Problem 20
Question
Find the solution set to each equation. $$\frac{x-3}{2}-\frac{1}{x-3}=\frac{8-3 x}{x-3}$$
Step-by-Step Solution
Verified Answer
The solution set is \(x = -3\).
1Step 1: Identify the common denominator
Notice that the denominators involved in the equation are 2 and \(x - 3\). The common denominator is \(2(x - 3)\).
2Step 2: Multiply both sides by the common denominator
To clear the fractions, multiply every term by \(2(x - 3)\). This gives:\[2(x-3) \times \frac{x-3}{2} - 2(x-3) \times \frac{1}{x-3} = 2(x-3) \times \frac{8-3x}{x-3}\]Simplify this to:\[(x - 3)^2 - 2 = 2(8 - 3x)\]
3Step 3: Expand and simplify
Expand \((x - 3)^2\) to get \(x^2 - 6x + 9\). Thus, now the equation becomes:\[x^2 - 6x + 9 - 2 = 16 - 6x\]Combine like terms on the left side:\[x^2 - 6x + 7 = 16 - 6x\]
4Step 4: Eliminate the common terms
Add \(6x\) to both sides to cancel out the \(-6x\) terms:\[x^2 + 7 = 16\]
5Step 5: Solve for \(x\)
Isolate \(x^2\) by subtracting 7 from both sides:\[x^2 = 9\]Take the square root of both sides to find \(x\):\[x = \text{±}3\]
6Step 6: Verify potential solutions
Since \(x = 3\) would make the original equation undefined (division by zero in \(x - 3)\), discard it. Therefore, the only valid solution is \(x = -3\).
Key Concepts
Common DenominatorMultiplying Both Sides of an EquationSquare Roots
Common Denominator
When solving rational equations, one essential step is finding a common denominator. This means identifying a single denominator that can incorporate all the individual denominators in the equation.
For example, in the equation \[ \frac{x-3}{2} - \frac{1}{x-3} = \frac{8 - 3x}{x-3}, \] the denominators are 2 and \( x - 3 \).
To clear these fractions and simplify the equation, we need to find a common denominator. Here, the common denominator is the product of 2 and \( x - 3 \), which is \( 2(x - 3) \).
By finding this common ground, we can transform the equation into a simpler form, making the equation easier to solve, ultimately aiding in clearing all the fractions.
For example, in the equation \[ \frac{x-3}{2} - \frac{1}{x-3} = \frac{8 - 3x}{x-3}, \] the denominators are 2 and \( x - 3 \).
To clear these fractions and simplify the equation, we need to find a common denominator. Here, the common denominator is the product of 2 and \( x - 3 \), which is \( 2(x - 3) \).
By finding this common ground, we can transform the equation into a simpler form, making the equation easier to solve, ultimately aiding in clearing all the fractions.
Multiplying Both Sides of an Equation
Once we have identified a common denominator, the next step is to multiply every term of the equation by this common denominator. This step allows us to eliminate fractions and simplify the equation further.
In our specific example, the expression \[ \frac{x-3}{2} - \frac{1}{x-3} = \frac{8 - 3x}{x-3} \] gets multiplied by the common denominator \( 2(x - 3) \), resulting in: \[ 2(x - 3) \times \frac{x-3}{2} - 2(x-3) \times \frac{1}{x-3} = 2(x-3) \times \frac{8 - 3x}{x-3}. \]
Simplifying this, we get: \[ (x - 3)^2 - 2 = 2(8 - 3x). \] This step-by-step process clears out the fractions making the equation easier to handle and solve in its simplified form. This technique is crucial for simplifying and solving many algebraic equations.
In our specific example, the expression \[ \frac{x-3}{2} - \frac{1}{x-3} = \frac{8 - 3x}{x-3} \] gets multiplied by the common denominator \( 2(x - 3) \), resulting in: \[ 2(x - 3) \times \frac{x-3}{2} - 2(x-3) \times \frac{1}{x-3} = 2(x-3) \times \frac{8 - 3x}{x-3}. \]
Simplifying this, we get: \[ (x - 3)^2 - 2 = 2(8 - 3x). \] This step-by-step process clears out the fractions making the equation easier to handle and solve in its simplified form. This technique is crucial for simplifying and solving many algebraic equations.
Square Roots
Square roots often appear in equations when you need to isolate a variable, particularly in quadratic equations. To solve these equations, taking the square root of both sides can be very helpful.
For example, let's continue with our simplified equation from the last step: \[ x^2 + 7 = 16. \]
To isolate \( x \), we subtract 7 from both sides, resulting in: \[ x^2 = 9. \]
Taking the square root of both sides, we get: \[ x = \pm 3. \]
However, we must validate the solutions. Because the original equation involves a term \( \frac{1}{x - 3} \), the solution \( x = 3 \) would make the denominator zero, leading to an undefined result. So, we discard \( x = 3 \) and accept \( x = -3 \) as our valid solution. Understanding when and how to apply square roots is essential for solving various equations accurately.
For example, let's continue with our simplified equation from the last step: \[ x^2 + 7 = 16. \]
To isolate \( x \), we subtract 7 from both sides, resulting in: \[ x^2 = 9. \]
Taking the square root of both sides, we get: \[ x = \pm 3. \]
However, we must validate the solutions. Because the original equation involves a term \( \frac{1}{x - 3} \), the solution \( x = 3 \) would make the denominator zero, leading to an undefined result. So, we discard \( x = 3 \) and accept \( x = -3 \) as our valid solution. Understanding when and how to apply square roots is essential for solving various equations accurately.
Other exercises in this chapter
Problem 19
Reduce each rational expression to its lowest terms. $$\frac{6}{57}$$
View solution Problem 20
$$\text { Solve each formula for the indicated variable.}$$ $$h=\frac{S-2 \pi r^{2}}{2 \pi r} \text { for } S$$
View solution Problem 20
Reduce each rational expression to its lowest terms. $$\frac{14}{91}$$
View solution Problem 21
Find the solution set to each equation. $$5+\frac{9}{x-2}=2+\frac{x+7}{x-2}$$
View solution