Problem 18
Question
Contain linear equations with constants in denominators. Solve each equation. $$\frac{x}{5}=\frac{x}{6}+1$$
Step-by-Step Solution
Verified Answer
The solution to the equation \(\frac{x}{5}=\frac{x}{6}+1\) is \(x = 30\)
1Step 1: Clear the fractions
Multiply each term of the equation by \(30\) (the least common multiple of \(5\) and \(6\)) to clear the fractions, forming: \(6x = 5x + 30\)
2Step 2: Arrange terms to isolate the variable
Subtract \(5x\) from each side of the equation to isolate \(x\) on one side, yielding: \(6x - 5x = 30\), therefore, \(x = 30\)
Key Concepts
Solving EquationsFractions in EquationsLeast Common MultipleIsolating Variables
Solving Equations
Solving equations is a fundamental skill in algebra. The goal is to find the value of the variable that makes the equation true. To achieve this, we manipulate the equation by performing operations such as addition, subtraction, multiplication, or division, keeping both sides of the equation balanced.
It is important that any operation done to one side of the equation is also performed on the other side. This ensures that the equation remains equal throughout the problem-solving process.
In complex equations, such as those involving multiple terms or fractions, breaking down the problem step by step is essential. This meticulous process lays the groundwork for understanding more intricate algebraic concepts.
It is important that any operation done to one side of the equation is also performed on the other side. This ensures that the equation remains equal throughout the problem-solving process.
In complex equations, such as those involving multiple terms or fractions, breaking down the problem step by step is essential. This meticulous process lays the groundwork for understanding more intricate algebraic concepts.
Fractions in Equations
Fractions often make linear equations look complicated, but they can be simplified effectively. When dealing with fractions in equations, our primary aim is to eliminate the fractions to simplify the solving process.
This can be achieved by identifying a common multiple of the denominators, which will be used to clear the fractions.
Multiplying every term of the equation by this common multiple removes the denominators, transforming the equation into a simpler form without fractions. This step is crucial as it makes the subsequent steps of the solution much easier and quicker to perform.
This can be achieved by identifying a common multiple of the denominators, which will be used to clear the fractions.
Multiplying every term of the equation by this common multiple removes the denominators, transforming the equation into a simpler form without fractions. This step is crucial as it makes the subsequent steps of the solution much easier and quicker to perform.
Least Common Multiple
The least common multiple (LCM) of numbers is the smallest number that is a multiple of each of those numbers. When solving equations that contain fractions, determining the LCM of the denominators is an important initial step.
This process involves finding a number that each denominator can divide into without leaving a remainder.
This process involves finding a number that each denominator can divide into without leaving a remainder.
- For example, with the denominators 5 and 6, the LCM is 30.
- By multiplying each term of the equation by this LCM, the denominators are canceled out.
Isolating Variables
Once the fractions are cleared in an equation, the next crucial step is isolating the variable. This means getting the variable alone on one side of the equation. Isolating the variable helps us see what value it must have for the equation to be true.
To do this, we need to rearrange the equation by moving terms around, typically using addition or subtraction. For instance, if we have terms involving the variable on both sides, we subtract or add those terms to isolate the variable on one side.
To do this, we need to rearrange the equation by moving terms around, typically using addition or subtraction. For instance, if we have terms involving the variable on both sides, we subtract or add those terms to isolate the variable on one side.
- Consider the rearranged equation: 6x - 5x = 30.
- Here, subtracting 5x from both sides isolates x on the left.
Other exercises in this chapter
Problem 18
Solve equation by the square root property. $$ 3 x^{2}-1=47 $$
View solution Problem 18
Including \(5 \%\) sales tax, an inn charges \(\$ 252\) per night. Find the inn's nightly cost before the tax is added.
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Find each product and write the result in standard form. $$ (-7-i)(-7+i) $$
View solution Problem 18
Graph each equation in Exercises \(13-28 .\) Let \(x=-3,-2,-1,0\) \(1,2,\) and 3. $$ y=2 x-4 $$
View solution