Solve for x -1/5+5/(6x)=-1/(4x)
The problem provided is a first-degree algebraic equation that requires solving for the variable 'x'. The equation contains fractional terms, where 'x' is present both in the denominators and the numerators, indicating that finding a common denominator or performing cross multiplication might be necessary to simplify the equation and isolate 'x'. The goal is to manipulate the equation in such a way that 'x' stands alone on one side of the equation, with all numerical values on the opposite side, thereby determining the value of 'x' that satisfies the equation.
$- \frac{1}{5} + \frac{5}{6 x} = - \frac{1}{4 x}$
Answer
Solution:
Step:1
To balance the equation, add $\frac{1}{5}$ to both sides: $\frac{5}{6x} = -\frac{1}{4x} + \frac{1}{5}$.
Step:2
Identify the Least Common Denominator (LCD) for the fractions involved.
Step:2.1
The LCD is equivalent to the Least Common Multiple (LCM) of the denominators $6x$, $4x$, and $5$.
Step:2.2
To find the LCM, separate the process into finding the LCM of the numbers $6$, $4$, $5$ and the variable parts $x^1$, $x^1$.
Step:2.3
The LCM is the smallest number that each of the numbers can divide into without a remainder. To find it:
Decompose each number into its prime factors.
For each prime factor, take the highest power that appears in any of the numbers.
Step:2.4
The prime factors of $6$ are $2$ and $3$: $2 \cdot 3$.
Step:2.5
The prime factors of $4$ are $2$ and $2$: $2 \cdot 2$.
Step:2.6
$5$ is a prime number, so its factors are $1$ and $5$ itself.
Step:2.7
The LCM of $6$, $4$, $5$ is obtained by multiplying the prime factors, choosing the highest power for each: $2 \cdot 2 \cdot 3 \cdot 5$.
Step:2.8
Perform the multiplication of $2 \cdot 2 \cdot 3 \cdot 5$.
Step:2.8.1
Multiply $2$ by $2$: $4 \cdot 3 \cdot 5$.
Step:2.8.2
Multiply $4$ by $3$: $12 \cdot 5$.
Step:2.8.3
Multiply $12$ by $5$: $60$.
Step:2.9
For the variable $x^1$, the LCM is simply $x$ since it appears once in each term.
Step:2.10
The LCM of $x^1$, $x^1$ is $x$.
Step:2.11
Combine the numeric LCM $60$ with the variable part to get the overall LCM: $60x$.
Step:3
Clear the fractions by multiplying each term in $\frac{5}{6x} = -\frac{1}{4x} + \frac{1}{5}$ by $60x$.
Step:3.1
Apply the multiplication: $\frac{5}{6x}(60x) = -\frac{1}{4x}(60x) + \frac{1}{5}(60x)$.
Step:3.2
Simplify the left side.
Step:3.2.1
Use the commutative property: $60 \cdot \frac{5}{6x} \cdot x = -\frac{1}{4x}(60x) + \frac{1}{5}(60x)$.
Step:3.2.2
Eliminate the common factor of $6$.
Step:3.2.2.1
Extract $6$ from $60$: $6(10) \cdot \frac{5}{6x} \cdot x = -\frac{1}{4x}(60x) + \frac{1}{5}(60x)$.
Step:3.2.2.2
Extract $6$ from $6x$: $6(10) \cdot \frac{5}{6(x)} \cdot x = -\frac{1}{4x}(60x) + \frac{1}{5}(60x)$.
Step:3.2.2.3
Cancel the common factor: $10 \cdot \frac{5}{x} \cdot x = -\frac{1}{4x}(60x) + \frac{1}{5}(60x)$.
Step:3.2.3
Combine $10$ and $\frac{5}{x}$: $\frac{10 \cdot 5}{x} \cdot x = -\frac{1}{4x}(60x) + \frac{1}{5}(60x)$.
Step:3.2.4
Multiply $10$ by $5$: $\frac{50}{x} \cdot x = -\frac{1}{4x}(60x) + \frac{1}{5}(60x)$.
Step:3.2.5
Cancel the common factor of $x$: $50 = -\frac{1}{4x}(60x) + \frac{1}{5}(60x)$.
Step:3.3
Simplify the right side.
Step:3.3.1
Simplify each term.
Step:3.3.1.1
Cancel the common factor of $4x$: $50 = -15 + \frac{1}{5}(60x)$.
Step:3.3.1.2
Multiply $-1$ by $15$: $50 = -15 + \frac{1}{5}(60x)$.
Step:3.3.1.3
Cancel the common factor of $5$: $50 = -15 + 12x$.
Step:4
Solve for $x$.
Step:4.1
Rewrite the equation: $-15 + 12x = 50$.
Step:4.2
Isolate terms with $x$ on one side: $12x = 50 + 15$.
Step:4.2.1
Add $15$ to both sides: $12x = 65$.
Step:4.3
Divide each side by $12$ to solve for $x$: $\frac{12x}{12} = \frac{65}{12}$.
Step:4.3.2
Simplify the left side: $x = \frac{65}{12}$.
Step:5
Express the solution in various forms.
Exact Form: $x = \frac{65}{12}$ Decimal Form: $x \approx 5.42$ Mixed Number Form: $x = 5 \frac{5}{12}$
Knowledge Notes:
Adding/Subtracting Fractions: To add or subtract fractions, one must find a common denominator, which is typically the least common multiple of the denominators.
Least Common Multiple (LCM): The LCM of two or more numbers is the smallest number that is a multiple of all the numbers. It can be found by listing the prime factors and taking the highest power of each that appears.
Prime Factorization: Decomposing a number into a product of prime numbers. For example, $6 = 2 \cdot 3$ and $4 = 2^2$.
Clearing Fractions: Multiplying each term by the LCD eliminates the fractions, simplifying the equation.
Solving Linear Equations: To solve for a variable, isolate the variable on one side of the equation by performing inverse operations, such as adding, subtracting, multiplying, or dividing both sides of the equation by the same number.
Equivalent Expressions: The same number can be expressed in different forms, such as an exact fraction, a decimal, or a mixed number.
