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Chapter 1: Problem 37

Solve the equation or inequality. Express the solutions in terms of intervals whenever possible. $$\frac{3 x+1}{5 x+7}=\frac{6 x+11}{10 x-3}$$

Short Answer

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

Step by step solution

01

Cross Multiply

To eliminate the fractions, we cross-multiply. Set \\[ (3x + 1)(10x - 3) = (6x + 11)(5x + 7) \]This gives us two new algebraic expressions without denominators.
02

Expand Both Sides

Now we need to expand both sides of the equation:- Left side: \[ 3x \cdot 10x + 3x \cdot (-3) + 1 \cdot 10x + 1 \cdot (-3) = 30x^2 - 9x + 10x - 3 \] Simplify to get: \[ 30x^2 + x - 3 \]- Right side: \[ 6x \cdot 5x + 6x \cdot 7 + 11 \cdot 5x + 11 \cdot 7 = 30x^2 + 42x + 55x + 77 \] Simplify to get: \[ 30x^2 + 97x + 77 \]
03

Set Up the Equation

After expanding, set both sides equal and simplify:\[ 30x^2 + x - 3 = 30x^2 + 97x + 77 \]Subtract \(30x^2\) from both sides to further simplify: \\[ x - 3 = 97x + 77 \]
04

Solve for x

Rearrange the terms to isolate \(x\) on one side:\[ x - 97x = 77 + 3 \]Combine like terms:\[ -96x = 80 \]Divide both sides by \(-96\) to solve for \(x\):\[ x = \frac{80}{-96} = -\frac{5}{6} \]
05

Check for Extraneous Solutions

Verify that \(x = -\frac{5}{6}\) is not an extraneous solution by checking the original denominators:- For \(5x + 7\), substitute \(x = -\frac{5}{6}\): \[ 5\left(-\frac{5}{6}\right) + 7 = \frac{-25}{6} + \frac{42}{6} = \frac{17}{6} eq 0 \]- For \(10x - 3\), substitute \(x = -\frac{5}{6}\): \[ 10\left(-\frac{5}{6}\right) - 3 = \frac{-50}{6} - \frac{18}{6} = \frac{-68}{6} eq 0 \]Since neither denominator equals zero, \( x = -\frac{5}{6} \) is a valid solution.

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Key Concepts

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

Cross-Multiplication
When solving rational equations, cross-multiplication is a powerful technique to eliminate fractions. It helps simplify the problem by creating a single variable equation.
  • Begin by multiplying the numerator of the first fraction by the denominator of the second fraction.
  • Then, multiply the numerator of the second fraction by the denominator of the first fraction.
  • After cross-multiplying, equate the two resulting expressions to form a new equation.
In our exercise, cross-multiplication allowed us to move from the fraction equation \( \frac{3x+1}{5x+7} = \frac{6x+11}{10x-3} \) to the algebraic expression \( (3x+1)(10x-3) = (6x+11)(5x+7) \). This step paved the way for a clearer path to simplify and solve the equation. Remember, this method works only when both sides of the equation are proper fractions.
Quadratic Expressions
Quadratic expressions often emerge when solving rational equations using cross-multiplication. Recognizing and simplifying these can simplify your path toward finding solutions.
  • A quadratic expression usually takes the form \( ax^2 + bx + c \).
  • These expressions can result from expanding products such as \((3x+1)(10x-3)\).
  • After expanding, always combine like terms to simplify; for example, \(30x^2 - 9x + 10x - 3\) simplifies to \(30x^2 + x - 3\).
Understanding the structure of quadratic expressions is crucial. They help identify the type of solutions (real or complex) and facilitate the process of solving equations by setting them to zero or simplifying further. In practice, manipulating these expressions can uncover terms we didn't initially see.
Extraneous Solutions
In mathematical equations, particularly rational ones, extraneous solutions can pop up unexpectedly. These are solutions that make the original equation undefined or don't satisfy it upon re-evaluation.
  • After solving, always substitute back into the original equation to check validity.
  • Pay special attention to variables in the denominators, as these could make expressions undefined if they turn zero.
  • In our example, checking \( x = -\frac{5}{6} \) against original denominators \( 5x + 7 \) and \( 10x - 3 \), neither equaled zero, confirming it's a valid solution.
Avoid the trap of assuming all calculated solutions are correct without verification. Extraneous solutions are crucial considerations that ensure sound and valid results.
Interval Notation
Interval notation is a way of representing a set of solutions or possible values of a variable in mathematics. Although our particular problem resulted in just one solution, understanding and using interval notation is important.
  • It expresses solution sets compactly, covering ranges of values.
  • Closed intervals use square brackets \([a, b]\), indicating endpoints are included.
  • Open intervals use parentheses \((a, b)\), meaning endpoints are not included.
In more complex scenarios, solutions might range across intervals or have multiple intervals. For example, \((- rac{5}{6}, \infty)\) would imply all values greater than \(-\frac{5}{6}\). Mastering interval notation enhances the ability to communicate mathematical solutions efficiently and correctly.

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Most popular questions from this chapter

Choose the equation that best describes the table of data. (Hint: Make assignments to \(\mathbf{Y}_{\mathbf{r}}-\mathbf{Y}_{\mathbf{4}}\) and examine a table of their values.) $$\begin{array}{|c|c|}\hline x & y \\\\\hline 1 & 0.8 \\\\\hline 2 & -0.4 \\\\\hline 3 & -1.6 \\\\\hline 4 & -2.8 \\\\\hline 5 & -4.0 \\\\\hline\end{array}$$ (1) \(y=-1.2 x+2\) (2) \(y=-1.2 x^{2}+2\) (3) \(y=0.8 \sqrt{x}\) (4) \(y=x^{3 / 4}-0.2\)

Replace the symbol \(\square\) with either \(=\) or \(\neq\) to make the resulting statement true, whenever the expression has meaning. Give a reason for your answer. $$ \left(a^{\prime}\right)^{2} \square a^{(r)} $$

Simplify the expression, and rationalize the denominator when appropriate. $$\sqrt[3]{3 t^{4} v^{2}} \sqrt[3]{-9 t^{-1} v^{4}}$$

When a tomado passes near a building, there is a rapid drop in the outdoor pressure and the indoor pressure does not have time to change. The resulting difference is capable of causing an outward pressure of \(1.4 \mathrm{Ib} / \mathrm{in}^{2}\) on the walls and ceiling of the building. (a) Calculate the force in pounds exerted on 1 square foot of a wall. (b) Estimate the tons of force exerted on a wall that is 8 feet high and 40 feet wide.

Rewrite the expression using rational exponents. $$\sqrt{a+\sqrt{b}}$$
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