Solving Linear Equations: Finding the Value of x

We are given a linear equation and asked to find the value of the variable 'x' that satisfies the equation.

The given equation is:

\(\frac{{5x}}{2} - \frac{5}{3}\left( {\frac{3}{2} + \frac{{4x}}{3}} \right) = \frac{5}{6}\)

To solve for x, we need to simplify the equation by performing the operations and isolating the terms containing x on one side of the equation.

Step 1: Simplify the expression inside the parenthesis.

First, distribute the \(\frac{5}{3}\) into the terms inside the parenthesis:

\(\frac{{5x}}{2} - \left(\frac{5}{3} \times \frac{3}{2}\right) - \left(\frac{5}{3} \times \frac{4x}{3}\right) = \frac{5}{6}\)

\(\frac{{5x}}{2} - \frac{15}{6} - \frac{20x}{9} = \frac{5}{6}\)

Simplify the fraction \(\frac{15}{6}\):

\(\frac{15}{6} = \frac{15 \div 3}{6 \div 3} = \frac{5}{2}\)

The equation becomes:

\(\frac{{5x}}{2} - \frac{5}{2} - \frac{20x}{9} = \frac{5}{6}\)

Step 2: Group the terms containing x on one side and constant terms on the other side.

Let's move the constant term \(-\frac{5}{2}\) to the right side of the equation by adding \(\frac{5}{2}\) to both sides:

\(\frac{{5x}}{2} - \frac{20x}{9} = \frac{5}{6} + \frac{5}{2}\)

Step 3: Combine the terms with x on the left side.

To combine \(\frac{5x}{2}\) and \(-\frac{20x}{9}\), we need a common denominator, which is the least common multiple (LCM) of 2 and 9. The LCM of 2 and 9 is 18.

Rewrite each fraction with the denominator 18:

\(\frac{5x}{2} = \frac{5x \times 9}{2 \times 9} = \frac{45x}{18}\)

\(\frac{20x}{9} = \frac{20x \times 2}{9 \times 2} = \frac{40x}{18}\)

Now, combine the terms on the left side:

\(\frac{45x}{18} - \frac{40x}{18} = \frac{45x - 40x}{18} = \frac{5x}{18}\)

So, the equation is now:

\(\frac{5x}{18} = \frac{5}{6} + \frac{5}{2}\)

Step 4: Combine the constant terms on the right side.

To combine \(\frac{5}{6}\) and \(\frac{5}{2}\), we need a common denominator, which is the LCM of 6 and 2. The LCM of 6 and 2 is 6.

Rewrite each fraction with the denominator 6:

\(\frac{5}{2} = \frac{5 \times 3}{2 \times 3} = \frac{15}{6}\)

Now, combine the terms on the right side:

\(\frac{5}{6} + \frac{15}{6} = \frac{5 + 15}{6} = \frac{20}{6}\)

So, the equation is now:

\(\frac{5x}{18} = \frac{20}{6}\)

Step 5: Solve for x.

To isolate x, we can multiply both sides of the equation by 18:

\(18 \times \frac{5x}{18} = 18 \times \frac{20}{6}\)

\(5x = 18 \times \frac{20}{6}\)

Simplify the right side:

\(5x = (18 \div 6) \times 20\)

\(5x = 3 \times 20\)

\(5x = 60\)

Now, divide both sides by 5 to find the value of x:

\(x = \frac{60}{5}\)

\(x = 12\)

Thus, the value of x that satisfies the given equation is 12.

Verification:

Let's substitute \(x=12\) back into the original equation to check if both sides are equal.

Original equation: \(\frac{{5x}}{2} - \frac{5}{3}\left( {\frac{3}{2} + \frac{{4x}}{3}} \right) = \frac{5}{6}\)

Left side (LHS): \(\frac{{5(12)}}{2} - \frac{5}{3}\left( {\frac{3}{2} + \frac{{4(12)}}{3}} \right)\)

LHS = \(\frac{60}{2} - \frac{5}{3}\left( {\frac{3}{2} + \frac{48}{3}} \right)\)

LHS = \(30 - \frac{5}{3}\left( {\frac{3}{2} + 16} \right)\)

Combine terms inside parenthesis: \(\frac{3}{2} + 16 = \frac{3}{2} + \frac{32}{2} = \frac{35}{2}\)

LHS = \(30 - \frac{5}{3}\left( {\frac{35}{2}} \right)\)

LHS = \(30 - \frac{5 \times 35}{3 \times 2}\)

LHS = \(30 - \frac{175}{6}\)

Find a common denominator (6) for 30:

\(30 = \frac{30 \times 6}{6} = \frac{180}{6}\)

LHS = \(\frac{180}{6} - \frac{175}{6}\)

LHS = \(\frac{180 - 175}{6} = \frac{5}{6}\)

The right side (RHS) of the original equation is \(\frac{5}{6}\).

Since LHS = RHS (\(\frac{5}{6} = \frac{5}{6}\)), our solution \(x=12\) is correct.

Revision Table: Key Concepts in Linear Equations

• Linear Equation: An equation in which the highest power of the variable is 1. It can be written in the form \(ax + b = c\), where a, b, and c are constants and \(a \neq 0\).
• Solving an Equation: The process of finding the value(s) of the variable that make the equation true.
• Combining Like Terms: Grouping terms that have the same variable raised to the same power (or constant terms) and combining their coefficients.
• Inverse Operations: Using the opposite operation to isolate the variable (e.g., using subtraction to undo addition, multiplication to undo division).
• Common Denominator: A common multiple of the denominators of several fractions, used to add or subtract fractions.

Additional Information: Techniques for Solving Linear Equations

Solving linear equations involves a series of steps aimed at isolating the variable. Here are some common techniques:

• Simplifying Expressions: Remove parentheses by distributing, combine like terms on each side of the equation.
• Clearing Fractions or Decimals: Multiply both sides of the equation by the least common multiple (LCM) of the denominators (or a power of 10 for decimals) to eliminate fractions or decimals. This can make the equation easier to work with.
• Isolating the Variable Term: Use addition or subtraction to move all terms containing the variable to one side of the equation and all constant terms to the other side.
• Solving for the Variable: Use multiplication or division to get the variable by itself. If the variable is multiplied by a coefficient, divide both sides by that coefficient.
• Checking the Solution: Substitute the found value of the variable back into the original equation to ensure it satisfies the equation.

These steps and techniques are fundamental to solving various types of algebraic equations.