Problem 58 - Elementary Algebra

Question

Solve the equation \(5(2 x-1)-4(x+7)=0\).

Step-by-Step Solution

Answer

Answer: \(x = \frac{11}{2}\)

Step 1: Expand the equation

To expand the equation, distribute the coefficients 5 and -4 to the terms inside the parentheses. This results in: \(10x - 5 - 4x - 28 = 0\)

Step 2: Simplify the equation

Combine like terms to simplify the equation. This means adding the x terms (10x and -4x) and the constant terms (-5 and -28): \(6x - 33 = 0\)

Step 3: Solve for x

To solve for x, we need to isolate the variable. Add 33 to both sides of the equation, then divide by 6: \(6x = 33\) \(x = \frac{33}{6}\) Now, simplify the fraction to get the value of x: \(x = \frac{11}{2}\) Therefore, the solution to the equation \(5(2x-1)-4(x+7)=0\) is \(x = \frac{11}{2}\).

Key Concepts

Understanding Linear Equations

Linear equations are mathematical expressions that describe a relationship where each variable is raised to the first power. They take the form of \(ax + b = 0\), where \(a\) and \(b\) are constants, and \(x\) is the variable we want to solve. Linear equations are popular in algebra because they appear in many real-life situations, such as calculating budgets, or determining speeds and distances.

A key characteristic of linear equations is that they graph as straight lines when plotted on a coordinate plane. The solution to a linear equation is the value of \(x\) that makes the equation true. Finding this requires manipulation of the equation according to algebraic rules. Mastering linear equations is fundamental for progressing to more complex algebraic topics.

Algebraic Manipulation Techniques

Algebraic manipulation involves using various operations to rearrange equations and expressions. It's a fundamental skill in solving equations, particularly linear ones. Let's explore some core techniques used in algebraic manipulation:

• **Expanding:** This involves distributing a number across terms within brackets. For example, expanding \(5(2x - 1)\) gives \(10x - 5\).
• **Combining Like Terms:** This process involves adding or subtracting terms that contain the same variable. In the equation from our exercise, \(10x - 4x\) becomes \(6x\).
• **Isolating the Variable:** To solve for \(x\), keep it on one side of the equation by moving other terms to the opposite side. For instance, to solve \(6x - 33 = 0\), add 33 to both sides and then divide by 6.

Each manipulation step simplifies the equation further, making it easier to find the value of the variable. Practicing these methods boosts confidence and proficiency in handling algebraic equations.

The Step-by-Step Solution Process

Understanding a step-by-step approach to solving linear equations is crucial for mastering algebra. Let's review the solution process used in the given exercise:

• **Step 1 - Expanding:** Start by distributing the numbers outside the parentheses to each term inside. For example, in the equation \(5(2x - 1) - 4(x + 7) = 0\), apply distribution to get \(10x - 5 - 4x - 28 = 0\).
• **Step 2 - Simplifying:** Next, combine like terms. Notice how \(10x\) and \(-4x\) add up to \(6x\), and \(-5\) with \(-28\) gives \(-33\). Now the equation reads \(6x - 33 = 0\).
• **Step 3 - Solving:** Finally, isolate \(x\) by performing arithmetic operations. Add 33 to both sides to yield \(6x = 33\), then divide by 6 to find \(x = \frac{33}{6}\), which simplifies to \(x = \frac{11}{2}\).

Each step is important and builds on the previous one. Mastering this process through practice allows you to tackle more complex problems efficiently.