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Basics of Algebra

4. Fractions Addition and Multiplication


Addition Rule: To add fractions with different denominators, find a common denominator, add the numerators, and keep the common denominator.

\[ \frac{1}{3} + \frac{2}{5} = \frac{5}{15} + \frac{6}{15} = \frac{11}{15} \]

Multiplication Rule: To multiply fractions, multiply the numerators together and the denominators together.

\[ \frac{1}{3} \times \frac{2}{5} = \frac{1 \times 2}{3 \times 5} = \frac{2}{15} \]

These rules also apply to fractions with variables.

Rules for variables:

\[\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{ad}{bd} + \dfrac{cb}{bd} = \dfrac{ad + cb}{bd}\] \[\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{ac}{bd}\]
\[ \text{Simplify: } \frac{4}{{x-1}} + \frac{1}{{x+1}} \]

To simplify the expression, we need to find a common denominator and combine the fractions. The common denominator in this case is \((x-1)(x+1)\).

\[ \frac{4}{{x-1}} + \frac{1}{{x+1}} = \frac{{4(x+1)}}{{(x-1)(x+1)}} + \frac{{1(x-1)}}{{(x-1)(x+1)}} \]

Now, we can combine the fractions by adding the numerators:

\[ \frac{{4(x+1) + 1(x-1)}}{{(x-1)(x+1)}} = \frac{{4x + 4 + x - 1}}{{(x-1)(x+1)}} \]

Simplifying the numerator:

\[ \frac{{5x + 3}}{{(x-1)(x+1)}} \]

Therefore, the simplified expression is: \( \dfrac{{5x + 3}}{{(x-1)(x+1)}} \)

Dividing by Fractions


Dividing one fraction by another is the same as multiplying by the reciprocal.
\[ \frac{4}{3} \div \frac{1}{2} = \frac{4}{3} \times \frac{2}{1} = \frac{8}{3} \]

This concept applies to fractions with variables as well.
\[ \frac{{x+1}}{{x^2+3}} \div \frac{{x+3}}{{x}} = \frac{{x+1}}{{x^2+3}} \times \frac{{x}}{{x+3}} \] \[ \dfrac{ \dfrac{{x+1}}{{x^2+3}}} { \dfrac{{x+3}}{{x}}} = \dfrac{{x+1}}{{x^2+3}} \times \dfrac{{x}}{{x+3}} \]

Solving equations with fractions


To solve the equation \( \dfrac{3}{4}x = \dfrac{2}{3} \), we multiply both sides by the reciprocal of \( \dfrac{3}{4} \), which is \( \dfrac{4}{3} \). This gives us:

\[ \dfrac{3}{4}x = \dfrac{2}{3}\] \[ \Rightarrow x = \dfrac{2}{3} \times \dfrac{4}{3} = \dfrac{8}{9} \]

For the equation \( \dfrac{3}{4}x + \dfrac{5}{3} = \dfrac{10}{3} \), we can simplify it by subtracting \( \dfrac{5}{3} \) from both sides and then multiplying both sides by the reciprocal of \( \dfrac{3}{4} \), which is \( \dfrac{4}{3} \). This leads to:

\[ \dfrac{3}{4}x + \dfrac{5}{3} = \dfrac{10}{3} \] \[\Rightarrow \dfrac{3}{4} x = \dfrac{10}{3} - \dfrac{5}{3} = \dfrac{5}{3} \] \[\Rightarrow x = \dfrac{5}{3} \times \dfrac{4}{3} = \dfrac{20}{9} \]

Therefore, the solutions to the equations are \( x = \dfrac{8}{9} \) and \( x = \dfrac{20}{9} \) respectively.

Factoring out the Common Factor


Sometimes, when we have fractions in the numerator and denominator, we need to simplify the expression by factoring out the common factor.

To factor out the common factor \(x\) in the numerator and denominator of the expression \(\dfrac{x^3 + 3x}{4x^2+2x}\), we identify the terms that contain \(x\) and divide them by \(x\).

\[ \frac{{x^3 + 3x}}{{4x^2+2x}} = \frac{{x(x^2 + 3)}}{{x(4x + 2)}} \]

Next, we can cancel out the common factor \(x\) in the numerator and denominator, which simplifies the expression:

\[ \frac{{x(x^2 + 3)}}{{x(4x + 2)}} = \frac{{x^2 + 3}}{{4x + 2}} \]

Therefore, after factoring out the common factor \(x\) and canceling it out, we are left with the simplified expression \(\dfrac{{x^2 + 3}}{{4x + 2}}\).

Simplify the expression by factoring out the common \(a^2\):

\[ \text{Simplify: } \frac{{4a^3 + a^2}}{{3a^5 - a^2}} \]

To factor out the common \(a^2\), we can rewrite it as:

\[\frac{{a^2(4a + 1)}}{{a^2(3a^3 - 1)}}\]

Canceling out the common factor \(a^2\), the simplified expression is:

\[\frac{{4a + 1}}{{3a^3 - 1}}\]

Cross-Multiplication


Cross multiplication is an powerful technique for solving equations that involve fractions. To solve the equation \( \dfrac{x}{3} = \dfrac{x+1}{5} \) using cross multiplication, we multiply the numerator of the first fraction by the denominator of the second fraction and vice versa, setting them equal to each other

So we get:
\[ 5x = 3(x+1) \]

Simplifying further, combining like terms and dividing both sides by 2:

\[ 5x = 3x + 3 \] \[ 2x = 3 \] \[ x = \frac{3}{2} \]

Therefore, the solution to the equation is \( x = \frac{3}{2} \).

*** Solve x/x-3 = 3/4 *** Solve the A = B/mC example on page 62