Problem 5 - Elementary Algebra

Question

Solve the following linear inequalities. $$ 7(4 s-3)<2 s+8 $$

Step-by-Step Solution

Answer

Answer: The solution to the linear inequality is s < 29/26.

Step 1: Expand the brackets

Multiply 7 with both terms inside the bracket: $$ 28s - 21 < 2s + 8 $$

Step 2: Move all terms to one side

Subtract 2s from both sides and add 21 to both sides of the inequality: $$ 28s - 2s < 8 + 21 $$

Step 3: Simplify the inequality

Combine the like terms on both sides of the inequality: $$ 26s < 29 $$

Step 4: Isolate the variable s

Divide both sides by 26 to isolate s: $$ s < \frac{29}{26} $$ So, the solution for the given inequality is: $$ s < \frac{29}{26} $$

Key Concepts

Understanding Algebra in Inequalities

Algebra is a branch of mathematics that uses symbols to represent numbers in equations and inequalities. In an inequality, these symbols help us to find a range of values that a variable can take to make the inequality true. Algebraic inequalities involve comparing expressions using symbols like <, >, ≤, or ≥. This becomes a powerful tool when solving problems because it focuses on relationships, not just individual numbers.
For instance, when you see an inequality like \(7(4s - 3) < 2s + 8\), the goal is to determine what values of \(s\) make the inequality valid. Here, we replace the simple number play with variables and operators to express more complex relationships.
Understanding algebra in inequalities involves grasping how to manipulate these symbols and expressions to isolate variables and solve for unknowns.

Solving Linear Inequalities

Solving linear inequalities is very similar to solving linear equations, with a few added rules to consider due to the inequality sign.
Just as in equations, you'll perform operations to both sides to isolate the variable. However, keep in mind that when you multiply or divide both sides by a negative number, the inequality sign flips direction.
In the provided inequality, \(7(4s - 3) < 2s + 8\), we don't need to worry about flipping the sign as we won't multiply or divide by a negative.

• First, expand the expressions.
• Next, move variables to one side and constants to the other.
• Simplify and solve for the variable.

This systematic approach makes sure you find the correct range of solutions.

Expanding Brackets in Algebra

When you see an expression with brackets, it often means you should expand it to simplify the equation or inequality. This involves distributing the term outside the brackets across each term inside.
In our exercise, we start by expanding the expression \(7(4s - 3)\). Here is how this step works:
Multiply 7 by each element inside the bracket, resulting in \(7 \times 4s = 28s\) and \(7 \times -3 = -21\). This leads to the expanded expression: \(28s - 21\).
Expanding brackets is crucial as it breaks down complex expressions, making it easier to solve for the variable. It transforms the inequality from its original form into one that's simpler and more workable, allowing you to isolate the variable more easily in later steps.

Variable Isolation in Solving Inequalities

Variable isolation is the process of rearranging an equation or inequality to get the variable by itself on one side. This is a cornerstone of solving inequalities because it directly allows us to find the solution set for the variable.
In the example \(28s - 21 < 2s + 8\), we aim to isolate \(s\) by moving all terms involving \(s\) to one side and constants to the opposite side.

• Subtract \(2s\) from both sides to group \(s\) terms together.
• Add 21 to both sides to handle the constants.

This operation leads to \(26s < 29\), which is more straightforward to solve.
Finally, dividing both sides by 26 isolates \(s\) completely, giving us \(s < \frac{29}{26}\).
Mastering variable isolation involves careful manipulation and understanding of the flexibility in mathematical operations.