Math Problem Statement

Solve this

Solve this

Solution

The expression in the image is:

y=x+32x1y = \frac{x + 3}{2x - 1}

To solve for yy, this equation is already solved for yy in terms of xx. Therefore, yy is a function of xx given by the above expression.

To evaluate this function for a specific value of xx:

  1. Substitute the value of xx into the equation.
  2. Simplify the expression to find the corresponding value of yy.

For example, if you want to evaluate yy at x=2x = 2:

y=2+32(2)1=541=53y = \frac{2 + 3}{2(2) - 1} = \frac{5}{4 - 1} = \frac{5}{3}

So, y=53y = \frac{5}{3} when x=2x = 2.

Would you like more details on how to work with this type of function, or do you have any specific questions?

Related Questions:

  1. What is the value of yy when x=0x = 0 in this function?
  2. How does the function y=x+32x1y = \frac{x + 3}{2x - 1} behave as xx approaches infinity?
  3. Are there any values of xx for which this function is undefined?
  4. How do you find the inverse of this function?
  5. What are the asymptotes of the function y=x+32x1y = \frac{x + 3}{2x - 1}?

Tip:

When dealing with rational functions, it's crucial to identify points where the function might be undefined, typically where the denominator equals zero.