Problem 29
Question
Simplify. $$\frac{12}{x}-\frac{5}{2 x}$$
Step-by-Step Solution
Verified Answer
The simplified form of the given expression is \( \frac{7}{x} \)
1Step 1: Factor out common terms
Factor out 2 from the latter fraction to make both fractions have the same denominator. The expression becomes: \( \frac{12}{x} - \frac{5}{2}*\frac{2}{2x} = \frac{12}{x} - \frac{5}{x} \)
2Step 2: Subtract fractions with the same denominator
Subtract the fractions. Since both fractions now have the same denominator, you can simply subtract the numerators. Your result is: \( \frac{12 - 5}{x} \)
3Step 3: Simplify
Simplify your fraction by calculating the subtraction in the numerator: \( \frac{7}{x} \)
Key Concepts
Subtracting FractionsCommon DenominatorSimplify Expressions
Subtracting Fractions
When dealing with algebraic expressions, subtracting fractions is a common task that requires attention to detail in handling numerators and denominators. Imagine slicing up a cake into equal sized pieces to ensure everyone has a fair share. Similarly, when subtracting fractions, we need to make sure each fraction is broken into parts that are the same size as each other; this way, we can easily see how much is being taken away.
In our exercise, we started with two fractions, \(\frac{12}{x}\) and \(\frac{5}{2x}\). To effectively subtract one from the other, it is crucial to express both fractions with the same denominator. Once the denominators are matched, we can then combine or subtract the numerators as if we were dealing with whole numbers. This process is akin to ensuring that all cake pieces are of an identical size before removing a few to serve to guests.
In our exercise, we started with two fractions, \(\frac{12}{x}\) and \(\frac{5}{2x}\). To effectively subtract one from the other, it is crucial to express both fractions with the same denominator. Once the denominators are matched, we can then combine or subtract the numerators as if we were dealing with whole numbers. This process is akin to ensuring that all cake pieces are of an identical size before removing a few to serve to guests.
Common Denominator
Finding a common denominator is like finding a universal language that lets two different-speaking individuals communicate. In the world of fractions, it's about adjusting the bottom parts (the denominators) so they are the same, allowing for direct comparison or combination of the fractions.
In the original exercise, the fractions \(\frac{12}{x}\) and \(\frac{5}{2x}\) have denominators that are related but not identical. We noticed that the second denominator is twice the first one. To communicate in this 'universal language,' we multiplied the numerator and denominator of the second fraction by 2, effectively making the denominator \(2x\) become \(x\), just like the denominator of the first fraction. This transformation does not change the value of the fraction, much like translating a sentence into another language does not change its meaning. Once the denominators matched, we proceeded with the subtraction.
In the original exercise, the fractions \(\frac{12}{x}\) and \(\frac{5}{2x}\) have denominators that are related but not identical. We noticed that the second denominator is twice the first one. To communicate in this 'universal language,' we multiplied the numerator and denominator of the second fraction by 2, effectively making the denominator \(2x\) become \(x\), just like the denominator of the first fraction. This transformation does not change the value of the fraction, much like translating a sentence into another language does not change its meaning. Once the denominators matched, we proceeded with the subtraction.
Simplify Expressions
The art of simplifying algebraic expressions lies in making them as straightforward and 'clean' as possible. Just like an editor pares down a draft into a polished piece of writing, we trim down mathematical expressions to their core, shedding any unnecessary complexity. This makes them more comprehensible and easier to work with in subsequent calculations.
After subtracting the fractions in our exercise, we arrived at \(\frac{12 - 5}{x}\). Simplification involved combining the numerators, because they represented the total number of 'parts' left after subtraction—similar to counting the remaining pieces of cake. The result was the much neater fraction \(\frac{7}{x}\), akin to a clear, concise sentence that conveys the intended message without any extra fluff.
After subtracting the fractions in our exercise, we arrived at \(\frac{12 - 5}{x}\). Simplification involved combining the numerators, because they represented the total number of 'parts' left after subtraction—similar to counting the remaining pieces of cake. The result was the much neater fraction \(\frac{7}{x}\), akin to a clear, concise sentence that conveys the intended message without any extra fluff.
Other exercises in this chapter
Problem 28
Simplify. $$\frac{x-7+\frac{5}{x-1}}{x-3+\frac{1}{x-1}}$$
View solution Problem 29
One technician can wire a security alarm in \(4 \mathrm{h}\), whereas it takes \(6 \mathrm{h}\) for a second technician to do the same job. After working alone
View solution Problem 29
Solve. $$\frac{4}{5 y-1}=\frac{2}{2 y-1}$$
View solution Problem 29
For Exercises 21 to \(32,\) solve for \(y\). $$2 x+7 y=14$$
View solution