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Chapter 7: Problem 40

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

Short Answer

Expert verified
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 by step solution

01

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.
02

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.
03

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 \(y^{2}-4y+7 = 0\).
04

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 \(y = \frac{4 \pm \sqrt{(-4)^{2}-4*7}}{2}\).
05

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 \(y = 2\pm i\sqrt{3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥攖he discriminant鈥攊s 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.

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Most popular questions from this chapter

A snowstorm causes a bus driver to decrease the usual average rate along a 60 -mile route by 15 miles per hour. As a result, the bus takes two hours longer than usual to complete the route. At what average rate does the bus usually cover the 60 -mile route?

denominators are opposites, or additive inverses. Add or subtract as indicated. Simplify the result, if possible. $$\frac{10}{x-2}-\frac{6}{2-x}$$

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. $$\frac{x-5}{6} \cdot \frac{3}{5-x}=\frac{1}{2}\( for any value of \)x$ except 5$$

Add or subtract as indicated. Simplify the result, if possible. $$\frac{x}{x+6}-1$$

denominators are opposites, or additive inverses. Add or subtract as indicated. Simplify the result, if possible. $$\frac{x-2}{x^{2}-25}-\frac{x-2}{25-x^{2}}$$
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