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MA158 — Spring 2026
When working with fractional expressions, we apply the same arithmetic rules that we use with numerical fractions. The key is to remember that algebraic fractions behave exactly like numerical fractions, but with algebraic expressions in the numerators and denominators.
(a) Simplify \(\displaystyle\frac{1}{3} - \frac{2}{9}\)
To subtract fractions, we need a common denominator. The least common denominator of 3 and 9 is 9:
\[\frac{1}{3} - \frac{2}{9} = \frac{1 \cdot 3}{3 \cdot 3} - \frac{2}{9} = \frac{3}{9} - \frac{2}{9} = \frac{1}{9}\](b) Simplify \(\displaystyle\frac{2}{7} \cdot \frac{3}{5}\)
When multiplying fractions, we multiply the numerators together and the denominators together:
\[\frac{2}{7} \cdot \frac{3}{5} = \frac{2 \cdot 3}{7 \cdot 5} = \frac{6}{35}\](c) Simplify \(\displaystyle\frac{1}{6} \cdot \frac{4}{7}\)
When multiplying fractions, we multiply the numerators together and the denominators together:
\[\frac{1}{6} \cdot \frac{4}{7} = \frac{1 \cdot 4}{6 \cdot 7} = \frac{4}{42} = \frac{2}{21}\](d) Simplify \(\displaystyle\frac{2}{7} \div \frac{3}{5}\)
When dividing fractions, we multiply by the reciprocal of the divisor:
\[\frac{2}{7} \div \frac{3}{5} = \frac{2}{7} \cdot \frac{5}{3} = \frac{2 \cdot 5}{7 \cdot 3} = \frac{10}{21}\]When working with fractions, we can cancel common factors in the numerator and denominator. This is a crucial simplification technique:
\[\frac{ab}{ac} = \frac{b}{c}\]Important: This cancellation is only valid when \(a \neq 0\) and \(c \neq 0\). We can only cancel factors that are multiplied, not terms that are added or subtracted.
Recall the difference of squares factoring formula:
\[a^2 - b^2 = (a-b)(a+b)\](a) Can you cancel in \(\displaystyle\frac{3(4+2)}{3(5+7)}\)?
Yes! The factor 3 appears in both the numerator and denominator, so we can cancel it:
\[\frac{3(4+2)}{3(5+7)} = \frac{\cancel{3}(4+2)}{\cancel{3}(5+7)} = \frac{6}{12} = \frac{1}{2}\](b) Can you cancel in \(\displaystyle\frac{3+4}{3+5}\)?
Not in its current form. The 3 is not a factor—it's a term being added. We cannot cancel terms that are added or subtracted. Let's check if it simplifies:
\[\frac{3+4}{3+5} = \frac{7}{8}\]This fraction does not simplify further, so we are done.
(c) Can you cancel in \(\displaystyle\frac{2x(x+2)}{5x(x-7)}\)?
Yes! The factor \(x\) appears in both the numerator and denominator:
\[\frac{2x(x+2)}{5x(x-7)} = \frac{2\cancel{x}(x+2)}{5\cancel{x}(x-7)} = \frac{2(x+2)}{5(x-7)} = \frac{2x+4}{5x-35}\](d) Can you cancel in \(\displaystyle\frac{2(x+2)}{5(x+2)}\)?
Yes! The factor \((x+2)\) appears in both the numerator and denominator:
\[\frac{2(x+2)}{5(x+2)} = \frac{2\cancel{(x+2)}}{5\cancel{(x+2)}} = \frac{2}{5}\]When simplifying more complex rational expressions, we often need to factor the numerator and denominator before we can identify common factors to cancel. This is where the factoring techniques from Lesson 2 become essential.
(a) Simplify \(\displaystyle\frac{m-7}{m^2-49}\)
The denominator is a difference of squares, so we can factor it using the formula \(a^2 - b^2 = (a-b)(a+b)\):
\[\frac{m-7}{m^2-49} = \frac{m-7}{(m-7)(m+7)}\]Now we can cancel the common factor \((m-7)\):
\[= \frac{\cancel{(m-7)}}{\cancel{(m-7)}(m+7)} = \frac{1}{m+7}\](b) Simplify \(\displaystyle\frac{3x^2-5x-2}{x^2-4} \cdot \frac{2x-2}{x-3}\)
Step 1: Factor each polynomial. For the numerator of the first fraction, we need to factor \(3x^2-5x-2\):
Using the ac-method: We need two numbers that multiply to \(3(-2) = -6\) and add to \(-5\). These numbers are \(-6\) and \(1\):
\[\begin{align} 3x^2-5x-2 &= 3x^2 + x - 6x - 2\\ &= x(3x+1) - 2(3x+1)\\ &= (x-2)(3x+1) \end{align}\]Step 2: Factor \(x^2-4\) using difference of squares:
\[x^2-4 = (x-2)(x+2)\]Step 3: Put everything together and cancel common factors:
\[\frac{(x-2)(3x+1)}{(x-2)(x+2)} \cdot \frac{2(x-1)}{x-3}\] \[= \frac{\cancel{(x-2)}(3x+1)}{\cancel{(x-2)}(x+2)} \cdot \frac{2(x-1)}{x-3}\] \[= \frac{(3x+1) \cdot 2(x-1)}{(x+2)(x-3)}\] \[= \frac{2(3x+1)(x-1)}{(x+2)(x-3)}\]We can expand the numerator and denominator using tables:
Expanding \((3x+1)(x-1)\):
| \(x\) | \(-1\) | |
| \(3x\) | \(3x^2\) | \(-3x\) |
| \(1\) | \(x\) | \(-1\) |
So \((3x+1)(x-1) = 3x^2 - 3x + x - 1 = 3x^2 - 2x - 1\)
Expanding \((x+2)(x-3)\):
| \(x\) | \(-3\) | |
| \(x\) | \(x^2\) | \(-3x\) |
| \(2\) | \(2x\) | \(-6\) |
So \((x+2)(x-3) = x^2 - 3x + 2x - 6 = x^2 - x - 6\)
Therefore:
\[= \frac{2(3x^2-2x-1)}{x^2-x-6} = \frac{6x^2-4x-2}{x^2-x-6}\](c) Simplify \(\displaystyle\frac{4y^2-81}{2y^2-23y+63} \div \frac{y^2-y-6}{y^2-5y-14}\)
Step 1: Factor each polynomial:
For \(4y^2-81\), this is a difference of squares:
\[4y^2-81 = (2y)^2 - 9^2 = (2y-9)(2y+9)\]For \(y^2-y-6\), we factor as:
\[y^2-y-6 = (y-3)(y+2)\]For \(2y^2-23y+63\), we use the ac-method with \(ac = 2(63) = 126 = 2 \cdot 3^2 \cdot 7\):
We need two numbers that multiply to 126 and add to \(-23\). These are \(-9\) and \(-14\):
\[\begin{align} 2y^2-23y+63 &= 2y^2-9y-14y+63\\ &= y(2y-9)-7(2y-9)\\ &= (y-7)(2y-9) \end{align}\]For \(y^2-5y-14\), we factor as:
\[y^2-5y-14 = (y-7)(y+2)\]Step 2: Write the expression in factored form:
\[= \frac{(2y-9)(2y+9)}{(y-7)(2y-9)} \div \frac{(y-3)(y+2)}{(y-7)(y+2)}\]Step 3: Convert division to multiplication by flipping the second fraction:
\[= \frac{(2y-9)(2y+9)}{(y-7)(2y-9)} \cdot \frac{(y-7)(y+2)}{(y-3)(y+2)}\]Step 4: Cancel common factors:
\[= \frac{\cancel{(2y-9)}(2y+9)}{\cancel{(y-7)}\cancel{(2y-9)}} \cdot \frac{\cancel{(y-7)}\cancel{(y+2)}}{(y-3)\cancel{(y+2)}}\] \[= \frac{2y+9}{1} \cdot \frac{1}{y-3} = \frac{2y+9}{y-3}\]Final Answer: \(\displaystyle\frac{2y+9}{y-3}\)
When we need to add or subtract rational expressions, we must first find a common denominator, just as we do with numerical fractions. After finding the common denominator, we combine the numerators and simplify the result.
(a) Simplify \(\displaystyle\frac{\frac{9}{x} + \frac{x}{4}}{\frac{2}{x} - 5}\)
Step 1: Identify the denominators. In the numerator, we have denominators \(x\) and \(4\). In the denominator, we have \(x\) and \(1\) (since 5 can be written as \(\frac{5}{1}\)).
The GCF of all denominators is \(4x\).
Step 2: Multiply numerator and denominator by \(4x\) to clear all fractions:
\[= \frac{\left(\frac{9}{x} + \frac{x}{4}\right)(4x)}{\left(\frac{2}{x} - \frac{5}{1}\right)(4x)}\] \[= \frac{\frac{9 \cdot 4x}{x} + \frac{x \cdot 4x}{4}}{\frac{2 \cdot 4x}{x} - \frac{5 \cdot 4x}{1}}\] \[= \frac{36 + x^2}{8 - 20x}\]Final Answer: \(\displaystyle\frac{36 + x^2}{8 - 20x}\)
(b) Simplify \(\displaystyle\frac{\frac{1}{x+3} - 2}{\frac{6}{x+3} + 7}\)
Step 1: Identify the denominators. We have \((x+3)\) and \(1\) (from the constants 2 and 7, which can be written as \(\frac{2}{1}\) and \(\frac{7}{1}\)).
The GCF of all denominators is \((x+3)\).
Step 2: Multiply numerator and denominator by \((x+3)\):
\[= \frac{\left(\frac{1}{x+3} - \frac{2}{1}\right)(x+3)}{\left(\frac{6}{x+3} + \frac{7}{1}\right)(x+3)}\] \[= \frac{\frac{1 \cdot (x+3)}{x+3} - \frac{2(x+3)}{1}}{\frac{6(x+3)}{x+3} + \frac{7(x+3)}{1}}\] \[= \frac{1 - 2(x+3)}{6 + 7(x+3)}\] \[= \frac{1 - 2x - 6}{6 + 7x + 21}\] \[= \frac{-2x - 5}{7x + 27}\]Final Answer: \(\displaystyle\frac{-2x-5}{7x+27}\)
When simplifying complex fractions (fractions within fractions), multiply both the numerator and denominator by the LCD of all the individual fractions within the complex fraction. This clears all the internal fractions and allows us to simplify the resulting expression.