Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 5 · Chapter companion part
Simplify algebraic fractions by factors, retain excluded values and build equivalent common-denominator forms.
Class 8 English Version: part A of the chapter 5 companion. Eight sections, 30 written tasks and a notebook project. Selected companion coverage, not complete textbook coverage; use all four parts. Numerical/coefficient answers are checked; equivalent symbolic forms, domains and proofs need guardian or teacher review.
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An algebraic fraction P/Q represents division of expressions and requires Q ≠ 0. P is the numerator; Q is the denominator. A zero numerator is allowed when Q is nonzero: 0/5 = 0. Division by zero and 0/0 are undefined. For (x + 1)/(x − 4), exclude x = 4 before simplifying or substituting. These are expression restrictions, not an instruction to solve an equation.
In 6x(x + 2)/[9x(x − 1)], x is a complete factor of both numerator and denominator. For x ≠ 0,1 the reduced fraction is 2(x + 2)/[3(x − 1)]. In (x + 2)/x, x is not a factor of the whole numerator: division gives 1 + 2/x, not 2. Cancellation means dividing top and bottom by the same nonzero factor; it never means erasing matching letters wherever they appear.
For 12x³y²/(18x²y), the coefficient ratio is 2/3 and the shared x²y cancels, giving 2xy/3 for x,y ≠ 0. For 6x²/(9x⁵), reduce to 2/(3x³), retaining x ≠ 0. Expand powers of products correctly: (2x)³ = 8x³. The highest common factor belongs to whole products; subtract exponents only when dividing powers with the same base.
Use x² − 9 = (x − 3)(x + 3). Then (x² − 9)/(x − 3) reduces to x + 3 for x ≠ 3. For (x² − 5x + 6)/(x² − 4), factor the top as (x − 2)(x − 3) and the bottom as (x − 2)(x + 2); the result is (x − 3)/(x + 2), still with x ≠ 2,−2. Lowest form has no common polynomial factor left.
Treat b + c as one term when factoring a² − b² − 2bc − c² = a² − (b + c)². Its factors are (a − b − c)(a + b + c). Also a² + 2ab + b² − c² = (a + b − c)(a + b + c). Their quotient reduces to (a − b − c)/(a + b − c), but both original denominator factors must remain nonzero. Higher-degree factors use the same square/cube identities from chapter 4.
The graph or value rule for (x² − 9)/(x − 3) agrees with x + 3 wherever the original is defined. At x = 3 the original is 0/0, not 6. A number-line drawing marks an excluded input with an open circle; it is a domain guide, not a plotted fraction graph. Record all distinct excluded roots, even when a factor was repeated or cancelled.
A common denominator is divisible by each denominator. For 6x²y and 9xy³, the least positive numerical content is 18 and the needed powers are x² and y³, so use D = 18x²y³. The respective multipliers are 3y² and 2x. For x² − 1 and x² − x, factor first: D = x(x − 1)(x + 1). The original exclusions are x ≠ −1,0,1.
If D is the chosen common denominator, rewrite P/Q as P(D/Q)/D; multiply top and bottom equally. For 1/(x − 1) and 2/(x + 1), the common forms are (x + 1)/(x² − 1) and 2(x − 1)/(x² − 1). Reversed factors differ by a sign: 2 − x = −(x − 2). The printed page 80 activity has a² − 4b, not a² − 4b²; do not assume it equals (a − 2b)(a + 2b). A safe common denominator for that printed pair is (a + 2b)(a² − 4b).
For (x² − 7x + 10)/(x² − 25), factor to (x−5)(x−2)/[(x−5)(x+5)]. Cancel x−5 only for x≠5, giving (x−2)/(x+5), with x≠5,−5. At x=7 both forms give 5/12; at x=5 the original remains undefined.
A simplified formula without its original restrictions can describe additional inputs. Keep the excluded values beside the final fraction.
The printed page 80 activity uses a²−4b. The companion explains a safe product denominator instead of silently changing b to b².
Restrictions are found before cancellation. These are original teaching examples.
| Original expression | Reduced form | Original restriction |
|---|---|---|
| (x²−9)/(x−3) | x+3 | x≠3 |
| (x²−5x+6)/(x²−4) | (x−3)/(x+2) | x≠2,−2 |
| 6x²/(9x⁵) | 2/(3x³) | x≠0 |
| (x−2)/(2−x) | −1 | x≠2 |
At x=3 the first original is 0/0, so its value is undefined. The displayed x+3 is only an equivalent value rule on the original domain.
Take denominators 6x²y and 9xy³, with x,y nonzero.
| Factor/content | First | Second | Chosen in D |
|---|---|---|---|
| Numerical content | 6 | 9 | 18 |
| Power of x | 2 | 1 | 2 |
| Power of y | 1 | 3 | 3 |
| Multiplier to reach D | 3y² | 2x | D=18x²y³ |
Check 6x²y×3y²=18x²y³ and 9xy³×2x=18x²y³. Multiply each entire numerator by the same respective multiplier.
Factor numerator and denominator before cancelling.
(x²−5x+6)/(x²−4) = (x−2)(x−3)/[(x−2)(x+2)] = (x−3)/(x+2), x≠2,−2
At x=2 the shorter formula exists but the original fraction does not. An equality of values here retains the original domain.
Use standard square and cube identities from the preceding chapter.
[(x³−y³)(x²−xy+y²)]/[(x²−y²)(x³+y³)] = (x²+xy+y²)/(x+y)²
Retain x²−y²≠0 and x³+y³≠0; the reduced denominator alone does not describe every original restriction.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original symbolic schematic. Box widths organise factors, not numerical magnitudes or lengths.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original domain schematic, equally spaced integer ticks; the real line continues beyond the displayed interval.
Show your working on paper. Check your answer, then choose Next.
2
5 − 3 = 2, so substitution is allowed.
4
The denominator is zero exactly at x = 4.
4
The reduced form is 2xy/3; 2 × 3 × 2 / 3 = 4.
3
Divide coefficients by 5 and cancel ab: 3a/(2b), with a,b nonzero.
3
The denominator retains x^(5−2) = x³; x ≠ 0.
8
(25 − 9)/(5 − 3) = 16/2 = 8.
1
Only x = 3 makes its original denominator zero.
7
6/21 reduces to 2/7.
-1
The factor subtracts the complete sum b + c.
18
The least common multiple of 6 and 9 is 18.
2
Use the larger exponent, max(2,1) = 2.
3
Use max(1,3) = 3.
4
The missing factor is x, which equals 4.
-1
3/(−3) = −1; reversed differences have opposite signs.
2
The two excluded values are 2 and −3; repeated factors do not create extra distinct values.
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Numerator 2x − 1, denominator x + 5, and x ≠ −5. A numerator of zero is allowed if the denominator is nonzero.
Identify both whole expressions and derive the excluded value.
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3x²/(4y), with x ≠ 0 and y ≠ 0. Divide by 6x²y²; x remains excluded even after disappearing from the denominator.
Show the common factor and retain both original restrictions.
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36x/y, with x,y ≠ 0. Numerical ratio is 16×16×243/(27×64) = 36; powers leave x/y.
Expand product powers before reducing coefficients.
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(x − 2)/(x + 5), with x ≠ 5,−5; factor the numerator as (x − 5)(x − 2).
Check both original roots, not only the remaining denominator.
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(x − 3)/(x − 5), with x ≠ 4,5, by cancelling x − 4.
Expand both factorizations and preserve the cancelled exclusion.
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(a − b − c)/(a + b − c), with a + b − c ≠ 0 and a + b + c ≠ 0. Both polynomials are differences of squares after grouping.
Name b+c and a+b as complete grouped expressions.
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(x² + xy + y²)/(x + y)² on the original domain. Factor cubes and squares; original factors x−y, x+y and x²−xy+y² must be nonzero.
Do not drop original exclusions or claim an equality at (0,0).
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a(a² + ab + b²)/[b(a² + b²)(a² − ab + b²)]. Use (a+b)² and the cube factors, then cancel (a+b)²(a−b). Retain (a³+b³)(a⁴b−b⁵) ≠ 0.
Check cancellation by expanding factors and state the original denominator condition.
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x is not a factor of the whole numerator. The expression is 1 + 6/x for x ≠ 0; at x = 3 it is 3, not 6.
Use both structural reasoning and a permitted counterexample.
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They agree for x ≠ 3; the original is undefined at 3 while x+3 is defined there. Cancellation preserves values on the original domain, not the missing input.
Mark an open circle at 3 on a domain line and label its meaning.
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Use 6x²y³: the numerators become 4y² and 5x. Require x,y ≠ 0.
Multiply entire numerators by their missing factors.
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D=x(x−1)(x+1); numerators x and 2(x+1). Require x ≠ −1,0,1.
Inventory all factors and keep parentheses around x+1.
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Use ybn with y,b,n nonzero. The equivalent numerators are xbn, ayn and myb. It is a valid symbolic common denominator; special numerical choices may admit a smaller one.
Divide the chosen denominator by each original denominator before rewriting.
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(a−2b)(a+2b)=a²−4b², which is different. A valid common denominator is (a+2b)(a²−4b); retain both nonzero conditions. Do not silently insert a missing square.
State what is actually printed and distinguish an assumption from an identity.
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It equals −1 for x ≠ y because y−x=−(x−y). At equality both numerator and denominator are zero, so the original is undefined.
Explain the sign and the excluded equality.
Make a factor-and-domain notebook with two monomial fractions, two polynomial fractions and one common-denominator pair.
Keep your work on paper or in a local file. This page does not upload it or automatically grade the project.
This original Kidemy practice follows the topic order of 2026 Class 8 Mathematics · English Version, chapter 5, printed pages 77–100. It does not replace the complete chapter or its practical teaching.
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