Problem 75 - Intermediate Algebra

Question

Let \(f(x)=5 x+14\) and \(g(x)=2 x+8 .\) Find all values of \(x\) for which \(f(x)>29\) and \(g(x)<20 .\)

Step-by-Step Solution

Answer

The values of \( x \) are \( 3 < x < 6 \).

Step 1: Set up the inequality for f(x)

To find when \( f(x) \) is greater than 29, we write the inequality \( 5x + 14 > 29 \).

Step 2: Solve the inequality for f(x)

Subtract 14 from both sides: \( 5x > 15 \). Then, divide both sides by 5: \( x > 3 \).

Step 3: Set up the inequality for g(x)

To find when \( g(x) \) is less than 20, we write the inequality \( 2x + 8 < 20 \).

Step 4: Solve the inequality for g(x)

Subtract 8 from both sides: \( 2x < 12 \). Then, divide both sides by 2: \( x < 6 \).

Step 5: Combine the inequalities

Combine the two inequalities: \( x > 3 \) and \( x < 6 \). This results in \( 3 < x < 6 \).

Key Concepts

Inequality Solving

In the realm of mathematics, solving inequalities is akin to unraveling a puzzle where we find the value or range of values that satisfy a particular condition. Inequalities are expressions involving the symbols \(<, >, \leq,\) or \(\geq\). They allow us to determine when one expression is less than, greater than, or sometimes equal to another.

Consider an example with the function \(f(x) = 5x + 14\). We are interested in finding when \(f(x) > 29\). Our approach starts by treating the inequality just like a regular equation up until the last step:

• Subtract 14 from both sides to get \(5x > 15\).
• Divide each side by 5, resulting in \(x > 3\).

Remember, the inequality sign flips direction only when multiplying or dividing through by a negative number, which did not occur in our steps. Thus, the solution indicates all \(x\) values greater than 3 satisfy the inequality. Practice reinforces familiarity, so try varying the numbers to master different inequality types!

Systems of Inequalities

When dealing with systems of inequalities, we explore solutions where multiple inequalities must be satisfied simultaneously. This is common in problems involving multiple conditions.

Let's analyze a system comprising \(f(x) = 5x + 14\) and \(g(x) = 2x + 8\), where \(f(x) 69 > 29\) and \(g(x)<20\) must hold true. Already, we solved each separately, finding \(x > 3\) and \(x < 6\).

The key task is combining these into a single, comprehensive solution. We can display them as a compound inequality:

• From the lower bound, \(x > 3\)
• To the upper bound, \(x < 6\)

The solution is the interval \(3 < x < 6\). Here, \(x\) must be greater than 3 and less than 6, satisfying both conditions at once. Imagining these as number lines may help where you search where the overlaps or intersection occurs. This teaches us how every requirement whittles away portions of the number line, narrowing toward our solution set.

Algebraic Expressions

Algebraic expressions form the foundation of solving problems involving inequalities and systems. An algebraic expression is a combination of numbers, variables, operators (like +, -, *, /), and sometimes exponents.

Consider the expression \(5x + 14\). Here:

• \(5x\) indicates 5 times the unknown \(x\), a linear term.
• \(14\) is a constant, which shapes the line's position on a graph.

These expressions are solved by isolation of the variable — through inverse operations. As seen in both \(f(x) 69 ext{and }g(x)\), they translate real-world problems to mathematical models, aiding in seeking solutions systematically. Manipulating terms allows us to move through operations (addition, subtraction, multiplication, division), sharpening our grasp on algebra's power.

Understanding variables and expressions helps in seamlessly transitioning between different mathematical contexts, facilitating ascertaining precise solutions in complex situations like inequality solving.