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4.5 — Just Act Rational
Human review completed .
Multiply, divide, add and subtract rational expressions while preserving the conditions that make the original operation defined.
1. In class: A Solidify Understanding Task
Use the same fraction rules with polynomial expressions
A rational expression is a ratio of polynomials. Its original denominator must be nonzero. Before an operation, record every excluded input; after simplification, keep those restrictions. Review 4.4: canceling factors and retaining the original domain (draft pending human review).
For polynomial expressions A, B, C and D, the rules are:
| Operation | Rule | Conditions |
|---|---|---|
| Multiply | \(\frac AB\cdot\frac CD=\frac{AC}{BD}\) | \(B\ne0,\ D\ne0\) |
| Divide | \(\frac AB\div\frac CD=\frac AB\cdot\frac DC=\frac{AD}{BC}\) | \(B\ne0,\ D\ne0,\ C\ne0\) |
| Add / subtract with like denominators | \(\frac AB\pm\frac CB=\frac{A\pm C}{B}\) | \(B\ne0\) |
For division, the divisor must be both defined and nonzero. Its numerator becoming a denominator after reciprocation explains the extra restriction C≠0. Cancellation removes common factors of a full numerator and denominator, not pieces of sums.
Multiply, factor and cancel
Multiply the numerators together and the denominators together. Keep the result factored so common factors are visible. Simplifying before multiplying usually keeps the work smaller; simplifying afterward gives the same result on the original domain.
Worked example — Main 1, expression B
Simplify the product:
1. Record the excluded inputs. The original denominators are zero at \(x=-2\) and \(x=2\), so both inputs must stay excluded.
2. Multiply the numerators and multiply the denominators. Keep the factors visible:
3. Cancel the common factor \(x-2\). It appears once in the numerator and once in the denominator. The two factors of \(x+2\) remain in the denominator:
Answer:
Canceling \(x-2\) does not make \(x=2\) an allowed input: the original product is still undefined there.
Divide by multiplying by the reciprocal
Worked example — Main 3, expression B
Simplify the quotient:
1. Record the excluded inputs. The original denominators exclude \(x=-2\) and \(x=3\). We also cannot divide by 0: the second fraction equals 0 at \(x=-1\), so exclude that input too.
2. Multiply by the reciprocal of the second fraction. Keep the first fraction unchanged, replace division with multiplication, and flip the second fraction:
3. Multiply the numerators and multiply the denominators.
4. Cancel the common factors. Cancel one factor of \(x+1\) and one factor of \(x+2\) from the numerator and denominator. The denominator becomes 1, leaving:
Answer:
The simplified expression has no denominator, but all three exclusions still apply to the original division problem.
Add using a common denominator
Fractions can be added when they refer to the same size of part. For polynomial fractions, rewrite both using a common denominator. Multiply each numerator and denominator by the same missing factor, so each fraction keeps its value.
- Factor the denominators and record their zeros.
- Build the least common denominator (LCD) using each distinct factor to the greatest power needed by either denominator.
- Multiply each fraction’s numerator and denominator by its missing factors.
- Add the full numerators; keep the common denominator.
- Factor the new numerator and cancel any full common factors, keeping exclusions.
For example, denominators \((x-2)(x+2)\) and x+2 need LCD \((x-2)(x+2)\), because x+2 already occurs in it. You do not need a second copy of that factor.
Original problem: complete the addition organizer for \(\frac3{x+7}+\frac4{x-4}\).
Worked solution: the packet already shows multiplying each fraction by its missing factor. Fill the next two rows as follows:
| Missing procedure | Missing expression |
|---|---|
| Write equivalent fractions with the common denominator | \(\frac{3(x-4)}{(x+7)(x-4)}+\frac{4(x+7)}{(x+7)(x-4)}\) |
| Add numerators; keep the denominator | \(\frac{3(x-4)+4(x+7)}{(x+7)(x-4)}\) |
Answer: distribute to get \(3x-12+4x+28=7x+16\), so the supplied final form is \(\frac{7x+16}{(x+7)(x-4)}\), x≠−7,4. The numerical model is \(\frac23+\frac17=\frac{14}{21}+\frac3{21}=\frac{17}{21}\).
Worked example — Main 7, expression B
Add and simplify:
1. Record the excluded inputs and find a common denominator. The denominators exclude \(x=-3\) and \(x=2\). The least common denominator is \((x+3)(x-2)\).
2. Rewrite each fraction with that denominator. The first needs a factor of \(x-2\); the second needs a factor of \(x+3\). Multiply both the numerator and denominator by the missing factor so each fraction keeps its value:
3. Add the numerators and keep the common denominator.
4. Expand and combine like terms.
Answer:
Neither denominator factor cancels with the combined numerator, so this is simplified.
Once you’re comfortable: You can usually combine Steps 2 and 3 into one line: write the combined fraction directly, with \((2x-3)(x-2)+(x+5)(x+3)\) over the common denominator. You do not need to show the separate multipliers \((x-2)/(x-2)\) and \((x+3)/(x+3)\) every time. You can also expand inside the combined numerator instead of writing each product on its own line. Keep enough work to check your algebra, and still state the excluded inputs.
Subtract the entire second numerator
Find an LCD as for addition. Then put parentheses around the second numerator: subtracting it changes the sign of every term. In general, \(\frac AB-\frac CD=\frac{AD-CB}{BD}\) when B and D are nonzero; using an LCD may give a smaller denominator than BD.
Worked example — Main 9, expression A
Subtract and simplify:
1. Record the excluded inputs and find a common denominator. The denominators exclude \(x=-5\) and \(x=2\). The least common denominator is \((x+5)(x-2)\).
2. Rewrite each fraction with that denominator. Multiply the first fraction’s numerator and denominator by \(x-2\), and the second fraction’s numerator and denominator by \(x+5\):
3. Subtract the entire second numerator. Keep the common denominator:
4. Expand each product, then distribute the minus sign.
Subtracting the second product changes the sign of every term, including −20:
Answer: factor 2 from the numerator:
Neither denominator factor cancels with the numerator, so this is simplified.
Once you’re comfortable: You can usually combine Steps 2 and 3 into one line: write the combined fraction directly, with \((3x+1)(x-2)-(x-4)(x+5)\) over the common denominator. The separate missing-factor multipliers and equivalent-fraction lines can be omitted. You can also expand inside that numerator, but keep parentheses around the entire second product until you distribute the minus sign. That is an important safeguard against sign errors. Still state the excluded inputs.
2. Ready — Review right-triangle ratios
For a refresher, see Math II 6.8: sine, cosine and tangent. Relative to the chosen acute angle, \(\sin=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos=\frac{\text{adjacent}}{\text{hypotenuse}}\), and \(\tan=\frac{\text{opposite}}{\text{adjacent}}\). The hypotenuse is opposite the right angle. Switching from A to B exchanges the legs’ roles. Try Ready before checking.
Check your work — Ready 1–4, with correction alternatives for 2
| Problem / angle | sin | cos | tan |
|---|---|---|---|
| 1 · A | \(\frac45\) | \(\frac35\) | \(\frac43\) |
| 1 · B | \(\frac35\) | \(\frac45\) | \(\frac34\) |
| 3 · A | \(\frac{\sqrt2}2\) | \(\frac{\sqrt2}2\) | \(1\) |
| 3 · B | \(\frac{\sqrt2}2\) | \(\frac{\sqrt2}2\) | \(1\) |
| 4 · A | \(\frac5{13}\) | \(\frac{12}{13}\) | \(\frac5{12}\) |
| 4 · B | \(\frac{12}{13}\) | \(\frac5{13}\) | \(\frac{12}5\) |
Ready 2 — Conditional answers
Keeping the printed legs 1 and \(\sqrt2\), and correcting the hypotenuse to \(\sqrt3\):
| Angle | sin | cos | tan |
|---|---|---|---|
| A | \(\frac{\sqrt3}3\) | \(\frac{\sqrt6}3\) | \(\frac{\sqrt2}2\) |
| B | \(\frac{\sqrt6}3\) | \(\frac{\sqrt3}3\) | \(\sqrt2\) |
Alternatively, keeping the hypotenuse 3 and leg BC=1: leg CA must be \(\sqrt{3^2-1^2}=2\sqrt2\). Then
| Angle | sin | cos | tan |
|---|---|---|---|
| A | \(\frac13\) | \(\frac{2\sqrt2}3\) | \(\frac{\sqrt2}4\) |
| B | \(\frac{2\sqrt2}3\) | \(\frac13\) | \(2\sqrt2\) |
Check with your teacher which label should change. Ratios using all three original labels are not ratios of a valid right triangle. The other triangles satisfy \(9^2+12^2=15^2\), \(7^2+7^2=(7\sqrt2)^2\), and \(10^2+24^2=26^2\).
3. Set — Operate on rational expressions and analyze errors
Study the relevant operation examples above, then try Set. Keep original denominator restrictions even when the result is constant. For division, exclude zeros of the divisor too.
Check your work — Set 5–12
Set 5 — Angela’s work
5c is correct. She multiplies the fractions, cancels the common x−2 factor and expands the remaining product. The result is \(\frac{x^2+6x+5}{x^2+4x+4}\), x≠−2,2.
5a is incorrect: the first common-denominator line is correct, but she distributes incorrectly. \(5x(x-1)=5x^2-5x\) and \(2(x-3)=2x-6\), so
5b is incorrect: she cancels x+3 across separate terms of a subtraction. Use an LCD instead:
Original problem: simplify \(\frac{4x+8}{5x-20}\div\frac{x^2-3x-10}{x^2-4x}\).
Worked solution: original denominators exclude 4 and 0. The divisor is zero at 5 and −2, so exclude those too. Factor and reciprocate:
Answer: \(\frac{4x}{5(x-5)}\), x≠−2,0,4,5.
| Problem | Simplified result | Original exclusions |
|---|---|---|
| 6 | 2 | \(x\ne-1\) |
| 7 | \(\frac{3x^2-9x-2}{(x+2)(x-5)}\) | \(x\ne-2,5\) |
| 8 | \(\frac{(x+4)^2}{(x-4)(x+2)}\) | \(x\ne1,4,-2\) |
| 9 | \(\frac{4x}{5(x-5)}\) | \(x\ne-2,0,4,5\) |
| 10 | \(\frac{6x-8}{(x-2)(x+2)}\) | \(x\ne-2,2\) |
| 11 | 2 | \(x\ne4\) |
6: \(\frac{2x+6-4}{x+1}=\frac{2(x+1)}{x+1}=2\), retaining x≠−1.
7: the numerator is \(2x(x-5)+(x-1)(x+2)=3x^2-9x-2\).
8: factor both numerators and denominators:
Cancel one x+2 and one x−1 to obtain the table result.
10: the LCD is \((x-2)(x+2)\); combine numerator \(2x+4(x-2)=6x-8\).
11: since \(4-x=-(x-4)\), the subtraction becomes addition:
Set 12 — Comparable statements
a) Polynomials are rational expressions with denominator 1.
b) Rational-number fractions are undefined when their denominator equals zero.
c) Adding, subtracting or multiplying rational expressions, or dividing by a nonzero rational expression, produces a rational expression wherever the original operation is defined.
d) A rational expression is proper when the numerator polynomial’s degree is less than the denominator polynomial’s degree.
4. Go — Find undefined inputs
Set the denominator equal to zero and solve. A zero numerator is allowed when the denominator is nonzero. Review 4.4: original denominator restrictions (draft pending human review). Try Go before checking.
Check your work — Go 13–16
| Problem | Rule | Undefined input(s) | Reason |
|---|---|---|---|
| 13 | \(\frac{10}{x-4}\) | \(x=4\) | x−4 becomes zero. |
| 14 | \(\frac{22}{x}\) | \(x=0\) | The denominator is x. |
| 15 | \(\frac{x-7}{x+15}\) | \(x=-15\) | x+15 becomes zero; x=7 is allowed and gives output 0. |
| 16 | \(\frac{2x}5\) | None | The denominator 5 is always nonzero; every real input is allowed. |
Additional learning resources
| Resource | What it helps with and scope |
|---|---|
| Khan Academy — Adding rational expression: unlike denominators | Close match for this lesson: find a common denominator, multiply both parts by missing factors, then combine and simplify. |
| Khan Academy — Dividing rational expressions | Close match for this lesson: rewrite division as reciprocal multiplication and factor to simplify. |
Teacher/parent review of the source packet
Ready 2, p. 34: legs 1 and \(\sqrt2\) with hypotenuse 3 violate the Pythagorean theorem. If preserving the printed legs, change the hypotenuse to \(\sqrt3\); if preserving hypotenuse 3 and leg 1, change the other leg to \(2\sqrt2\). The key labels both alternatives; confirm the intended correction before grading.
Set 12, p. 36: closure under division requires a nonzero divisor. Add this condition to the supplied rational-number statement and the corresponding rational-expression statement.
Main 9, p. 33: replace “adding” in the instruction with “subtracting.” Domain exclusions must survive all cancellation; division additionally excludes zeros of the divisor.
Coverage
| Packet component | Support in this lesson |
|---|---|
| Main 1–4 · pp. 29–30 | Every missing multiplication/division procedure and expression row, supplied final forms checked, domain restrictions and both explanations. |
| Main 5–8 · pp. 31–32 | LCD procedure, every missing addition row, expansion, equivalence explanation and labeled extra example for further reduction. |
| Main 9–10 · p. 33 | Every subtraction row, complete sign distribution and strategy explanation; wording repaired. |
| Ready 1–4 · p. 34 | Linked trig review and all six ratios per valid triangle; inconsistent 2 flagged with two conditional correction keys. Source figures referenced without reproduction. |
| Set 5–12 · pp. 35–36 | All error analyses, six simplified results with original restrictions, worked division and four comparison statements. Division-by-zero qualification marked. |
| Go 13–16 · p. 36 | All undefined inputs and reasons. |
Data and brief prompts adapted from Mathematics Vision Project, Secondary Mathematics III, Module 4 student packet, Section 4.5, printed pp. 29–36, under CC BY 4.0. Explanations, solutions, and diagrams were prepared for this supplement.
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