Summary: Solving equations requires patience and practice. Eventually, the process becomes more intuitive with repeated application of these steps.

Algebraic Manipulation

Algebraic manipulation refers to the process of rearranging and simplifying expressions and equations to achieve a desired form or solution. It involves the application of various algebraic properties and operations such as the distributive property, combining like terms, and performing operations on both sides of an equation.

In the provided solution, we used the distributive property to simplify both sides of the equation by multiplying each term within the parentheses by the factor outside. Another key manipulation step was to align the variable terms on one side of the equation:

• Subtracting \(5c\) from both sides to get all \(c\)-terms together
• Subtracting 30 from both sides to bring the constants together

These manipulations simplified the equation to a point where we could clearly see the value of the variable. Mastery of algebraic manipulation is essential for solving not just linear equations, but also more complex algebraic expressions and systems of equations.