Problem 40 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 21-70: Solve the inequality, and express the solutions in terms of intervals whenever possible. $$ 2 x(6 x+5)<(3 x-2)(4 x+1) $$

Step-by-Step Solution

Answer

The solution is \((-\infty, -\frac{2}{15})\).

Step 1: Expand Both Sides of the Inequality

First, expand the expressions on both sides of the inequality. Start with the left side: \(2x(6x+5) = 12x^2 + 10x\). Next, expand the right side: \((3x-2)(4x+1) = 12x^2 + 3x - 8x - 2 = 12x^2 - 5x - 2\). This gives us:\[12x^2 + 10x < 12x^2 - 5x - 2\]

Step 2: Simplify the Inequality

Cancel out \(12x^2\) from both sides since they are equal. This simplifies the inequality to:\[10x < -5x - 2\]

Step 3: Collect Like Terms

Add \(5x\) to both sides to collect the \(x\)-terms on one side:\[10x + 5x < -2\] This simplifies to:\[15x < -2\]

Step 4: Solve for \(x\)

Divide both sides by \(15\) to solve for \(x\):\[x < -\frac{2}{15}\]

Step 5: Express the Solution in Interval Notation

The solution to the inequality in interval notation is:\((-\infty, -\frac{2}{15})\). This means \(x\) can be any value less than \(-\frac{2}{15}\).

Key Concepts

Solving Inequalities

In mathematics, solving inequalities is similar to solving equations, but instead of finding numbers that make two expressions equal, we find numbers that fit within a range specified by the inequality symbol. When solving inequalities like \( ax + b < cx + d \), we aim to isolate the variable on one side. As shown in the exercise, this is achieved through several steps:

• First, we expand both sides if necessary by removing parentheses and applying the distributive property.
• Next, we simplify the inequality by combining like terms.
• Finally, we solve for the variable by isolating it, often involving addition, subtraction, multiplication, or division.

It's crucial to remember that if you multiply or divide both sides by a negative number, the inequality symbol must be flipped. This rule exists because multiplying or dividing by a negative affects the orientation of the numbers relative to zero. When solving, verify your solution by substituting it back into the original inequality, ensuring that it maintains the inequality's truth.

Interval Notation

Interval notation provides a succinct way to express solution sets for inequalities. Instead of listing all possible solutions, we use intervals to describe their range. In the context of this exercise, we expressed the solution \( x < -\frac{2}{15} \) with interval notation as \((-\infty, -\frac{2}{15})\).
Using interval notation:

• An open interval \((a, b)\) represents all numbers greater than \(a\) and less than \(b\), not including \(a\) and \(b\) themselves.
• A closed interval \([a, b]\) includes \(a\) and \(b\).
• Infinity (\(\pm\infty\)) is always denoted with a parenthesis, \((\infty)\) or \((-\infty)\), as infinity is not a real number that can be attained.

Understanding this symbolism is important for writing mathematical solutions clearly and conveying complete information succinctly. In our scenario, \((-\infty, -\frac{2}{15})\) means that all numbers less than \(-\frac{2}{15}\) (but not including \(-\frac{2}{15}\) itself) are solutions for \(x\).

Polynomial Expansion

Polynomial expansion involves removing parentheses in expressions that are being multiplied, known as expanding. It requires applying the distributive property, a fundamental algebraic concept, and can clarify the terms you're working with.
In our exercise, we expanded both sides:

• The left side: \(2x(6x+5)\) became \(12x^2 + 10x\). This involved distributing \(2x\) into each term inside the parenthesis \((6x+5)\).
• The right side: \((3x-2)(4x+1)\) turned into \(12x^2 - 5x - 2\) by distributing each term \((3x)\) and \((-2)\) by each term \((4x)\) and \((+1)\), then combining like terms.

Knowing how to expand polynomials helps in solving inequalities because it transforms the problem into a more familiar equation without parentheses. This operation simplifies analysis and allows further manipulation, such as combining like terms and eventually isolating variables. Mastery of polynomial expansion is pivotal in algebra, leading up to more complex concepts in calculus and beyond.