Unit 8: Algebraic Fraction (बीजीय भिन्न) Formulae & Solutions
Solutions by Important Edu Notes Team
1. Important Formulas & Concepts
A. Factorization Identities
| Name | Formula |
|---|---|
| Difference of Squares | $$a^2 – b^2 = (a – b)(a + b)$$ |
| Square of Sum/Diff |
$$(a + b)^2 = a^2 + 2ab + b^2$$ $$(a – b)^2 = a^2 – 2ab + b^2$$ |
| Sum/Diff of Cubes |
$$a^3 + b^3 = (a + b)(a^2 – ab + b^2)$$ $$a^3 – b^3 = (a – b)(a^2 + ab + b^2)$$ |
B. Simplification Tricks
- Sign Reversal: $b – a = -(a – b)$. Essential when denominators are opposites.
- Cyclic Order: Maintain $(a-b), (b-c), (c-a)$ for symmetry.
- LCM Rule: Factorize denominators fully before taking LCM. Take the highest power of each unique factor.
- Surds: $(1-\sqrt{x})(1+\sqrt{x}) = 1 – x$.
2. Exercise 8.1 – Q1: Change into Simplest Form
Q1.a: Change into simplest form: $\frac{x^2 – 5x}{x^2 – 25}$
Numerator: $x^2 – 5x = x(x – 5)$
Denominator: $x^2 – 25 = (x – 5)(x + 5)$
$$= \frac{x(x – 5)}{(x – 5)(x + 5)}$$
$$= \frac{x}{x + 5}$$
$$= \frac{x}{x + 5}$$
$\frac{x}{x + 5}$
Q1.b: Change into simplest form: $\frac{x^2 – b^2}{(x + b)^2}$
Numerator: $x^2 – b^2 = (x – b)(x + b)$
Denominator: $(x + b)^2 = (x + b)(x + b)$
$$= \frac{(x – b)(x + b)}{(x + b)(x + b)}$$
$$= \frac{x – b}{x + b}$$
$$= \frac{x – b}{x + b}$$
$\frac{x – b}{x + b}$
Q1.c: Change into simplest form: $\frac{x^2 – 5x + 6}{x^2 – 7x + 12}$
Numerator: $x^2 – 5x + 6 = x^2 – 2x – 3x + 6 = x(x – 2) – 3(x – 2) = (x – 2)(x – 3)$
Denominator: $x^2 – 7x + 12 = x^2 – 3x – 4x + 12 = x(x – 3) – 4(x – 3) = (x – 3)(x – 4)$
$$= \frac{(x – 2)(x – 3)}{(x – 3)(x – 4)}$$
$$= \frac{x – 2}{x – 4}$$
$$= \frac{x – 2}{x – 4}$$
$\frac{x – 2}{x – 4}$
Q2.a: Simplify: $\frac{a}{a-b} + \frac{b}{b-a}$
$$= \frac{a}{a-b} + \frac{b}{-(a-b)}$$
$$= \frac{a}{a-b} – \frac{b}{a-b}$$
$$= \frac{a-b}{a-b}$$
$$= 1$$
$$= \frac{a}{a-b} – \frac{b}{a-b}$$
$$= \frac{a-b}{a-b}$$
$$= 1$$
1
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3. Exercise 8.1 – Q2: Simplify (Continued)
Q2.b: Simplify: $\frac{1}{b-c} – \frac{b+c}{b^2-c^2}$
$$= \frac{1}{b-c} – \frac{b+c}{(b-c)(b+c)}$$
$$= \frac{1}{b-c} – \frac{1}{b-c}$$
$$= 0$$
$$= \frac{1}{b-c} – \frac{1}{b-c}$$
$$= 0$$
0
Q2.c: Simplify: $\frac{1}{m+n} + \frac{n}{m^2-n^2}$
LCM is $m^2 – n^2 = (m-n)(m+n)$.
$$= \frac{1(m-n)}{(m+n)(m-n)} + \frac{n}{(m-n)(m+n)}$$
$$= \frac{m – n + n}{m^2 – n^2}$$
$$= \frac{m}{m^2 – n^2}$$
$$= \frac{m – n + n}{m^2 – n^2}$$
$$= \frac{m}{m^2 – n^2}$$
$\frac{m}{m^2 – n^2}$
Q2.d: Simplify: $\frac{m+n}{m-n} + \frac{m-n}{m+n}$
LCM is $(m-n)(m+n) = m^2 – n^2$.
$$= \frac{(m+n)^2 + (m-n)^2}{m^2 – n^2}$$
$$= \frac{(m^2 + 2mn + n^2) + (m^2 – 2mn + n^2)}{m^2 – n^2}$$
$$= \frac{2m^2 + 2n^2}{m^2 – n^2}$$
$$= \frac{2(m^2 + n^2)}{m^2 – n^2}$$
$$= \frac{(m^2 + 2mn + n^2) + (m^2 – 2mn + n^2)}{m^2 – n^2}$$
$$= \frac{2m^2 + 2n^2}{m^2 – n^2}$$
$$= \frac{2(m^2 + n^2)}{m^2 – n^2}$$
$\frac{2(m^2 + n^2)}{m^2 – n^2}$
Q2.e: Simplify: $\frac{1}{m-n} + \frac{1}{m+n}$
LCM is $m^2 – n^2$.
$$= \frac{1(m+n) + 1(m-n)}{m^2 – n^2}$$
$$= \frac{m + n + m – n}{m^2 – n^2}$$
$$= \frac{2m}{m^2 – n^2}$$
$$= \frac{m + n + m – n}{m^2 – n^2}$$
$$= \frac{2m}{m^2 – n^2}$$
$\frac{2m}{m^2 – n^2}$
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Q2.f: Simplify: $\frac{3}{x^2-4} + \frac{1}{(x-2)^2}$
LCM is $(x-2)^2(x+2)$.
$$= \frac{3}{(x-2)(x+2)} + \frac{1}{(x-2)(x-2)}$$
$$= \frac{3(x-2) + 1(x+2)}{(x-2)^2(x+2)}$$
$$= \frac{3x – 6 + x + 2}{(x-2)^2(x+2)}$$
$$= \frac{4x – 4}{(x-2)^2(x+2)}$$
$$= \frac{4(x – 1)}{(x-2)^2(x+2)}$$
$$= \frac{3(x-2) + 1(x+2)}{(x-2)^2(x+2)}$$
$$= \frac{3x – 6 + x + 2}{(x-2)^2(x+2)}$$
$$= \frac{4x – 4}{(x-2)^2(x+2)}$$
$$= \frac{4(x – 1)}{(x-2)^2(x+2)}$$
$\frac{4(x – 1)}{(x-2)^2(x+2)}$
Q2.g: Simplify: $\frac{a^3+b^3}{a^2-ab+b^2} + \frac{a^3-b^3}{a^2+ab+b^2}$
Using formula: $a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$.
$$= \frac{(a+b)(a^2-ab+b^2)}{a^2-ab+b^2} + \frac{(a-b)(a^2+ab+b^2)}{a^2+ab+b^2}$$
$$= (a + b) + (a – b)$$
$$= 2a$$
$$= (a + b) + (a – b)$$
$$= 2a$$
2a
Q2.h: Simplify: $\frac{4x^2+25y^2}{4x^2-25y^2} – \frac{2x-5y}{2x+5y}$
LCM is $4x^2 – 25y^2 = (2x-5y)(2x+5y)$.
$$= \frac{4x^2+25y^2 – (2x-5y)(2x-5y)}{4x^2-25y^2}$$
$$= \frac{4x^2+25y^2 – (4x^2 – 20xy + 25y^2)}{4x^2-25y^2}$$
$$= \frac{4x^2+25y^2 – 4x^2 + 20xy – 25y^2}{4x^2-25y^2}$$
$$= \frac{20xy}{4x^2-25y^2}$$
$$= \frac{4x^2+25y^2 – (4x^2 – 20xy + 25y^2)}{4x^2-25y^2}$$
$$= \frac{4x^2+25y^2 – 4x^2 + 20xy – 25y^2}{4x^2-25y^2}$$
$$= \frac{20xy}{4x^2-25y^2}$$
$\frac{20xy}{4x^2-25y^2}$
Q2.i: Simplify: $\frac{4x^3}{x^4+a^4} – \frac{8x^7}{x^8-a^8}$
$$= \frac{4x^3}{x^4+a^4} – \frac{8x^7}{(x^4-a^4)(x^4+a^4)}$$
$$LCM = x^8 – a^8$$
$$= \frac{4x^3(x^4-a^4) – 8x^7}{x^8-a^8}$$
$$= \frac{4x^7 – 4x^3a^4 – 8x^7}{x^8-a^8}$$
$$= \frac{-4x^7 – 4x^3a^4}{x^8-a^8}$$
$$= \frac{-4x^3(x^4+a^4)}{(x^4-a^4)(x^4+a^4)}$$
$$= \frac{-4x^3}{x^4-a^4}$$
$$= \frac{4x^3}{a^4-x^4}$$
$$LCM = x^8 – a^8$$
$$= \frac{4x^3(x^4-a^4) – 8x^7}{x^8-a^8}$$
$$= \frac{4x^7 – 4x^3a^4 – 8x^7}{x^8-a^8}$$
$$= \frac{-4x^7 – 4x^3a^4}{x^8-a^8}$$
$$= \frac{-4x^3(x^4+a^4)}{(x^4-a^4)(x^4+a^4)}$$
$$= \frac{-4x^3}{x^4-a^4}$$
$$= \frac{4x^3}{a^4-x^4}$$
$\frac{4x^3}{a^4-x^4}$
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Q2.j: Simplify: $\frac{3}{a+3} + \frac{4}{a-3} + \frac{9a}{2(9-a^2)}$
Note: $9-a^2 = -(a^2-9)$.
$$= \frac{3}{a+3} + \frac{4}{a-3} – \frac{9a}{2(a^2-9)}$$
$$LCM = 2(a+3)(a-3)$$
$$= \frac{3 \cdot 2(a-3) + 4 \cdot 2(a+3) – 9a}{2(a^2-9)}$$
$$= \frac{6a – 18 + 8a + 24 – 9a}{2(a^2-9)}$$
$$= \frac{5a + 6}{2(a^2-9)}$$
$$LCM = 2(a+3)(a-3)$$
$$= \frac{3 \cdot 2(a-3) + 4 \cdot 2(a+3) – 9a}{2(a^2-9)}$$
$$= \frac{6a – 18 + 8a + 24 – 9a}{2(a^2-9)}$$
$$= \frac{5a + 6}{2(a^2-9)}$$
$\frac{5a + 6}{2(a^2-9)}$
Q2.k: Simplify: $\frac{x}{x-y} – \frac{x}{x+y} + \frac{2xy}{x^2+y^2}$
$$= \frac{x(x+y) – x(x-y)}{x^2-y^2} + \frac{2xy}{x^2+y^2}$$
$$= \frac{x^2+xy-x^2+xy}{x^2-y^2} + \frac{2xy}{x^2+y^2}$$
$$= \frac{2xy}{x^2-y^2} + \frac{2xy}{x^2+y^2}$$
$$= \frac{2xy(x^2+y^2) + 2xy(x^2-y^2)}{x^4-y^4}$$
$$= \frac{2x^3y + 2xy^3 + 2x^3y – 2xy^3}{x^4-y^4}$$
$$= \frac{4x^3y}{x^4-y^4}$$
$$= \frac{x^2+xy-x^2+xy}{x^2-y^2} + \frac{2xy}{x^2+y^2}$$
$$= \frac{2xy}{x^2-y^2} + \frac{2xy}{x^2+y^2}$$
$$= \frac{2xy(x^2+y^2) + 2xy(x^2-y^2)}{x^4-y^4}$$
$$= \frac{2x^3y + 2xy^3 + 2x^3y – 2xy^3}{x^4-y^4}$$
$$= \frac{4x^3y}{x^4-y^4}$$
$\frac{4x^3y}{x^4-y^4}$
Q2.l: Simplify: $\frac{1}{x+2y} – \frac{1}{x-2y} + \frac{2x}{4y^2-x^2}$
Third term denominator $4y^2-x^2 = -(x^2-4y^2)$.
$$= \frac{1}{x+2y} – \frac{1}{x-2y} – \frac{2x}{x^2-4y^2}$$
$$LCM = x^2-4y^2 = (x-2y)(x+2y)$$
$$= \frac{(x-2y) – (x+2y) – 2x}{x^2-4y^2}$$
$$= \frac{x – 2y – x – 2y – 2x}{x^2-4y^2}$$
$$= \frac{-2x – 4y}{x^2-4y^2}$$
$$= \frac{-2(x+2y)}{(x-2y)(x+2y)}$$
$$= \frac{-2}{x-2y} = \frac{2}{2y-x}$$
$$LCM = x^2-4y^2 = (x-2y)(x+2y)$$
$$= \frac{(x-2y) – (x+2y) – 2x}{x^2-4y^2}$$
$$= \frac{x – 2y – x – 2y – 2x}{x^2-4y^2}$$
$$= \frac{-2x – 4y}{x^2-4y^2}$$
$$= \frac{-2(x+2y)}{(x-2y)(x+2y)}$$
$$= \frac{-2}{x-2y} = \frac{2}{2y-x}$$
$\frac{2}{2y-x}$
Q2.m: Simplify: $\frac{a}{(a-b)(a-c)} + \frac{b}{(b-a)(b-c)} + \frac{c}{(c-b)(c-a)}$
Make order cyclic: $(a-b), (b-c), (c-a)$.
$$= \frac{-a}{(a-b)(c-a)} + \frac{-b}{(a-b)(b-c)} + \frac{-c}{(b-c)(c-a)}$$
$$= – \left[ \frac{a(b-c) + b(c-a) + c(a-b)}{(a-b)(b-c)(c-a)} \right]$$
$$Numerator: ab-ac+bc-ab+ac-bc = 0$$
$$= 0$$
$$= – \left[ \frac{a(b-c) + b(c-a) + c(a-b)}{(a-b)(b-c)(c-a)} \right]$$
$$Numerator: ab-ac+bc-ab+ac-bc = 0$$
$$= 0$$
0
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Q2.n: Simplify: $\frac{y-z}{x^2-(y-z)^2} + \frac{z-x}{y^2-(z-x)^2} + \frac{x-y}{z^2-(x-y)^2}$
Factorize denominators as difference of squares: $A^2-B^2 = (A-B)(A+B)$.
Let $S = x+y+z$. Then $x-y+z = S-2y$ and $x+y-z = S-2z$.
$$Term 1 = \frac{y-z}{(S-2y)(S-2z)}$$
$$Term 2 = \frac{z-x}{(S-2z)(S-2x)}$$
$$Term 3 = \frac{x-y}{(S-2x)(S-2y)}$$
$$LCM = (S-2x)(S-2y)(S-2z)$$
$$Numerator = (y-z)(S-2x) + (z-x)(S-2y) + (x-y)(S-2z)$$
$$= S(y-z+z-x+x-y) – 2(x(y-z) + y(z-x) + z(x-y))$$
$$= S(0) – 2(xy-xz + yz-yx + zx-zy)$$
$$= 0 – 2(0) = 0$$
$$Term 2 = \frac{z-x}{(S-2z)(S-2x)}$$
$$Term 3 = \frac{x-y}{(S-2x)(S-2y)}$$
$$LCM = (S-2x)(S-2y)(S-2z)$$
$$Numerator = (y-z)(S-2x) + (z-x)(S-2y) + (x-y)(S-2z)$$
$$= S(y-z+z-x+x-y) – 2(x(y-z) + y(z-x) + z(x-y))$$
$$= S(0) – 2(xy-xz + yz-yx + zx-zy)$$
$$= 0 – 2(0) = 0$$
0
Q2.o: Simplify: $\frac{x^2-(a-b)^2}{(x+b)^2-a^2} + \frac{a^2-(x-b)^2}{(x+a)^2-b^2} + \frac{b^2-(x-a)^2}{(a+b)^2-x^2}$
$$Term 1: \frac{(x-a+b)(x+a-b)}{(x+b-a)(x+b+a)} = \frac{x+a-b}{x+a+b}$$
$$Term 2: \frac{(a-x+b)(a+x-b)}{(x+a-b)(x+a+b)} = \frac{a+b-x}{x+a+b}$$
$$Term 3: \frac{(b-x+a)(b+x-a)}{(a+b-x)(a+b+x)} = \frac{b+x-a}{x+a+b}$$
$$Sum = \frac{(x+a-b) + (a+b-x) + (b+x-a)}{x+a+b}$$
$$= \frac{x+a+b}{x+a+b}$$
$$= 1$$
$$Term 2: \frac{(a-x+b)(a+x-b)}{(x+a-b)(x+a+b)} = \frac{a+b-x}{x+a+b}$$
$$Term 3: \frac{(b-x+a)(b+x-a)}{(a+b-x)(a+b+x)} = \frac{b+x-a}{x+a+b}$$
$$Sum = \frac{(x+a-b) + (a+b-x) + (b+x-a)}{x+a+b}$$
$$= \frac{x+a+b}{x+a+b}$$
$$= 1$$
1
Q2.p: Simplify: $\frac{1}{p^2+7p+12} + \frac{2}{p^2+5p+6} – \frac{3}{p^2+6p+8}$
Factors: $(p+3)(p+4)$, $(p+2)(p+3)$, $(p+2)(p+4)$.
LCM: $(p+2)(p+3)(p+4)$.
$$= \frac{1(p+2) + 2(p+4) – 3(p+3)}{(p+2)(p+3)(p+4)}$$
$$= \frac{p+2 + 2p+8 – 3p-9}{(p+2)(p+3)(p+4)}$$
$$= \frac{3p+10 – 3p-9}{(p+2)(p+3)(p+4)}$$
$$= \frac{1}{(p+2)(p+3)(p+4)}$$
$$= \frac{p+2 + 2p+8 – 3p-9}{(p+2)(p+3)(p+4)}$$
$$= \frac{3p+10 – 3p-9}{(p+2)(p+3)(p+4)}$$
$$= \frac{1}{(p+2)(p+3)(p+4)}$$
$\frac{1}{(p+2)(p+3)(p+4)}$
Q2.q: Simplify: $\frac{x+3}{x^2+3x+9} + \frac{x-3}{x^2-3x+9} – \frac{54}{x^4+9x^2+81}$
Note: $(x^2+3x+9)(x^2-3x+9) = x^4+9x^2+81$.
$$= \frac{(x+3)(x^2-3x+9) + (x-3)(x^2+3x+9) – 54}{x^4+9x^2+81}$$
$$= \frac{(x^3+27) + (x^3-27) – 54}{x^4+9x^2+81}$$
$$= \frac{2x^3 – 54}{x^4+9x^2+81}$$
$$= \frac{(x^3+27) + (x^3-27) – 54}{x^4+9x^2+81}$$
$$= \frac{2x^3 – 54}{x^4+9x^2+81}$$
$\frac{2(x^3 – 27)}{x^4+9x^2+81}$
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Q2.r: Simplify: $\frac{1}{x^2-5x+6} + \frac{2}{4x-x^2-3} – \frac{3}{x^2-3x+2}$
Factors: $(x-2)(x-3)$, $-(x-1)(x-3)$, $(x-1)(x-2)$.
$$= \frac{1}{(x-2)(x-3)} – \frac{2}{(x-1)(x-3)} – \frac{3}{(x-1)(x-2)}$$
$$LCM: (x-1)(x-2)(x-3)$$
$$= \frac{1(x-1) – 2(x-2) – 3(x-3)}{(x-1)(x-2)(x-3)}$$
$$= \frac{x-1 – 2x+4 – 3x+9}{(x-1)(x-2)(x-3)}$$
$$= \frac{-4x+12}{(x-1)(x-2)(x-3)} = \frac{-4(x-3)}{(x-1)(x-2)(x-3)}$$
$$= \frac{-4}{(x-1)(x-2)}$$
$$LCM: (x-1)(x-2)(x-3)$$
$$= \frac{1(x-1) – 2(x-2) – 3(x-3)}{(x-1)(x-2)(x-3)}$$
$$= \frac{x-1 – 2x+4 – 3x+9}{(x-1)(x-2)(x-3)}$$
$$= \frac{-4x+12}{(x-1)(x-2)(x-3)} = \frac{-4(x-3)}{(x-1)(x-2)(x-3)}$$
$$= \frac{-4}{(x-1)(x-2)}$$
$\frac{-4}{(x-1)(x-2)}$
Q2.s: Simplify: $\frac{1}{1-b+b^2} – \frac{1}{1+b+b^2} – \frac{2b}{1-b^2+b^4}$
LCM: $1+b^2+b^4$.
$$= \frac{(1+b+b^2) – (1-b+b^2) – 2b}{1+b^2+b^4}$$
$$= \frac{2b – 2b}{1+b^2+b^4}$$
$$= 0$$
$$= \frac{2b – 2b}{1+b^2+b^4}$$
$$= 0$$
0
Q2.t: Simplify: $\frac{b+2}{1+b+b^2} – \frac{b-2}{1-b+b^2} – \frac{2b^2}{1+b^2+b^4}$
LCM: $1+b^2+b^4$.
$$Num: (b+2)(1-b+b^2) – (b-2)(1+b+b^2) – 2b^2$$
$$(b+2)(1-b+b^2) = b^3+b^2-b+2$$
$$(b-2)(1+b+b^2) = b^3-b^2-b-2$$
$$Subtracted: (2b^2 + 4) – 2b^2 = 4$$
$$= \frac{4}{1+b^2+b^4}$$
$$(b+2)(1-b+b^2) = b^3+b^2-b+2$$
$$(b-2)(1+b+b^2) = b^3-b^2-b-2$$
$$Subtracted: (2b^2 + 4) – 2b^2 = 4$$
$$= \frac{4}{1+b^2+b^4}$$
$\frac{4}{1+b^2+b^4}$
Q2.u: Simplify: $\frac{a+c}{a^2+ac+c^2} + \frac{a-c}{a^2-ac+c^2} + \frac{2c^3}{a^4+a^2c^2+c^4}$
LCM: $a^4+a^2c^2+c^4$.
$$Num: (a+c)(a^2-ac+c^2) + (a-c)(a^2+ac+c^2) + 2c^3$$
$$= (a^3+c^3) + (a^3-c^3) + 2c^3$$
$$= 2a^3 + 2c^3$$
$$= (a^3+c^3) + (a^3-c^3) + 2c^3$$
$$= 2a^3 + 2c^3$$
$\frac{2(a^3+c^3)}{a^4+a^2c^2+c^4}$
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4. Exercise 8.1 – Q3: Simplify (with Roots)
Q3.a: Simplify: $\frac{1}{4(1-\sqrt{x})} – \frac{1}{4(1+\sqrt{x})} + \frac{2\sqrt{x}}{4(1-x)}$
$$= \frac{1}{4} \left[ \frac{(1+\sqrt{x}) – (1-\sqrt{x})}{1-x} \right] + \frac{2\sqrt{x}}{4(1-x)}$$
$$= \frac{2\sqrt{x}}{4(1-x)} + \frac{2\sqrt{x}}{4(1-x)}$$
$$= \frac{4\sqrt{x}}{4(1-x)}$$
$$= \frac{\sqrt{x}}{1-x}$$
$$= \frac{2\sqrt{x}}{4(1-x)} + \frac{2\sqrt{x}}{4(1-x)}$$
$$= \frac{4\sqrt{x}}{4(1-x)}$$
$$= \frac{\sqrt{x}}{1-x}$$
$\frac{\sqrt{x}}{1-x}$
Q3.b: Simplify: $\frac{1}{8(1-\sqrt{x})} – \frac{1}{8(1+\sqrt{x})} + \frac{2\sqrt{x}}{8(1-x)}$
$$= \frac{1}{8} \left[ \frac{2\sqrt{x}}{1-x} \right] + \frac{2\sqrt{x}}{8(1-x)}$$
$$= \frac{4\sqrt{x}}{8(1-x)}$$
$$= \frac{\sqrt{x}}{2(1-x)}$$
$$= \frac{4\sqrt{x}}{8(1-x)}$$
$$= \frac{\sqrt{x}}{2(1-x)}$$
$\frac{\sqrt{x}}{2(1-x)}$
Q3.c: Simplify: $\frac{1}{(a+1)^2} + \frac{1}{(a-1)^2} – \frac{2}{a^2-1}$
$$= \frac{(a-1)^2 + (a+1)^2 – 2(a+1)(a-1)}{(a+1)^2(a-1)^2}$$
$$= \frac{2a^2+2 – (2a^2-2)}{(a^2-1)^2}$$
$$= \frac{4}{(a^2-1)^2}$$
$$= \frac{2a^2+2 – (2a^2-2)}{(a^2-1)^2}$$
$$= \frac{4}{(a^2-1)^2}$$
$\frac{4}{(a^2-1)^2}$
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5. Exercise 8.1 – Q4 & Q5: Find Variables
Q4: If $\frac{a}{2x+1} + \frac{1}{x+2} = \frac{4x+5}{2x^2+5x+2}$, find a.
$$\frac{a(x+2) + 1(2x+1)}{(2x+1)(x+2)} = \frac{4x+5}{(2x+1)(x+2)}$$
$$ax + 2a + 2x + 1 = 4x + 5$$
$$(a+2)x + (2a+1) = 4x + 5$$
$$Compare\ x: a+2=4 \Rightarrow a=2$$
$$Check\ constant: 2(2)+1 = 5 \quad (Correct)$$
$$ax + 2a + 2x + 1 = 4x + 5$$
$$(a+2)x + (2a+1) = 4x + 5$$
$$Compare\ x: a+2=4 \Rightarrow a=2$$
$$Check\ constant: 2(2)+1 = 5 \quad (Correct)$$
a = 2
Q5: If $\frac{a}{2x-3} + \frac{b}{3x+4} = \frac{x+7}{6x^2-x-12}$, find a and b.
$$\frac{a(3x+4) + b(2x-3)}{(2x-3)(3x+4)} = \frac{x+7}{(2x-3)(3x+4)}$$
$$3ax + 4a + 2bx – 3b = x + 7$$
$$(3a+2b)x + (4a-3b) = x + 7$$
$$System:$$
$$3a+2b = 1 \dots (i)$$
$$4a-3b = 7 \dots (ii)$$
$$Multiply (i) by 3, (ii) by 2:$$
$$9a+6b=3$$
$$8a-6b=14$$
$$Add: 17a=17 \Rightarrow a=1$$
$$Sub\ in\ (i): 3(1)+2b=1 \Rightarrow 2b=-2 \Rightarrow b=-1$$
$$3ax + 4a + 2bx – 3b = x + 7$$
$$(3a+2b)x + (4a-3b) = x + 7$$
$$System:$$
$$3a+2b = 1 \dots (i)$$
$$4a-3b = 7 \dots (ii)$$
$$Multiply (i) by 3, (ii) by 2:$$
$$9a+6b=3$$
$$8a-6b=14$$
$$Add: 17a=17 \Rightarrow a=1$$
$$Sub\ in\ (i): 3(1)+2b=1 \Rightarrow 2b=-2 \Rightarrow b=-1$$
a = 1, b = -1
Disclaimer: The solutions provided here are prepared by the Important Edu Notes Team and are based on the CDC curriculum.