Chapter 7: Problem 36
Solve each equation. See Examples I through 6. \(\frac{15}{x+4}=\frac{x-4}{x}\)
Short Answer
Expert verified
The solution is \( x = 16 \).
Step by step solution
01
Understand the Equation
We have the equation \( \frac{15}{x+4} = \frac{x-4}{x} \). It is a rational equation, meaning it involves fractions with polynomials in the numerator and denominator. Our goal is to solve for \( x \).
02
Cross Multiply
To remove the fractions, we cross multiply. This means we multiply the numerator of each fraction by the denominator of the other fraction:\[(15)(x) = (x-4)(x+4)\].
03
Expand the Right Hand Side
Expand \((x-4)(x+4)\) using the difference of squares formula:\((x-4)(x+4) = x^2 - 16\).
04
Formulate the Equation
After cross multiplying, our equation becomes:\[15x = x^2 - 16\].
05
Rearrange the Equation
Rearrange the equation to one side to form a quadratic equation:\[x^2 - 15x - 16 = 0\].
06
Factor the Quadratic
Try to factor the quadratic equation. We look for factors of -16 that add up to -15. These factors are -16 and 1, so the factored form is:\[(x-16)(x+1) = 0\].
07
Solve for x
Set each factor equal to zero and solve for \( x \):\[x - 16 = 0\] gives \( x = 16 \),\[x + 1 = 0\] gives \( x = -1 \).
08
Verify the Solutions
Check if the solutions are valid by substituting back into the original equation. \( x = 16 \) makes both sides equal, while \( x = -1 \) leads to division by zero; thus, it is not a valid solution. Only \( x = 16 \) is valid.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cross-Multiplication
When dealing with rational equations like \( \frac{15}{x+4} = \frac{x-4}{x} \), you're faced with fractions on both sides of the equation. Cross-multiplication is a method we use to eliminate these fractions and clear the equation for easier solving. In essence, this tactic reduces the problem to simpler arithmetic or algebraic operations.
By cross-multiplying, we multiply the numerator of each fraction by the denominator of the other fraction. For this equation, cross-multiplying involves the following steps:
By cross-multiplying, we multiply the numerator of each fraction by the denominator of the other fraction. For this equation, cross-multiplying involves the following steps:
- Multiply 15 by \( x \) (coming from the second fraction's denominator).
- Multiply \( x-4 \) by \( x+4 \) (coming from the first fraction's denominator).
Quadratic Equations
Once we've cross-multiplied and simplified the rational equation, we end up with a quadratic equation. Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \). In our example, after some rearrangement, we get the quadratic equation \( x^2 - 15x - 16 = 0 \).
Quadratic equations are significant because they exhibit unique behaviors and patterns that allow multiple solving techniques. To better understand them, it's crucial to break down their standard form:
Quadratic equations are significant because they exhibit unique behaviors and patterns that allow multiple solving techniques. To better understand them, it's crucial to break down their standard form:
- \( a \) is the coefficient of \( x^2 \), which determines the parabola's direction.
- \( b \) is the coefficient of \( x \), affecting the parabola's position along the x-axis.
- \( c \) is the constant term, giving the y-intercept.
Factoring Quadratics
Factoring quadratics is a crucial skill used to solve quadratic equations. When a quadratic like \( x^2 - 15x - 16 = 0 \) can be factored, it means you can express it as the product of two binomials. This makes finding the solutions (or roots) much more straightforward.
To factor a quadratic equation, you look for two numbers that multiply to give the constant term \( c \) (in this case, -16) and add up to give the middle coefficient \( b \) (here it's -15). The choice of numbers here is -16 and 1:
To factor a quadratic equation, you look for two numbers that multiply to give the constant term \( c \) (in this case, -16) and add up to give the middle coefficient \( b \) (here it's -15). The choice of numbers here is -16 and 1:
- \( (x - 16) \) and \( (x + 1) \)
- \( x - 16 = 0 \)
- \( x + 1 = 0 \)