Problem 92

Question

$89-96$ m State whether the given equation is true for all values of the variables. (Disregard any value that makes a denominator zero.) $$ \frac{x+1}{y+1}=\frac{x}{y} $$

Step-by-Step Solution

Verified

Answer

The equation is true when $ y = x $, with $ x eq 0 $ and $ x eq -1 $.

1Step 1: Subtract Fractions

First, we simplify and rewrite the equation by subtracting the right-hand side fraction from the left-hand side. This gives us:\[ \frac{x+1}{y+1} - \frac{x}{y} = 0 \]

2Step 2: Common Denominator

To subtract these fractions, find a common denominator, which is $y(y+1)$. Rewriting each fraction with this common denominator yields:\[ \frac{(x+1)y - x(y+1)}{y(y+1)} = 0 \]

3Step 3: Simplify the Numerator

Now, expand and simplify the numerator:\[ (x+1)y - x(y+1) = xy + y - xy - x = y - x \]So the equation becomes:\[ \frac{y - x}{y(y+1)} = 0 \]

4Step 4: Check Numerator Zero Condition

For the fraction to be zero, the numerator has to be zero while the denominator should not be zero. Therefore, we solve:\[ y - x = 0 \]or\[ y = x \]

5Step 5: Check Denominator Non-Zero Condition

The denominator $y(y+1)$ must not be zero. Hence, $y eq 0$ and $y eq -1$.This condition doesn't contradict $y = x$ as long as $x eq 0$ and $x eq -1$.

Key Concepts

FractionsCommon DenominatorNumerator SimplificationZero Condition

Fractions

In mathematics, a fraction represents a part of a whole number or a ratio between two numbers, typically arranged in a numerator over a denominator. In our given equation $\frac{x+1}{y+1} = \frac{x}{y}$, two fractions are equated. This means that the ratio of the two expressions on the left and right must be equal.

The numerator is the number above the fraction line and indicates how many parts we have.
The denominator is the number below the fraction line, showing total equal parts the whole is divided into.
A key aspect of understanding fractions is recognizing how to manipulate them to simplify equations or find solutions.

Fractions allow us to perform calculations without converting them into decimals, maintaining precision in mathematical operations.

Common Denominator

When dealing with equations involving fractions, a common denominator is often necessary to simplify the expression properly. In the original solution, the process of finding a common denominator is demonstrated. To subtract two fractions like $\frac{x+1}{y+1}$ and $\frac{x}{y}$, we need the denominators to match.

To find a common denominator, identify a common multiple of the current denominators, which in this case is $y(y+1)$.
Rewrite each fraction to have this common denominator.
In our example, $\frac{(x+1)y}{y(y+1)} - \frac{x(y+1)}{y(y+1)}$ is created, effectively combining and allowing subtraction of the two fractions.

Finding a common denominator enables us to combine fractions so we can manipulate them more easily, preserving the equality of the equation.

Numerator Simplification

Numerator simplification is crucial to reducing fractions to their simplest form, especially when solving equations. In our equation, simplifying the numerator helps us to isolate terms. The numerator initially is $(x+1)y - x(y+1)$.

First, distribute by applying the distributive property: $xy + y - xy - x$.
Then, combine like terms to get $y - x$.

This step simplifies the expression significantly, leading us to the necessity that the numerator itself equals zero to satisfy the equation condition. This principle is essential in solving fraction equations as it reveals relationships between variables directly and simplifies further calculation.

Zero Condition

The zero condition is a critical concept when dealing with equations involving fractions. For a fraction to equate to zero, its numerator must equal zero. At the same time, its denominator should not be zero because division by zero is undefined.

In our problem, the simplified equation becomes $\frac{y-x}{y(y+1)} = 0$.
The numerator $y-x$ must equal zero, leading to finding $y = x$.
The denominator condition: $y(y+1) eq 0$, ensures there's no division by zero and confirms $y eq 0$ and $y eq -1$.

This condition helps in ensuring that the solution is valid across the range of possible values, excluding those which make the equation undefined.

Problem 91

Problem 94

Other exercises in this chapter

Problem 91

$89-96$ m State whether the given equation is true for all values of the variables. (Disregard any value that makes a denominator zero.) $$ \frac{2}{4+x}=\fra

View solution

Problem 91

(a) Show that $a b=\frac{1}{2}\left[(a+b)^{2}-\left(a^{2}+b^{2}\right)\right] .$ (b) Show that \(\left(a^{2}+b^{2}\right)^{2}-\left(a^{2}-b^{2}\right)^{2}=4 a

View solution

Problem 94

$89-96$ m State whether the given equation is true for all values of the variables. (Disregard any value that makes a denominator zero.) $$ 2\left(\frac{a}{b}

View solution

Problem 94

Mowing a Field A square field in a certain state park is mowed around the edges every week. The rest of the field is kept unmowed to serve as a habitat for bird

View solution