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Magazine Chapter 6: Problem 32
Solve the given equations and check the results. $$\frac{2}{2 x^{2}+5 x-3}-\frac{1}{4 x-2}+\frac{3}{2 x+6}=0$$
Short Answer
Expert verified
The equation is solved with \(x = \frac{2}{5}\).
Step by step solution
01
Factor Denominators
First, we need to factor all the denominators to determine the common denominator. The first term's denominator is \(2x^2 + 5x - 3\). We factor it using the quadratic factoring method, giving us \((2x - 1)(x + 3)\). The second term's denominator, \(4x - 2\), is factored as \(2(2x - 1)\). The third term's denominator, \(2x + 6\), factors as \(2(x + 3)\).
02
Determine the Common Denominator
The common denominator for these terms will be the product of the individual factors: \(2(2x - 1)(x + 3)\). This will be used to simplify the equation by expressing each fraction with this denominator.
03
Rewrite Each Fraction
Convert each fraction so that they all share the common denominator. - The first fraction: Multiply its numerator by 2 to get \(\frac{4}{2(2x - 1)(x + 3)}\).- The second fraction: Multiply its numerator by \((x + 3)\) to get \(\frac{x + 3}{2(2x - 1)(x + 3)}\).- The third fraction: Multiply its numerator by \((2x - 1)\) to get \(\frac{3(2x - 1)}{2(2x - 1)(x + 3)}\).
04
Combine Fractions
Now, write the equation with the common denominator:\[ \frac{4 - (x+3) + 3(2x-1)}{2(2x-1)(x+3)} = 0 \]Simplify the numerator: \[ 4 - x - 3 + 6x - 3 = 5x - 2 \]
05
Solve the Numerator for Zero
Since the equation equals zero, the numerator must be zero. Solve:\[ 5x - 2 = 0 \]Add 2 to both sides to get:\[ 5x = 2 \]Divide both sides by 5:\[ x = \frac{2}{5} \]
06
Check for Validity of x
Substitute \(x = \frac{2}{5}\) back into the original equation to ensure the denominators don't become zero. - \(2x - 1 = 0\) leads to \(x = \frac{1}{2}\), not valid (denominator cannot be zero at this x).- \(x + 3 = 0\) leads to \(x = -3\), not valid.Since \(x = \frac{2}{5}\) does not match these values, it is valid. Therefore, \(x = \frac{2}{5}\) is correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Factoring
Quadratic factoring is a technique used to simplify algebraic expressions where the highest degree of the variable is two. This process involves rewriting the quadratic expression in the form \[ ax^2 + bx + c = (mx + p)(nx + q)\] The goal is to find pairs
- that multiply to give the product of the coefficient of \(x^2\) (i.e., \(a\)) and the constant term (i.e., \(c\)),
- and add up to give the middle term's coefficient \(b\).
Common Denominator
The concept of finding a common denominator is crucial in solving equations that involve fractions, as it allows for combining terms into a single expression. To find a common denominator, look for the least common multiple of all denominators.
For example, consider the fractions in the original exercise:
For example, consider the fractions in the original exercise:
- The first term's denominator, \(2x^2 + 5x - 3\), factors to \((2x - 1)(x + 3)\).
- The second term's denominator, \(4x - 2\), factors to \(2(2x - 1)\).
- The third term鈥檚 denominator, \(2x + 6\), simplifies to \(2(x + 3)\).
Equation Solving Steps
Solving an algebraic equation often involves several steps to isolate the variable of interest. After establishing a common denominator, the following steps simplify the process:
- Rewrite each fraction: Convert each fraction so that they have the same common denominator. For example, by multiplying numerators accordingly:
- the first fraction's numerator becomes \(4\)
- the second fraction's numerator becomes \(x + 3\)
- the third fraction's numerator becomes \(3(2x-1)\)
- Combine fractions: Combine the fractions into a single equation using the common denominator. This results in a new equation of the form:\[\frac{4 - (x+3) + 3(2x-1)}{2(2x-1)(x+3)} = 0\]
- Simplify and solve the numerator: Since the entire fraction is equal to zero, set and solve the numerator for zero:\[5x - 2 = 0\]
- Verify solutions: Substitute back into the original equation to ensure none of the denominators equal zero at this value, which is crucial to validate the solution.
