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Chapter 1: Problem 95

Solve each inequality analytically. Write the solution set in interval notation. Support the answer graphically. $$0.6 x-2(0.5 x+0.2) \leq 0.4-0.3 x$$

Short Answer

Expert verified
The solution is \([-8, \infty)\).

Step by step solution

01

Distribute and Simplify

First, distribute the \(-2\) into the expression \((0.5x + 0.2)\). This results in \(0.6x - 2 \times 0.5x - 2 \times 0.2\). Simplifying this gives \(0.6x - 1.0x - 0.4\). The inequality becomes:\[0.6x - 1.0x - 0.4 \leq 0.4 - 0.3x\] which simplifies further to \[-0.4x - 0.4 \leq 0.4 - 0.3x\].
02

Combine Like Terms

To simplify further, add \(0.3x\) to both sides of the inequality:\[-0.4x + 0.3x - 0.4 \leq 0.4\]This simplifies to \[-0.1x - 0.4 \leq 0.4\].
03

Isolate the Variable

Add \(0.4\) to both sides to isolate the terms with \(x\):\[-0.1x - 0.4 + 0.4 \leq 0.4 + 0.4\]This simplifies to:\[-0.1x \leq 0.8\].
04

Solve for x

Now, divide both sides by \(-0.1\) to solve for \(x\). Remember that when dividing by a negative number, the inequality sign flips:\[x \geq -8\].
05

Express the Solution Set

The solution set in interval notation is:\[[-8, \, \infty)\].
06

Graph the Solution

On a number line, graph by drawing a solid circle at \(-8\) and shading to the right towards positive infinity, indicating all values greater than or equal to \(-8\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Interval Notation
Interval notation is a way of writing subsets of the real number line. It uses brackets and parentheses to describe the range of values that satisfy an inequality.
  • Brackets, like \(\left[a, b\right]\), indicate that the endpoints \(a\) and \(b\) are included in the interval. This is known as a closed interval.
  • Parentheses, like \(\left(a, b\right)\), mean the endpoints are not included. This is referred to as an open interval.
  • For half-open or half-closed intervals, a combination of a bracket and a parenthesis is used, such as \(\left[a, b\right)\) or \(\left(a, b\right]\).
In the context of the inequality \(x \geq -8\), the solution includes all numbers from \(-8\) up to infinity. Since \(-8\) is part of the solution, we use a closed bracket. Therefore, the interval notation is \([-8, \infty)\). Infinity always gets a parenthesis since it's not a number you can reach.
Graphical Solutions
Graphical solutions help visualize the solutions to inequalities on a number line or Cartesian coordinate plane. This approach makes it easier to understand which numbers satisfy the inequality.When dealing with the inequality \(x \geq -8\):
  • First, locate \(-8\) on the number line, and draw a solid circle over it. The solid circle indicates that \(-8\) is included in the solution set.
  • Next, shade the line to the right of \(-8\) to indicate all numbers greater than \(-8\). This shading represents the idea that there are infinitely many solutions stretching towards positive infinity.
Graphical solutions serve as a convenient way to check calculations and provide an intuitive grasp of the problem. When teaching or learning, it reinforces the abstract idea of an inequality with a concrete visual representation.
Distributive Property
The distributive property is a fundamental algebraic concept that helps simplify expressions and solve equations or inequalities. It states the following for any numbers \(a, b, \text{and}\, c\):\[a(b + c) = ab + ac\]In the given exercise, the distributive property is applied to the term \(-2(0.5x + 0.2)\) to expand the expression:
  • Distribute \(-2\) throughout the terms in the parentheses:
    • First, multiply \(-2\) by \(0.5x\), resulting in \(-1.0x\).
    • Then, multiply \(-2\) by \(0.2\), resulting in \(-0.4\).
  • This process transforms the expression into \(-1.0x - 0.4\).
Using the distributive property makes complex expressions simpler and is particularly useful in solving inequalities, as seen here. It clears the way to handle other operations like combining like terms and isolating the variable.

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Most popular questions from this chapter

Solve each equation and inequality analytically. Use interval notation to write the solution set for each inequality. (a) \(1-2 x=0\) (b) \(1-2 x \leq 0\) (c) \(1-2 x \geq 0\)

Solve each equation analytically. Check it analytically, and then support the solution graphically. $$\frac{7}{3}(2 x-1)=\frac{1}{5} x+\frac{2}{5}(4-3 x)$$

Triangles can be classified by their sides. (a) An isosceles triangle has at least two sides of equal length. Determine whether the triangle with vertices \((0,0),(3,4),\) and \((7,1)\) is isosceles. (b) An equilateral triangle has all sides of equal length. Determine whether the triangle with vertices \((-1,-1),(2,3),\) and \((-4,3)\) is equilateral. (c) Determine whether a triangle having vertices \((-1,0),(1,0)\) and \((0, \sqrt{3})\) is isosceles, equilateral, or neither. (d) Determine whether a triangle having vertices \((-3,3),(-2,5)\) and \((-1,3)\) is isosceles, equilateral, or neither.

Solve each equation and inequality analytically. Use interval notation to write the solution set for each inequality. (a) \(5 x+10=0\) (b) \(5 x+10>0\) (c) \(5 x+10<0\)

Find the zero of the function \(f\). \(f(x)=-8 x+0.5(2 x+8)\)
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