Problem 104 - Algebra and Trigonometry

Question

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

Answer

The equation is only true when \(y = x\).

Step 1: Cross-Multiply to Eliminate Fractions

To determine if the given equation is true for all values of the variables, start by cross-multiplying both fractions. This means multiplying the numerator of the first fraction by the denominator of the second fraction and setting it equal to the numerator of the second fraction multiplied by the denominator of the first. This gives us: \[(x + 1)\cdot y = x\cdot (y + 1)\]

Step 2: Distribute Terms

Next, distribute the multiplication over addition in each expression:\[xy + y = xy + x\]

Step 3: Simplify the Equation

Now, simplify the equation by subtracting \(xy\) from both sides:\[y = x\]

Step 4: Consider All Values of x and y

The simplification \(y = x\) shows that the original equation is only true when \(y\) equals \(x\). It is not true for all possible values of \(x\) and \(y\), but only when \(y = x\). The equation does not apply in cases where these variables are not equal.

Key Concepts

Cross-Multiplication

In algebra, cross-multiplication is a powerful tool used to eliminate fractions in equations. It simplifies the process of solving equations involving fractions by getting rid of the denominators.
For the given equation \(\frac{x+1}{y+1}=\frac{x}{y}\), cross-multiplication involves the following steps:

• Multiply the numerator of the first fraction \((x + 1)\) by the denominator of the second fraction \((y)\), resulting in \((x + 1) \cdot y\).
• Multiply the numerator of the second fraction \((x)\) by the denominator of the first fraction \((y + 1)\), resulting in \(x \cdot (y + 1)\).

The equation now becomes \((x + 1) \cdot y = x \cdot (y + 1)\).
This method effectively "crosses over" the terms on the diagonal to form a new equation without fractions, making further algebraic manipulation easier.

Variable Values

When solving equations, it's crucial to determine the values that variables can take. In the example equation, we needed to check if it holds true for all values of \(x\) and \(y\).
After simplifying, the result was \(y = x\). This tells us that the equation is true only when the values of \(y\) and \(x\) are equal.
However, some values are disregarded: those that make the denominator zero. In the original equation, we must avoid \(y = -1\) as it would make \(y + 1 = 0\), creating an undefined fraction.

• Check for any values that could make a denominator zero.
• Identify special conditions, like \(y = x\), where the equation holds true.

Always remember, not all equations are true for all variable values; hence, careful examination is needed to identify valid variable conditions.

Distributive Property

The distributive property is a fundamental concept in algebra that deals with distributing multiplication over addition. In our simplified equation, we apply this property:

• We distribute \(y\) across \((x + 1)\) resulting in \(xy + y\).
• We also distribute \(x\) across \((y + 1)\) resulting in \(xy + x\).

This step transforms the equation into \(xy + y = xy + x\).
Using the distributive property allows us to expand and simplify expressions, making it easier to isolate terms and solve equations.
It's a versatile tool that is essential for balancing equations and solving for variables efficiently.