Question

Determine whether each point is a solution of the equation. Equation Points \(y=\frac{x+1}{5-x}\) (a) \(\left(1, \frac{1}{2}\right)\) (b) \((0,1)\)

Short answer

Point (a) \(\left(1, \frac{1}{2}\right)\) is a solution of the given equation and point (b) \((0,1)\) is not a solution of the equation.

Step-by-step solution

Substitution of point (a)

Substitute \(x=1\) and \(y=\frac{1}{2}\) in the equation \(y=\frac{x+1}{5-x}\). If the left side \(y\) equals the right side of the equation, the point is a solution. If they don't match, the point is not a solution.

Check equality for point (a)

After substitution, we get \(\frac{1}{2}=\frac{1+1}{5-1}\), which after simplifying becomes \(\frac{1}{2}= \frac{1}{2}\). This shows that left side equals right side. Point (a) is a solution.

Substitution of point (b)

Substitute \(x=0\) and \(y=1\) in the equation \(y=\frac{x+1}{5-x}\). If the left side \(y\) equals the right side of the equation, the point is a solution. Otherwise, the point is not a solution.

Check equality for point (b)

After substitution, we get \(1=\frac{0+1}{5-0}\), which after simplification becomes \(1= \frac{1}{5}\). Here the left side does not equal the right side. Hence point (b) is not a solution.

Key concepts

Substitution Method

The substitution method is a foundational tool in algebra used to determine if a particular point lies on the curve represented by an equation.

To use this method, simply replace the variables in the equation with the coordinates of a given point. For instance, if you have the equation \( y = \frac{x+1}{5-x} \), and you are given a point \( (a, b) \), you would replace \( x \) with \( a \) and \( y \) with \( b \). After substitution, if the equation holds true (the left side equates to the right side), the point is indeed a solution.

It's critical for learners to remember that each variable must be substituted with its corresponding value from the point's coordinates, and the equation must be simplified to check for the equality. Moreover, substituting correctly helps to avoid common errors and ensures that the method yields reliable results.

Checking Solutions for Equations

When confronted with the task of verifying solutions for equations, particularly those involving rational expressions, the process is not just about finding equivalence after substitution. It's a method to ensure the integrity and correctness of your solution.

After substituting the variables with your potential solution, simplifying the expression on each side of the equation allows you to compare values effectively. For example, when you substituted \( x = 1 \) and \( y = \frac{1}{2} \) and arrived at \( \frac{1}{2} = \frac{2}{4} \), a simplification to \( \frac{1}{2} = \frac{1}{2} \) confirms the solution is valid. Contrastingly, the point (b) fails this test, signalling an incorrect solution.

Through this process, students not only validate answers but also hone their simplification skills—a critical component of algebraic proficiency.

Solving Rational Expressions

Rational expressions involve ratios of polynomials and can seem daunting at first. When solving them, it's imperative to understand their properties, such as restrictions in the denominator—values that make the denominator zero are not permitted since division by zero is undefined.

The process often requires finding a common denominator, simplifying the expression, and then solving for the variable. In the given exercise, simplifying the rational expression after substitution is crucial to determine the validity of a solution. As you may encounter complex rational expressions, breaking them down step by step and simplifying as you go ensures accuracy and a thorough understanding of the involved mathematical principles.

Remember to carefully analyze and simplify your rational expressions for consistency and clarity in your solutions.