Detailed Summary
This section focuses on methods to simplify and solve equations, particularly those containing fractions and multiple terms. Simplifying equations can often involve opening brackets, rearranging terms, and finding a common denominator. The examples provided guide students through these processes, showing how to multiply both sides of an equation by the least common multiple (LCM) of denominators to eliminate fractions, and how to effectively combine like terms for clarity. Through systematic steps, students learn to transform complicated expressions into a linear form that is easier to handle.
Example 1: Solve
$$\frac{4x + 7}{3} = \frac{6x + 5}{6}$$
Solution: Multiply both sides of the equation by 6,
$$6 \cdot \left(\frac{4x + 7}{3}\right) = 6 \cdot \left(\frac{6x + 5}{6}\right)$$
or
$$2(4x + 7) = 6x + 5$$
or
$$8x + 14 = 6x + 5$$
or
$$2x = -9$$
or
$$x = -\frac{9}{2}$$
(required solution)
Example 2: Solve
$$\frac{7x + 2}{4} = \frac{5x - 1}{3}$$
Solution: Multiply both sides of the equation by 12,
$$12 \cdot \left(\frac{7x + 2}{4}\right) = 12 \cdot \left(\frac{5x - 1}{3}\right)$$
or
$$3(7x + 2) = 4(5x - 1)$$
or
$$21x + 6 = 20x - 4$$
or
$$x = -10$$
(required solution)
Check:
LHS = $$\frac{7(-10) + 2}{4} = \frac{-70 + 2}{4} = \frac{-68}{4} = -17$$
RHS = $$\frac{5(-10) - 1}{3} = \frac{-50 - 1}{3} = \frac{-51}{3} = -17$$
Thus, LHS = RHS = (as required)
Example : Solve \( 4x - 3(2x + 5) = 3(4 - x) + 14 \)
Solution:
Let’s open the brackets.
\[
\text{LHS} = 4x - 6x - 15 \quad \text{(expanding the left side)}
\]
\[
\text{RHS} = 12 - 3x + 14 \quad \text{(expanding the right side)}
\]
The equation is \[
-2x - 15 = 26 - 3x \]
\[
\text{(transposing terms)}
\]
Check: \( \text{LHS} = -2x - 15 \)
\[
\text{RHS} = 26 - 3x \quad \text{(as required)}
\]
Therefore, required solution is: \( x = 1 \)