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Every question from the end of every unit, with hints and answers to open. After you check an answer, mark how you did, and come back to the ones you didn’t get.

29 questions

Unit 8 · Algebraic Fraction

Question 1

a.

x2−5xx2−25 \dfrac{x^2 - 5x}{x^2 - 25}

Answer

x(x−5)(x+5)(x−5)= \dfrac{x(x - 5)}{(x + 5)(x - 5)} = xx+5 \dfrac{x}{x + 5}

b.

a2−b2(a−b)2 \dfrac{a^2 - b^2}{(a - b)^2}

Answer

(a+b)(a−b)(a−b)(a−b)= \dfrac{(a + b)(a - b)}{(a - b)(a - b)} = a+ba−b \dfrac{a + b}{a - b}

c.

x2−5x+6x2−7x+12 \dfrac{x^2 - 5x + 6}{x^2 - 7x + 12}

Answer

(x−2)(x−3)(x−3)(x−4)= \dfrac{(x - 2)(x - 3)}{(x - 3)(x - 4)} = x−2x−4 \dfrac{x - 2}{x - 4}

Question 2

a.

aa−b+bb−a \dfrac{a}{a - b} + \dfrac{b}{b - a}

Answer

bb−a=−ba−b \dfrac{b}{b - a} = -\dfrac{b}{a - b} , so

aa−b−ba−b=a−ba−b=1 \frac{a}{a - b} - \frac{b}{a - b} = \frac{a - b}{a - b} = 1

1

b.

1b−c−b+cb2−c2 \dfrac{1}{b - c} - \dfrac{b + c}{b^2 - c^2}

Answer

The second fraction reduces: b+c(b−c)(b+c)=1b−c \dfrac{b + c}{(b - c)(b + c)} = \dfrac{1}{b - c} . So the difference is 1b−c−1b−c \dfrac{1}{b - c} - \dfrac{1}{b - c} : 0.

c.

1m−n+1m+n \dfrac{1}{m - n} + \dfrac{1}{m + n}

Answer

(m+n)+(m−n)(m−n)(m+n)=2mm2−n2 \frac{(m + n) + (m - n)}{(m - n)(m + n)} = \frac{2m}{m^2 - n^2}

2mm2−n2 \dfrac{2m}{m^2 - n^2}

d.

m+nm−n+m−nm+n \dfrac{m + n}{m - n} + \dfrac{m - n}{m + n}

Answer

(m+n)2+(m−n)2m2−n2=2m2+2n2m2−n2 \begin{aligned} &\frac{(m + n)^2 + (m - n)^2}{m^2 - n^2} \\ &= \frac{2m^2 + 2n^2}{m^2 - n^2} \end{aligned}

2(m2+n2)m2−n2 \dfrac{2(m^2 + n^2)}{m^2 - n^2}

e.

1m+n+nm2−n2 \dfrac{1}{m + n} + \dfrac{n}{m^2 - n^2}

Answer

LCM (m + n)(m − n):

(m−n)+nm2−n2=mm2−n2 \frac{(m - n) + n}{m^2 - n^2} = \frac{m}{m^2 - n^2}

mm2−n2 \dfrac{m}{m^2 - n^2}

f.

3x2−4+1(x−2)2 \dfrac{3}{x^2 - 4} + \dfrac{1}{(x - 2)^2}

Answer

x2−4=(x+2)(x−2) x^2 - 4 = (x + 2)(x - 2) , so the LCM is (x+2)(x−2)2 (x + 2)(x - 2)^2 .

3(x−2)+(x+2)(x+2)(x−2)2=4x−4(x+2)(x−2)2 \begin{aligned} &\frac{3(x - 2) + (x + 2)}{(x + 2)(x - 2)^2} \\ &= \frac{4x - 4}{(x + 2)(x - 2)^2} \end{aligned}

4(x−1)(x+2)(x−2)2 \dfrac{4(x - 1)}{(x + 2)(x - 2)^2}

g.

a3+b3a2−ab+b2+a3−b3a2+ab+b2 \dfrac{a^3 + b^3}{a^2 - ab + b^2} + \dfrac{a^3 - b^3}{a^2 + ab + b^2}

Answer

a3+b3=(a+b)(a2−ab+b2) a^3 + b^3 = (a + b)(a^2 - ab + b^2) and a3−b3=(a−b)(a2+ab+b2) a^3 - b^3 = (a - b)(a^2 + ab + b^2) , so each fraction reduces:

(a+b)+(a−b)=2a (a + b) + (a - b) = 2a

2a

h.

4x2+25y24x2−25y2−2x−5y2x+5y \dfrac{4x^2 + 25y^2}{4x^2 - 25y^2} - \dfrac{2x - 5y}{2x + 5y}

Answer

4x2−25y2=(2x−5y)(2x+5y) 4x^2 - 25y^2 = (2x - 5y)(2x + 5y) , so the second fraction needs (2x − 5y):

4x2+25y2−(2x−5y)24x2−25y2=4x2+25y2−4x2+20xy−25y24x2−25y2 \begin{aligned} &\frac{4x^2 + 25y^2 - (2x - 5y)^2}{4x^2 - 25y^2} \\ &= \frac{4x^2 + 25y^2 - 4x^2 + 20xy - 25y^2}{4x^2 - 25y^2} \end{aligned}

20xy4x2−25y2 \dfrac{20xy}{4x^2 - 25y^2}

i.

4x3x4+a4−8x7x8−a8 \dfrac{4x^3}{x^4 + a^4} - \dfrac{8x^7}{x^8 - a^8}

Answer

x8−a8=(x4+a4)(x4−a4) x^8 - a^8 = (x^4 + a^4)(x^4 - a^4) , the LCM.

4x3(x4−a4)−8x7x8−a8=−4x7−4a4x3x8−a8=−4x3(x4+a4)(x4+a4)(x4−a4)=−4x3x4−a4 \begin{aligned} &\frac{4x^3(x^4 - a^4) - 8x^7}{x^8 - a^8} \\ &= \frac{-4x^7 - 4a^4x^3}{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} \end{aligned}

4x3a4−x4 \dfrac{4x^3}{a^4 - x^4}

j.

xx−y−xx+y+2xyx2+y2 \dfrac{x}{x - y} - \dfrac{x}{x + y} + \dfrac{2xy}{x^2 + y^2}

Answer

The first two:

x(x+y)−x(x−y)x2−y2=2xyx2−y2 \frac{x(x + y) - x(x - y)}{x^2 - y^2} = \frac{2xy}{x^2 - y^2}

Then, with LCM x4−y4 x^4 - y^4 :

2xy(x2+y2)+2xy(x2−y2)x4−y4=2xy⋅2x2x4−y4 \begin{aligned} &\frac{2xy(x^2 + y^2) + 2xy(x^2 - y^2)}{x^4 - y^4} \\ &= \frac{2xy \cdot 2x^2}{x^4 - y^4} \end{aligned}

4x3yx4−y4 \dfrac{4x^3y}{x^4 - y^4}

k.

3a+3+4a−3+9a2(9−a2) \dfrac{3}{a + 3} + \dfrac{4}{a - 3} + \dfrac{9a}{2(9 - a^2)}

Answer

9a2(9−a2)=−9a2(a2−9) \dfrac{9a}{2(9 - a^2)} = -\dfrac{9a}{2(a^2 - 9)} ; the LCM is 2(a + 3)(a − 3).

6(a−3)+8(a+3)−9a2(a2−9)=6a−18+8a+24−9a2(a2−9) \begin{aligned} &\frac{6(a - 3) + 8(a + 3) - 9a}{2(a^2 - 9)} \\ &= \frac{6a - 18 + 8a + 24 - 9a}{2(a^2 - 9)} \end{aligned}

5a+62(a2−9) \dfrac{5a + 6}{2(a^2 - 9)}

l.

1x+2y−1x−2y+2x4y2−x2 \dfrac{1}{x + 2y} - \dfrac{1}{x - 2y} + \dfrac{2x}{4y^2 - x^2}

Answer

2x4y2−x2=−2xx2−4y2 \dfrac{2x}{4y^2 - x^2} = -\dfrac{2x}{x^2 - 4y^2} ; the LCM is (x + 2y)(x − 2y).

(x−2y)−(x+2y)−2xx2−4y2=−2x−4yx2−4y2=−2(x+2y)(x+2y)(x−2y)=−2x−2y \begin{aligned} &\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} \end{aligned}

22y−x \dfrac{2}{2y - x}

m.

a(a−b)(a−c)+b(b−a)(b−c)+c(c−b)(c−a) \dfrac{a}{(a - b)(a - c)} + \dfrac{b}{(b - a)(b - c)} + \dfrac{c}{(c - b)(c - a)}

Answer

Turn the brackets so that only (a − b), (b − c) and (a − c) appear: (b − a)(b − c) = −(a − b)(b − c) and (c − b)(c − a) = (b − c)(a − c) (two signs changed). The sum becomes

a(a−b)(a−c)−b(a−b)(b−c)+c(b−c)(a−c) \frac{a}{(a - b)(a - c)} - \frac{b}{(a - b)(b - c)} + \frac{c}{(b - c)(a - c)}

with LCM (a − b)(b − c)(a − c). Its numerator:

a(b−c)−b(a−c)+c(a−b)=ab−ac−ab+bc+ac−bc=0 \begin{aligned} &a(b - c) - b(a - c) + c(a - b) \\ &= ab - ac - ab + bc + ac - bc = 0 \end{aligned}

0

n.

y−zx2−(y−z)2+z−xy2−(z−x)2+x−yz2−(x−y)2 \dfrac{y - z}{x^2 - (y - z)^2} + \dfrac{z - x}{y^2 - (z - x)^2} + \dfrac{x - y}{z^2 - (x - y)^2}

Answer

Each denominator is a difference of squares. Write P = x + y − z, Q = y + z − x and R = z + x − y:

x2−(y−z)2=(x−y+z)(x+y−z)=RPy2−(z−x)2=(y−z+x)(y+z−x)=PQz2−(x−y)2=(z−x+y)(z+x−y)=QR \begin{aligned} x^2 - (y - z)^2 &= (x - y + z)(x + y - z) = RP \\ y^2 - (z - x)^2 &= (y - z + x)(y + z - x) = PQ \\ z^2 - (x - y)^2 &= (z - x + y)(z + x - y) = QR \end{aligned}

The LCM is PQR, and the numerator is (y − z)Q + (z − x)R + (x − y)P:

(y−z)(y+z−x)=y2−z2−xy+xz(z−x)(z+x−y)=z2−x2−yz+xy(x−y)(x+y−z)=x2−y2−xz+yz \begin{aligned} (y - z)(y + z - x) &= y^2 - z^2 - xy + xz \\ (z - x)(z + x - y) &= z^2 - x^2 - yz + xy \\ (x - y)(x + y - z) &= x^2 - y^2 - xz + yz \end{aligned}

These three add up to 0. 0

o.

x2−(a−b)2(x+b)2−a2+a2−(x−b)2(x+a)2−b2+b2−(x−a)2(a+b)2−x2 \dfrac{x^2 - (a - b)^2}{(x + b)^2 - a^2} + \dfrac{a^2 - (x - b)^2}{(x + a)^2 - b^2} + \dfrac{b^2 - (x - a)^2}{(a + b)^2 - x^2}

Answer

Reduce each fraction first: both its top and its bottom are differences of squares.

(x−a+b)(x+a−b)(x+b−a)(x+b+a)=x+a−bx+a+b(a−x+b)(a+x−b)(x+a−b)(x+a+b)=a+b−xx+a+b(b−x+a)(b+x−a)(a+b−x)(a+b+x)=x+b−ax+a+b \begin{aligned} \frac{(x - a + b)(x + a - b)}{(x + b - a)(x + b + a)} &= \frac{x + a - b}{x + a + b} \\ \frac{(a - x + b)(a + x - b)}{(x + a - b)(x + a + b)} &= \frac{a + b - x}{x + a + b} \\ \frac{(b - x + a)(b + x - a)}{(a + b - x)(a + b + x)} &= \frac{x + b - a}{x + a + b} \end{aligned}

They have the same denominator, and the numerators add up to x + a + b:

(x+a−b)+(a+b−x)+(x+b−a)x+a+b=1 \frac{(x + a - b) + (a + b - x) + (x + b - a)}{x + a + b} = 1

1

p.

1p2+7p+12+2p2+5p+6−3p2+6p+8 \dfrac{1}{p^2 + 7p + 12} + \dfrac{2}{p^2 + 5p + 6} - \dfrac{3}{p^2 + 6p + 8}

Answer

The denominators are (p + 3)(p + 4), (p + 2)(p + 3) and (p + 2)(p + 4); the LCM is (p + 2)(p + 3)(p + 4).

(p+2)+2(p+4)−3(p+3)(p+2)(p+3)(p+4)=p+2+2p+8−3p−9(p+2)(p+3)(p+4) \begin{aligned} &\frac{(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)} \end{aligned}

1(p+2)(p+3)(p+4) \dfrac{1}{(p + 2)(p + 3)(p + 4)}

q.

x+3x2+3x+9+x−3x2−3x+9−54x4+9x2+81 \dfrac{x + 3}{x^2 + 3x + 9} + \dfrac{x - 3}{x^2 - 3x + 9} - \dfrac{54}{x^4 + 9x^2 + 81}

Answer

(x2+3x+9)(x2−3x+9)=(x2+9)2−9x2=x4+9x2+81 (x^2 + 3x + 9)(x^2 - 3x + 9) = (x^2 + 9)^2 - 9x^2 = x^4 + 9x^2 + 81 , the LCM. The numerator of the first two is (x3+27)+(x3−27)=2x3 (x^3 + 27) + (x^3 - 27) = 2x^3 , a sum and a difference of cubes. So

2x3−54x4+9x2+81=2(x−3)(x2+3x+9)(x2+3x+9)(x2−3x+9) \begin{aligned} &\frac{2x^3 - 54}{x^4 + 9x^2 + 81} \\ &= \frac{2(x - 3)(x^2 + 3x + 9)}{(x^2 + 3x + 9)(x^2 - 3x + 9)} \end{aligned}

2(x−3)x2−3x+9 \dfrac{2(x - 3)}{x^2 - 3x + 9}

r.

1x2−5x+6+24x−x2−3−3x2−3x+2 \dfrac{1}{x^2 - 5x + 6} + \dfrac{2}{4x - x^2 - 3} - \dfrac{3}{x^2 - 3x + 2}

Answer

x2−5x+6=(x−2)(x−3) x^2 - 5x + 6 = (x - 2)(x - 3) ; 4x−x2−3=−(x−1)(x−3) 4x - x^2 - 3 = -(x - 1)(x - 3) ; x2−3x+2=(x−1)(x−2) x^2 - 3x + 2 = (x - 1)(x - 2) . The LCM is (x − 1)(x − 2)(x − 3).

(x−1)−2(x−2)−3(x−3)(x−1)(x−2)(x−3)=−4x+12(x−1)(x−2)(x−3)=−4(x−1)(x−2) \begin{aligned} &\frac{(x - 1) - 2(x - 2) - 3(x - 3)}{(x - 1)(x - 2)(x - 3)} \\ &= \frac{-4x + 12}{(x - 1)(x - 2)(x - 3)} \\ &= \frac{-4}{(x - 1)(x - 2)} \end{aligned}

43x−x2−2 \dfrac{4}{3x - x^2 - 2} , the same as −4x2−3x+2 -\dfrac{4}{x^2 - 3x + 2} .

s.

b+21+b+b2−b−21−b+b2−2b21+b2+b4 \dfrac{b + 2}{1 + b + b^2} - \dfrac{b - 2}{1 - b + b^2} - \dfrac{2b^2}{1 + b^2 + b^4}

Answer

(1+b+b2)(1−b+b2)=(1+b2)2−b2=1+b2+b4 (1 + b + b^2)(1 - b + b^2) = (1 + b^2)^2 - b^2 = 1 + b^2 + b^4 , the LCM.

(b+2)(1−b+b2)=b3+b2−b+2(b−2)(1+b+b2)=b3−b2−b−2 \begin{aligned} (b + 2)(1 - b + b^2) &= b^3 + b^2 - b + 2 \\ (b - 2)(1 + b + b^2) &= b^3 - b^2 - b - 2 \end{aligned}

Their difference is 2b2+4 2b^2 + 4 , so the whole is 2b2+4−2b21+b2+b4 \dfrac{2b^2 + 4 - 2b^2}{1 + b^2 + b^4} .

41+b2+b4 \dfrac{4}{1 + b^2 + b^4}

t.

11−b+b2−11+b+b2−2b1−b2+b4 \dfrac{1}{1 - b + b^2} - \dfrac{1}{1 + b + b^2} - \dfrac{2b}{1 - b^2 + b^4}

Answer

The first two:

(1+b+b2)−(1−b+b2)1+b2+b4=2b1+b2+b4 \frac{(1 + b + b^2) - (1 - b + b^2)}{1 + b^2 + b^4} = \frac{2b}{1 + b^2 + b^4}

Then, since (1+b2+b4)(1−b2+b4)=(1+b4)2−b4=1+b4+b8 (1 + b^2 + b^4)(1 - b^2 + b^4) = (1 + b^4)^2 - b^4 = 1 + b^4 + b^8 :

2b(1−b2+b4)−2b(1+b2+b4)1+b4+b8=2b(−2b2)1+b4+b8 \begin{aligned} &\frac{2b(1 - b^2 + b^4) - 2b(1 + b^2 + b^4)}{1 + b^4 + b^8} \\ &= \frac{2b(-2b^2)}{1 + b^4 + b^8} \end{aligned}

−4b31+b4+b8 \dfrac{-4b^3}{1 + b^4 + b^8}

u.

a+ca2+ac+c2+a−ca2−ac+c2+2c3a4+a2c2+c4 \dfrac{a + c}{a^2 + ac + c^2} + \dfrac{a - c}{a^2 - ac + c^2} + \dfrac{2c^3}{a^4 + a^2c^2 + c^4}

Answer

The LCM of the first two, (a2+ac+c2)(a2−ac+c2)=a4+a2c2+c4 (a^2 + ac + c^2)(a^2 - ac + c^2) = a^4 + a^2c^2 + c^4 , is the third denominator. The first two numerators give (a3+c3)+(a3−c3)=2a3 (a^3 + c^3) + (a^3 - c^3) = 2a^3 :

2a3+2c3a4+a2c2+c4=2(a+c)(a2−ac+c2)(a2+ac+c2)(a2−ac+c2) \begin{aligned} &\frac{2a^3 + 2c^3}{a^4 + a^2c^2 + c^4} \\ &= \frac{2(a + c)(a^2 - ac + c^2)}{(a^2 + ac + c^2)(a^2 - ac + c^2)} \end{aligned}

2(a+c)a2+ac+c2 \dfrac{2(a + c)}{a^2 + ac + c^2}

Question 3

a.

14(1−x)−14(1+x)+2x4(1−x) \dfrac{1}{4(1 - \sqrt x)} - \dfrac{1}{4(1 + \sqrt x)} + \dfrac{2\sqrt x}{4(1 - x)}

Answer

(1−x)(1+x)=1−x (1 - \sqrt x)(1 + \sqrt x) = 1 - x . The first two:

(1+x)−(1−x)4(1−x)=2x4(1−x) \frac{(1 + \sqrt x) - (1 - \sqrt x)}{4(1 - x)} = \frac{2\sqrt x}{4(1 - x)}

Add the third: 4x4(1−x) \dfrac{4\sqrt x}{4(1 - x)} . x1−x \dfrac{\sqrt x}{1 - x}

b.

18(1−x)−18(1+x)+2x8(1−x) \dfrac{1}{8(1 - \sqrt x)} - \dfrac{1}{8(1 + \sqrt x)} + \dfrac{2\sqrt x}{8(1 - x)}

Answer

As in a, the first two give 2x8(1−x) \dfrac{2\sqrt x}{8(1 - x)} , and adding the third gives 4x8(1−x) \dfrac{4\sqrt x}{8(1 - x)} . x2(1−x) \dfrac{\sqrt x}{2(1 - x)}

c.

1(a+1)2+1(a−1)2−2a2−1 \dfrac{1}{(a + 1)^2} + \dfrac{1}{(a - 1)^2} - \dfrac{2}{a^2 - 1}

Answer

The LCM is (a+1)2(a−1)2=(a2−1)2 (a + 1)^2(a - 1)^2 = (a^2 - 1)^2 .

(a−1)2+(a+1)2−2(a2−1)(a2−1)2=2a2+2−2a2+2(a2−1)2 \begin{aligned} &\frac{(a - 1)^2 + (a + 1)^2 - 2(a^2 - 1)}{(a^2 - 1)^2} \\ &= \frac{2a^2 + 2 - 2a^2 + 2}{(a^2 - 1)^2} \end{aligned}

4(a2−1)2 \dfrac{4}{(a^2 - 1)^2}

Questions 4 and 5

4.

If a2x+1+1x+2=4x+52x2+5x+2 \dfrac{a}{2x + 1} + \dfrac{1}{x + 2} = \dfrac{4x + 5}{2x^2 + 5x + 2} , find the value of a.

Hint

2x2+5x+2=(2x+1)(x+2) 2x^2 + 5x + 2 = (2x + 1)(x + 2) : add the left side and compare the numerators.

Answer

a(x+2)+(2x+1)(2x+1)(x+2)=4x+5(2x+1)(x+2) \dfrac{a(x + 2) + (2x + 1)}{(2x + 1)(x + 2)} = \dfrac{4x + 5}{(2x + 1)(x + 2)} , so (a + 2)x + (2a + 1) = 4x + 5 for every x.
The x terms: a + 2 = 4; the numbers: 2a + 1 = 5. Both give a = 2.

5.

If a2x−3+b3x+4=x+76x2−x−12 \dfrac{a}{2x - 3} + \dfrac{b}{3x + 4} = \dfrac{x + 7}{6x^2 - x - 12} , find the values of a and b.

Answer

6x2−x−12=(2x−3)(3x+4) 6x^2 - x - 12 = (2x - 3)(3x + 4) , so a(3x + 4) + b(2x − 3) = x + 7, that is, (3a + 2b)x + (4a − 3b) = x + 7.
3a + 2b = 1 and 4a − 3b = 7. Multiply the first by 3 and the second by 2 and add: 17a = 17, so a = 1, and 2b = 1 − 3 = −2.
a = 1, b = −1. Check: 12x−3−13x+4=(3x+4)−(2x−3)6x2−x−12=x+76x2−x−12 \dfrac{1}{2x - 3} - \dfrac{1}{3x + 4} = \dfrac{(3x + 4) - (2x - 3)}{6x^2 - x - 12} = \dfrac{x + 7}{6x^2 - x - 12} ✓.