Problem 8
Question
Solve the equation. \(\frac{2}{2 x+5}+\frac{3}{2 x-5}=\frac{10 x+5}{4 x^{2}-25}\)
Step-by-Step Solution
Verified Answer
The solution is all real numbers except \(x = \pm \frac{5}{2}\).
1Step 1: Identify the Common Denominator
The denominators of the fractions are \(2x+5\), \(2x-5\), and \(4x^2 - 25\). Notice that \(4x^2 - 25\) can be factored as \((2x+5)(2x-5)\). This is the common denominator for the equation.
2Step 2: Clear the Fractions
Multiply every term by the common denominator \((2x+5)(2x-5)\) to eliminate the fractions. The equation becomes:\[ 2(2x-5) + 3(2x+5) = 10x + 5 \]
3Step 3: Simplify Each Side of the Equation
Distribute and simplify the expressions on both sides:For the left side: \[ 2(2x-5) = 4x - 10 \] \[ 3(2x+5) = 6x + 15 \] Combining these, the left side is: \[ 4x - 10 + 6x + 15 = 10x + 5 \]The right side is already simplified.
4Step 4: Solve for x
With the simplified equation \(10x + 5 = 10x + 5\), subtract \(10x\) from both sides to simplify further:\[ 5 = 5 \]This shows that the equation is true for all values of \(x\) except those that make the original denominator zero.
5Step 5: Determine Appropriate Domain
The original denominators \((2x+5)\) and \((2x-5)\) imply restrictions. Solve:\(2x+5 eq 0\) gives \(x eq -\frac{5}{2}\).\(2x-5 eq 0\) gives \(x eq \frac{5}{2}\).Therefore, the solution is all real numbers except \(x = \pm \frac{5}{2}\).
Key Concepts
Rational EquationsDomain of a FunctionFactoringExcluded Values
Rational Equations
Rational equations involve fractions where the numerator and/or the denominator is a polynomial. These equations look intimidating because they have variables in the denominator. However, they can be solved using techniques similar to regular polynomial equations.
When working with rational equations, the first step is to find a common denominator. This common denominator allows us to eliminate the fractions so that we can more easily work with the equation.
Once the denominators are cleared out, the problem usually becomes a simpler linear or quadratic equation. Remembering that these steps hinge on understanding and manipulating fractions is key.
When working with rational equations, the first step is to find a common denominator. This common denominator allows us to eliminate the fractions so that we can more easily work with the equation.
Once the denominators are cleared out, the problem usually becomes a simpler linear or quadratic equation. Remembering that these steps hinge on understanding and manipulating fractions is key.
Domain of a Function
The domain of a function refers to all possible input values (usually represented as 'x') for which the function is defined. For rational equations, this involves being careful about the values that make the equation undefined.
Since division by zero is undefined, solving the denominator equal to zero will help us find these so-called restricted or excluded values.
When solving the equation \(\frac{2}{2x+5} + \frac{3}{2x-5} = \frac{10x+5}{4x^2-25}\), we must exclude any \(x\) that makes \((2x+5)\) or \((2x-5)\) zero because they are in the denominator.
Since division by zero is undefined, solving the denominator equal to zero will help us find these so-called restricted or excluded values.
When solving the equation \(\frac{2}{2x+5} + \frac{3}{2x-5} = \frac{10x+5}{4x^2-25}\), we must exclude any \(x\) that makes \((2x+5)\) or \((2x-5)\) zero because they are in the denominator.
Factoring
Factoring is an essential algebraic skill, especially when it comes to solving rational equations. In the given equation, notice that the polynomial \(4x^2 - 25\) can be factored.
Factoring involves writing the polynomial as a product of its simpler components. Here, \(4x^2 - 25\) factors into \((2x+5)(2x-5)\). Recognizing these factorable patterns, like the difference of squares in this case, is crucial for simplifying equations.
After factoring, it becomes much easier to combine fractions and detect common terms, helping us to progress with solving the equation more smoothly.
Factoring involves writing the polynomial as a product of its simpler components. Here, \(4x^2 - 25\) factors into \((2x+5)(2x-5)\). Recognizing these factorable patterns, like the difference of squares in this case, is crucial for simplifying equations.
After factoring, it becomes much easier to combine fractions and detect common terms, helping us to progress with solving the equation more smoothly.
Excluded Values
Excluded values are values that make any rational expression in the equation undefined. In simpler terms, they are the values that produce division by zero situations.
To determine these values, set the denominator(s) equal to zero and solve for \(x\). In our example, \(2x+5 = 0\) and \(2x-5 = 0\) give us the excluded values \(x = -\frac{5}{2}\) and \(x = \frac{5}{2}\), respectively.
These values are not permissible as solutions in the final answer because they would cause division by zero, which can lead to undefined outcomes. Ensuring that these are excluded is a vital part of reaching the correct solution for any rational equation.
To determine these values, set the denominator(s) equal to zero and solve for \(x\). In our example, \(2x+5 = 0\) and \(2x-5 = 0\) give us the excluded values \(x = -\frac{5}{2}\) and \(x = \frac{5}{2}\), respectively.
These values are not permissible as solutions in the final answer because they would cause division by zero, which can lead to undefined outcomes. Ensuring that these are excluded is a vital part of reaching the correct solution for any rational equation.
Other exercises in this chapter
Problem 8
Express the number in the form \(a / b,\) where \(a\) and \(b\) are integers. $$9^{1 / 2}$$
View solution Problem 8
Express the statement as an inequality. (a) \(b\) is positive. (b) \(s\) is nonpositive. (c) \(w\) is greater than or equal to \(-4\). (d) \(c\) is between \(\f
View solution Problem 8
Simplify the expression, and rationalize the denominator when appropriate. $$x^{-2}-y^{-1}$$
View solution Problem 8
Express as a polynomial. $$(5 x-4 y)^{2}$$
View solution