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How to Solve the Linear Equation with Fractions: $\\frac{2}{3}(x-1)-\\frac{1}{5}(2x-3)=1$
Mathematics
Grade 8 (Junior High School)
Question Content
Solve the equation: $\\frac{2}{3}(x-1)-\\frac{1}{5}(2x-3)=1$. If the equation has no solution, write No solution.
Correct Answer
$x=3$
Detailed Solution Steps
1
Step 1: Find the least common denominator (LCD) of the fractions. The denominators are 3 and 5, so the LCD is 15.
2
Step 2: Multiply every term in the equation by 15 to clear the fractions: $15\\times\\frac{2}{3}(x-1) - 15\\times\\frac{1}{5}(2x-3) = 15\\times1$. This simplifies to $10(x-1) - 3(2x-3) = 15$.
3
Step 3: Use the distributive property to expand the parentheses: $10x - 10 - 6x + 9 = 15$.
4
Step 4: Combine like terms on the left side: $(10x-6x)+(-10+9)=15$, which simplifies to $4x - 1 = 15$.
5
Step 5: Isolate the variable term by adding 1 to both sides: $4x = 15 + 1$, so $4x=16$.
6
Step 6: Solve for $x$ by dividing both sides by 4: $x=\\frac{16}{4}=3$.
Knowledge Points Involved
1
Clearing Fractions in Linear Equations
This method involves multiplying every term in an equation by the least common denominator (LCD) of all fractions to eliminate denominators, making the equation easier to solve. It is used when solving linear equations with fractional coefficients to avoid working with fractions directly.
2
Distributive Property
The distributive property states that $a(b+c)=ab+ac$. It is used to expand expressions with parentheses, allowing you to multiply a single term across multiple terms inside parentheses. This is essential for simplifying algebraic expressions after clearing fractions.
3
Combining Like Terms
Like terms are terms that have the same variable raised to the same power. Combining like terms involves adding or subtracting the coefficients of these terms to simplify an algebraic expression. This step is used to reduce the number of terms in an equation before isolating the variable.
4
Isolating the Variable in Linear Equations
To solve for a variable, you perform inverse operations (addition/subtraction, multiplication/division) on both sides of the equation to get the variable alone on one side. This is the final step in solving linear equations, giving the value of the unknown variable.
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