Problem 40 - Introductory Algebra for College Students

Question

Solve each rational equation. $$\frac{3}{2 y-2}+\frac{1}{2}=\frac{2}{y-1}$$

Step-by-Step Solution

Answer

Therefore, this equation doesn't have real solutions. It only has complex solutions which are \(y = 2\pm i\sqrt{3}\) provided complex solutions are allowed in the curriculum.

Step 1: Make Denominators Same

The given equation is \(\frac{3}{2 y-2}+\frac{1}{2}=\frac{2}{y-1}\). Before starting to solve, it's easier to make the denominators similar. We can do this by multiplying the whole equation by \((2y - 2)(y-1)\) to eliminate the denominators.

Step 2: Multiply Across

After multiplying across, the given equation will get rid of the fractions and become \(3(y-1) + \frac{1}{2}(2y-2)(y-1) = 2(2y-2)\). This simplifies further to \(3y-3 + y^{2}-y = 4y-4\). And now we can put it in quadratic form.

Step 3: Simplify The Equation

To simplify, allow us to collect like terms and express the equation as a quadratic equation. The equation will now become \(y^{2}-y-4y+3+4 = 0\), and further simplifying comes to \(y^{2}-4y+7 = 0\).

Step 4: Solve The Quadratic Equation

Now we solve the quadratic equation \(y^{2}-4y+7 = 0\). Since this format is \(ax^{2}+bx+c\), we can solve for \(y\) by using the quadratic formula \(y = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\). After plugging values, we find \(y = \frac{4 \pm \sqrt{(-4)^{2}-4*7}}{2}\).

Step 5: Final Answer

By solving this, we find the resulting values of y which are \(\frac{4\pm\sqrt{16-28}}{2} = \frac{4\pm\sqrt{-12}}{2}\). But, there is a negative under the square root, which means there is no solution in the real number system. If complex solutions are allowed, one can express the solutions as complex numbers that involve an \(i\), here it would be \(y = 2\pm i\sqrt{3}\).

Key Concepts

Rational Equations

Rational equations are a type of algebraic equation where each term has a fraction. These fractions have variables in the denominator. In the given exercise, we start with

• \(\frac{3}{2y-2}+\frac{1}{2}=\frac{2}{y-1}\)

To make solving the equation easier, the first step is to eliminate the denominators. This is done by multiplying all terms by a common denominator. In this case, it's

• \((2y-2)(y-1)\).

This method removes fractions and simplifies the equation into a standard polynomial form that is easier to handle.

Quadratic Formula

The quadratic formula is used to find the solutions of a quadratic equation. A quadratic equation has the standard form

• \(ax^2 + bx + c = 0\)

In our exercise, upon simplifying the rational equation, we get

• \(y^2 - 4y + 7 = 0\)

We can use the quadratic formula:

• \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -4\), and \(c = 7\)

Plug these values in to find the solutions. This formula is especially helpful when the equation does not easily factor.

Complex Solutions

In the context of quadratic equations, complex solutions occur when the term under the square root—the discriminant—is negative. In our solved example, the discriminant is

• \(b^2 - 4ac = 16 - 28 = -12\)

A negative discriminant indicates that the solutions are not real numbers but complex numbers. Complex numbers include the imaginary unit \(i\), where

• \(i^2 = -1\)

Thus, the solutions can be written as

• \(y = 2 \pm i\sqrt{3}\)

This part of algebra marks the entry into the world of complex numbers, where solutions have both real and imaginary components.