/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Solve each equation. $$|5-6 x|... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 6

Solve each equation. $$|5-6 x|=0$$

Short Answer

Expert verified
The solution is \(x = \frac{5}{6}\).

Step by step solution

01

Understanding Absolute Value Equation

We have the equation \(|5-6x| = 0\). An absolute value of a number equals zero only when the number inside the absolute value is zero.
02

Set the Inner Expression to Zero

Since the absolute value is zero, we set the inner expression to zero: \(5 - 6x = 0\).
03

Solve for x

To solve for \(x\), subtract 5 from both sides:\(-6x = -5\).Then, divide both sides by -6:\(x = \frac{-5}{-6} = \frac{5}{6}\).

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Key Concepts

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

Solving Equations
Solving an equation is like finding the answer to a mystery. It means figuring out what value the variable represents to make the equation true.
When we solve any equation, we aim to isolate the variable on one side of the equation. This lets us see directly what value the variable must be.
Think of variables like empty boxes we need to fill, and solving the equation tells us exactly what goes inside each box. The original problem gave us an equation involving absolute values. Here, our task was to determine what value makes the expression inside the absolute value zero. This is because the absolute value of zero is zero. We'll get that value by performing algebraic operations that simplify the equation step by step.
Absolute Value
In mathematics, absolute value is like a magical wrapper that turns everything inside it into a non-negative number. It measures the distance a number is from zero on the number line.
For instance, the absolute values of both 3 and -3 are 3. They are equal in distance away from 0 on the number line, despite being on opposite sides.When dealing with equations, if the absolute value of a number is zero, it means the number itself must be zero. That鈥檚 because zero is the only number that has no distance from itself on the number line.
So in the problem, \[ |5 - 6x| = 0 \] we knew immediately that \[5 - 6x\] must be zero.This immediately simplifies our task because we focus just on what makes the expression inside zero.
Linear Equations
Linear equations are like the straight paths of algebra. They create straight lines when graphed on a coordinate plane. Typically, a linear equation takes the form \[ ax + b = 0 \]where \[ a \] and \[ b \]are constants. They only have the variable \[ x \]to the first power.The equation in our problem, \[ 5 - 6x = 0 \],is a linear equation.
To solve linear equations:
  • First, simplify each side of the equation, if necessary.
  • Then, isolate \[ x \]by performing the same operation on both sides until \[ x \]stands alone.
In this case, we subtracted 5 and divided by -6 to find \[ x = \frac{5}{6} \].
Linear equations are straightforward to manipulate and solve, which makes them an excellent introduction to algebra.

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Most popular questions from this chapter

In this exercise we investigate the effect of the constant \(c\) upon the roots of the quadratic equation \(x^{2}-6 x+c=0\) We do this by looking at the \(x\) -intercepts of the graphs of the corresponding equations \(y=x^{2}-6 x+c\) (a) Set a viewing rectangle that extends from 0 to 5 in the \(x\) -direction and from -2 to 3 in the \(y\) -direction. Then (on the same set of axes) graph the equations \(y=x^{2}-6 x+c\) with \(c\) running from 8 to 10 at increments of \(0.25 .\) In other words, graph the equations \(y=x^{2}-6 x+8, y=x^{2}-6 x+8.25, y=x^{2}-6 x+8.50\) and so on, up through \(y=x^{2}-6 x+10\) (b) Note from the graphs in part (a) that, initially, as \(c\) increases, the \(x\) -intercepts draw closer and closer together. For which value of \(c\) do the two \(x\) -intercepts seem to merge into one? (c) Use algebra as follows to check your observation in part (b). Using that value of \(c\) for which there appears to be only one intercept, solve the quadratic equation \(x^{2}-6 x+c=0 .\) How many roots do you obtain? (d) Some of the graphs in part (a) have no \(x\) -intercepts. What are the corresponding values of \(c\) in these cases? Pick any one of these values of \(c\) and use the quadratic formula to solve the equation \(x^{2}-6 x+c=0 .\) What happens?

Solve the inequalities Suggestion: A calculator may be useful for approximating key numbers. $$\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\left(x+\frac{3}{2}\right)<0$$

Solve the inequalities. Suggestion: A calculator may be useful for approximating key numbers. $$\frac{1}{x-2}-\frac{1}{x-1} \geq \frac{1}{6}$$

Let \(a, b,\) and \(c\) be nonnegative numbers. Follow steps (a) through (e) to show that $$\sqrt[3]{a b c} \leq \frac{a+b+c}{3}$$ with equality holding if and only if \(a=b=c\). This result is known as the arithmetic-geometric mean inequality for three numbers. (Applications are developed in the projects at the ends of Sections 4.6 and \(4.7 .\) ) (a) By multiplying out the right-hand side, show that the following equation holds for all real numbers \(A, B\) and \(C\) $$\begin{aligned} 3 A B C=& A^{3}+B^{3}+C^{3}-\frac{1}{2}(A+B+C) \\ & \times\left[(A-B)^{2}+(B-C)^{2}+(C-A)^{2}\right] \quad \quad (1) \end{aligned}$$ (b) Now assume for the remainder of this exercise that \(A\) \(B,\) and \(C\) are nonnegative numbers. Use equation (1) to explain why $$3 A B C \leq A^{3}+B^{3}+C^{3}\quad \quad (2)$$ (c) Make the following substitutions in inequality ( 2 ): \(A^{3}=a, B^{3}=b,\) and \(C^{3}=c .\) Show that the result can be written $$\sqrt[3]{a b c} \leq \frac{a+b+c}{3} \quad \quad (3)$$ (d) Assuming that \(a=b=c,\) show that inequality (3) becomes an equality. (e) Assuming \(\sqrt[3]{a b c}=\frac{a+b+c}{3},\) show that \(a=b=c\) Hint: In terms of \(A, B,\) and \(C,\) the assumption becomes \(A B C=\frac{A^{3}+B^{3}+C^{3}}{3} .\) Use this to substitute for \(A B C\) on the left-hand side of equation (1). Then use the resulting equation to deduce that \(A=B=C,\) and consequently \(a=b=c\)

Solve the inequalities. Suggestion: A calculator may be useful for approximating key numbers. $$\frac{x^{2}-x-1}{x^{2}+x-1}>0$$
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