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Chapter 3: Problem 100

If \(f(x)=\frac{5}{6} x-\frac{3}{4},\) find the value \((s)\) of \(x\) so that \(f(x)=-\frac{7}{16}\)

Short Answer

Expert verified
The value of \(x\) is \( \frac{3}{8} \).

Step by step solution

01

- Set Up the Equation

Given the function \( f(x) = \frac{5}{6}x - \frac{3}{4} \) and the condition \( f(x) = -\frac{7}{16} \), we set up the equation: \( \frac{5}{6}x - \frac{3}{4} = -\frac{7}{16} \).
02

- Isolate the Linear Term

To isolate the term involving \(x\), add \( \frac{3}{4} \) to both sides of the equation: \( \frac{5}{6}x = -\frac{7}{16} + \frac{3}{4} \).
03

- Find a Common Denominator

Convert \( \frac{3}{4} \) to a fraction with a denominator of 16: \( \frac{3}{4} = \frac{12}{16} \). The equation now reads: \( \frac{5}{6}x = -\frac{7}{16} + \frac{12}{16} \).
04

- Simplify the Right Side

Combine the fractions on the right side: \( \frac{5}{6}x = \frac{12}{16} - \frac{7}{16} = \frac{5}{16} \).
05

- Solve for x

To solve for \(x\), multiply both sides by the reciprocal of \( \frac{5}{6} \), which is \( \frac{6}{5} \): \( x = \frac{5}{16} \times \frac{6}{5} = \frac{6}{16} = \frac{3}{8} \).

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Key Concepts

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

Solving Linear Equations
Solving linear equations is a foundational skill in algebra. A linear equation typically has the form: y = mx + bwhere m and b are constants. To solve a linear equation, follow these steps:
  • Isolate the variable (x): Your goal is to get x by itself on one side of the equation. You do this through operations like addition, subtraction, multiplication, and division.
  • Combine like terms and simplify: If the equation has terms that can be combined, such as constant numbers, do so to simplify the equation.
  • Check your solution: Substitute your solution back into the original equation to ensure it works.
Simplifying and isolating x might involve additional small steps to handle fractions or rearrange terms, but the core idea is always to get x alone.
Function Notation
Function notation is a shorthand way of representing functions. It uses the format f(x) to denote a function named f with x as the input variable. For example, f(x) = mx + b can describe a linear function, where m represents the slope and b the y-intercept. Here are a few points to note:
  • Evaluating functions: To find f(a) for some value a, replace x with a in the function鈥檚 formula.
  • Equality in functions: Setting f(x) equal to a value lets you solve for x, as in the exercise above where f(x) = -7/16.
  • Visual interpretations: Functions can be graphed on a coordinate plane, showing how y changes with x.
Understanding function notation makes it easier to work with and solve various types of algebraic problems.
Fractions in Equations
Handling fractions in equations can seem tricky, but there are strategies to make it easier:
  • Common denominators: When adding or subtracting fractions, convert them to have a common denominator. For example, in the exercise, \[ \frac{3}{4} \] became \[ \frac{12}{16} \] to match \[ \frac{7}{16} \]. This simplifies combining the fractions.
  • Reciprocals: To clear fractions involving the variable, use the reciprocal. In the exercise, multiplying both sides by \[ \frac{6}{5} \]removed the fractional coefficient of x.
  • Avoiding Errors: Be careful with signs and arithmetic operations. Double-check each step.
Mastering these concepts ensures fractions won鈥檛 be a stumbling block in linear equations.
Algebraic Manipulation
Algebraic manipulation involves rearranging equations to isolate variables or simplify expressions. Key techniques include:
  • Combining like terms: Simplify the equation by merging terms with the same variable or constant.
  • Using properties of equality: Add, subtract, multiply, or divide both sides of the equation by the same number to maintain equality.
  • Simplifying expressions: Break down complex expressions into simpler components to make manipulation easier, like converting \[ \frac{3}{4} \] to \[ \frac{12}{16} \].
These skills are crucial not just in linear equations but across diverse topics in mathematics. They provide a toolkit for tackling increasingly complex problems.

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Most popular questions from this chapter

Find the midpoint of the line segment connecting the points (-2,1) and \(\left(\frac{3}{5},-4\right)\)

The slope of the secant line containing the two points \((x, f(x))\) and \((x+h, f(x+h))\) on the graph of a function \(y=f(x)\) may be given as \(m_{\mathrm{sec}}=\frac{f(x+h)-f(x)}{(x+h)-x}=\frac{f(x+h)-f(x)}{h} \quad h \neq 0\) (a) Express the slope of the secant line of each function in terms of \(x\) and \(h\). Be sure to simplify your answer. (b) Find \(m_{\text {sec }}\) for \(h=0.5,0.1\), and 0.01 at \(x=1 .\) What value does \(m_{\text {sec }}\) approach as h approaches \(0 ?\) (c) Find an equation for the secant line at \(x=1\) with \(h=0.01\). (d) Use a graphing utility to graph fand the secant line found in part ( \(c\) ) in the same viewing window. Problems \(85-92\) require the following discussion of a secant line. The slope of the secant line containing the two points \((x, f(x))\) and \((x+h, f(x+h))\) on the graph of a function \(y=f(x)\) may be given as \(m_{\mathrm{sec}}=\frac{f(x+h)-f(x)}{(x+h)-x}=\frac{f(x+h)-f(x)}{h} \quad h \neq 0\) (a) Express the slope of the secant line of each function in terms of \(x\) and \(h\). Be sure to simplify your answer. (b) Find \(m_{\text {sec }}\) for \(h=0.5,0.1\), and 0.01 at \(x=1 .\) What value does \(m_{\text {sec }}\) approach as h approaches \(0 ?\) (c) Find an equation for the secant line at \(x=1\) with \(h=0.01\). (d) Use a graphing utility to graph fand the secant line found in part ( \(c\) ) in the same viewing window. \(f(x)=-3 x+2\)

Simplify \(\sqrt[3]{16 x^{5} y^{6} z}\)

Use a graphing utility. Graph \(y=x^{3}\). Then on the same screen graph \(y=(x-1)^{3}+2\). Could you have predicted the result?

Suppose (1,3) is a point on the graph of \(y=f(x)\) (a) What point is on the graph of \(y=f(x+3)-5 ?\) (b) What point is on the graph of \(y=-2 f(x-2)+1 ?\) (c) What point is on the graph of \(y=f(2 x+3) ?\)
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