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Chapter 13

Fundamental Concepts of Algebra — Exercise 13(E)

Class - 7 Concise Mathematics Selina

Exercise 13(E)

Question 1

Simplify:

x2+x4\dfrac{x}{2} + \dfrac{x}{4}

Answer

Simplifying,

⇒x2+x4⇒2x+x4[ L.C.M. of 2 and 4 = 4 ]⇒3x4\Rightarrow \dfrac{x}{2} + \dfrac{x}{4}\\[1em] \Rightarrow \dfrac{2x + x}{4}[\text{ L.C.M. of 2 and 4 = 4 }]\\[1em] \Rightarrow \dfrac{3x}{4}

Hence, x2+x4=3x4\bm{\dfrac{x}{2} + \dfrac{x}{4} = \dfrac{3x}{4}}.

Question 2

Simplify:

a10+2a5\dfrac{a}{10} + \dfrac{2a}{5}

Answer

Simplifying,

⇒a10+2a5⇒a+4a10[ L.C.M. of 10 and 5 = 10 ]⇒5a10⇒a2\Rightarrow \dfrac{a}{10} + \dfrac{2a}{5}\\[1em] \Rightarrow \dfrac{a + 4a}{10}[\text{ L.C.M. of 10 and 5 = 10 }]\\[1em] \Rightarrow \dfrac{5a}{10}\\[1em] \Rightarrow \dfrac{a}{2}

Hence, a10+2a5=a2\bm{\dfrac{a}{10} + \dfrac{2a}{5} = \dfrac{a}{2}}.

Question 3

Simplify:

y4+3y5\dfrac{y}{4} + \dfrac{3y}{5}

Answer

Simplifying,

⇒y4+3y5⇒5y+12y20[ L.C.M. of 4 and 5 = 20 ]⇒17y20\Rightarrow \dfrac{y}{4} + \dfrac{3y}{5}\\[1em] \Rightarrow \dfrac{5y + 12y}{20}[\text{ L.C.M. of 4 and 5 = 20 }]\\[1em] \Rightarrow \dfrac{17y}{20}

Hence, y4+3y5=17y20\bm{\dfrac{y}{4} + \dfrac{3y}{5} = \dfrac{17y}{20}}.

Question 4

Simplify:

x2−x8\dfrac{x}{2} - \dfrac{x}{8}

Answer

Simplifying,

⇒x2−x8⇒4x−x8[ L.C.M. of 2 and 8 = 8 ]⇒3x8\Rightarrow \dfrac{x}{2} - \dfrac{x}{8}\\[1em] \Rightarrow \dfrac{4x - x}{8}[\text{ L.C.M. of 2 and 8 = 8 }]\\[1em] \Rightarrow \dfrac{3x}{8}

Hence, x2−x8=3x8\bm{\dfrac{x}{2} - \dfrac{x}{8} = \dfrac{3x}{8}}.

Question 5

Simplify:

3y4−y5\dfrac{3y}{4} - \dfrac{y}{5}

Answer

Simplifying,

⇒3y4−y5⇒15y−4y20[ L.C.M. of 4 and 5 = 20 ]⇒11y20\Rightarrow \dfrac{3y}{4} - \dfrac{y}{5}\\[1em] \Rightarrow \dfrac{15y - 4y}{20}[\text{ L.C.M. of 4 and 5 = 20 }]\\[1em] \Rightarrow \dfrac{11y}{20}

Hence, 3y4−y5=11y20\bm{\dfrac{3y}{4} - \dfrac{y}{5} = \dfrac{11y}{20}}.

Question 6

Simplify:

2p3−3p5\dfrac{2p}{3} - \dfrac{3p}{5}

Answer

Simplifying,

⇒2p3−3p5⇒10p−9p15[ L.C.M. of 3 and 5 = 15 ]⇒p15\Rightarrow \dfrac{2p}{3} - \dfrac{3p}{5}\\[1em] \Rightarrow \dfrac{10p - 9p}{15}[\text{ L.C.M. of 3 and 5 = 15 }]\\[1em] \Rightarrow \dfrac{p}{15}

Hence, 2p3−3p5=p15\bm{\dfrac{2p}{3} - \dfrac{3p}{5} = \dfrac{p}{15}}.

Question 7

Simplify:

k2+k3+2k5\dfrac{k}{2} + \dfrac{k}{3} + \dfrac{2k}{5}

Answer

Simplifying,

⇒k2+k3+2k5⇒15k+10k+12k30[ L.C.M. of 2, 3 and 5 = 30 ]⇒37k30\Rightarrow \dfrac{k}{2} + \dfrac{k}{3} + \dfrac{2k}{5}\\[1em] \Rightarrow \dfrac{15k + 10k + 12k}{30}[\text{ L.C.M. of 2, 3 and 5 = 30 }]\\[1em] \Rightarrow \dfrac{37k}{30}

Hence, k2+k3+2k5=37k30\bm{\dfrac{k}{2} + \dfrac{k}{3} + \dfrac{2k}{5} = \dfrac{37k}{30}}.

Question 8

Simplify:

2x5+3x4−3x5\dfrac{2x}{5} + \dfrac{3x}{4} - \dfrac{3x}{5}

Answer

Simplifying,

⇒2x5+3x4−3x5⇒8x+15x−12x20[ L.C.M. of 5, 4 and 5 = 20 ]⇒11x20\Rightarrow \dfrac{2x}{5} + \dfrac{3x}{4} - \dfrac{3x}{5}\\[1em] \Rightarrow \dfrac{8x + 15x - 12x}{20}[\text{ L.C.M. of 5, 4 and 5 = 20 }]\\[1em] \Rightarrow \dfrac{11x}{20}

Hence, 2x5+3x4−3x5=11x20\bm{\dfrac{2x}{5} + \dfrac{3x}{4} - \dfrac{3x}{5} = \dfrac{11x}{20}}.

Question 9

Simplify:

4a7−2a3+a7\dfrac{4a}{7} - \dfrac{2a}{3} + \dfrac{a}{7}

Answer

Simplifying,

⇒4a7−2a3+a7⇒12a−14a+3a21[ L.C.M. of 7, 3 and 7 = 21 ]⇒a21\Rightarrow \dfrac{4a}{7} - \dfrac{2a}{3} + \dfrac{a}{7}\\[1em] \Rightarrow \dfrac{12a - 14a + 3a}{21}[\text{ L.C.M. of 7, 3 and 7 = 21 }]\\[1em] \Rightarrow \dfrac{a}{21}

Hence, 4a7−2a3+a7=a21\bm{\dfrac{4a}{7} - \dfrac{2a}{3} + \dfrac{a}{7} = \dfrac{a}{21}}.

Question 10

Simplify:

2b5−7b15+13b3\dfrac{2b}{5} - \dfrac{7b}{15} + \dfrac{13b}{3}

Answer

Simplifying,

⇒2b5−7b15+13b3⇒6b−7b+65b15[ L.C.M. of 5, 15 and 3 = 15 ]⇒64b15\Rightarrow \dfrac{2b}{5} - \dfrac{7b}{15} + \dfrac{13b}{3}\\[1em] \Rightarrow \dfrac{6b - 7b + 65b}{15}[\text{ L.C.M. of 5, 15 and 3 = 15 }]\\[1em] \Rightarrow \dfrac{64b}{15}

Hence, 2b5−7b15+13b3=64b15\bm{\dfrac{2b}{5} - \dfrac{7b}{15} + \dfrac{13b}{3} = \dfrac{64b}{15}}.

Question 11

Simplify:

6k7−(8k9−k3)\dfrac{6k}{7} - \left(\dfrac{8k}{9} - \dfrac{k}{3}\right)

Answer

Simplifying,

⇒6k7−(8k9−k3)⇒6k7−(8k−3k9)[ Terms inside the bracket are simplified first ]⇒6k7−5k9⇒54k−35k63[ L.C.M. of 7 and 9 = 63 ]⇒19k63\Rightarrow \dfrac{6k}{7} - \left(\dfrac{8k}{9} - \dfrac{k}{3}\right)\\[1em] \Rightarrow \dfrac{6k}{7} - \left(\dfrac{8k - 3k}{9}\right)[\text{ Terms inside the bracket are simplified first }]\\[1em] \Rightarrow \dfrac{6k}{7} - \dfrac{5k}{9}\\[1em] \Rightarrow \dfrac{54k - 35k}{63}[\text{ L.C.M. of 7 and 9 = 63 }]\\[1em] \Rightarrow \dfrac{19k}{63}

Hence, 6k7−(8k9−k3)=19k63\bm{\dfrac{6k}{7} - \left(\dfrac{8k}{9} - \dfrac{k}{3}\right) = \dfrac{19k}{63}}.

Question 12

Simplify:

3a8+4a5−(a2+2a5)\dfrac{3a}{8} + \dfrac{4a}{5} - \left(\dfrac{a}{2} + \dfrac{2a}{5}\right)

Answer

Simplifying,

⇒3a8+4a5−(a2+2a5)⇒3a8+4a5−(5a+4a10)[ Terms inside the bracket are simplified first ]⇒3a8+4a5−9a10⇒15a+32a−36a40[ L.C.M. of 8, 5 and 10 = 40 ]⇒11a40\Rightarrow \dfrac{3a}{8} + \dfrac{4a}{5} - \left(\dfrac{a}{2} + \dfrac{2a}{5}\right)\\[1em] \Rightarrow \dfrac{3a}{8} + \dfrac{4a}{5} - \left(\dfrac{5a + 4a}{10}\right)[\text{ Terms inside the bracket are simplified first }]\\[1em] \Rightarrow \dfrac{3a}{8} + \dfrac{4a}{5} - \dfrac{9a}{10}\\[1em] \Rightarrow \dfrac{15a + 32a - 36a}{40}[\text{ L.C.M. of 8, 5 and 10 = 40 }]\\[1em] \Rightarrow \dfrac{11a}{40}

Hence, 3a8+4a5−(a2+2a5)=11a40\bm{\dfrac{3a}{8} + \dfrac{4a}{5} - \left(\dfrac{a}{2} + \dfrac{2a}{5}\right) = \dfrac{11a}{40}}.

Question 13

Simplify:

x+x2+x3x + \dfrac{x}{2} + \dfrac{x}{3}

Answer

Simplifying,

⇒x+x2+x3⇒6x+3x+2x6[ L.C.M. of 1, 2 and 3 = 6 ]⇒11x6\Rightarrow x + \dfrac{x}{2} + \dfrac{x}{3}\\[1em] \Rightarrow \dfrac{6x + 3x + 2x}{6}[\text{ L.C.M. of 1, 2 and 3 = 6 }]\\[1em] \Rightarrow \dfrac{11x}{6}

Hence, x+x2+x3=11x6\bm{x + \dfrac{x}{2} + \dfrac{x}{3} = \dfrac{11x}{6}}.

Question 14

Simplify:

y5+y−19y15\dfrac{y}{5} + y - \dfrac{19y}{15}

Answer

Simplifying,

⇒y5+y−19y15⇒3y+15y−19y15[ L.C.M. of 5, 1 and 15 = 15 ]⇒−y15\Rightarrow \dfrac{y}{5} + y - \dfrac{19y}{15}\\[1em] \Rightarrow \dfrac{3y + 15y - 19y}{15}[\text{ L.C.M. of 5, 1 and 15 = 15 }]\\[1em] \Rightarrow -\dfrac{y}{15}

Hence, y5+y−19y15=−y15\bm{\dfrac{y}{5} + y - \dfrac{19y}{15} = -\dfrac{y}{15}}.

Question 15

Simplify:

x5+x+12\dfrac{x}{5} + \dfrac{x + 1}{2}

Answer

Simplifying,

⇒x5+x+12⇒2x+5(x+1)10[ L.C.M. of 5 and 2 = 10 ]⇒2x+5x+510⇒7x+510\Rightarrow \dfrac{x}{5} + \dfrac{x + 1}{2}\\[1em] \Rightarrow \dfrac{2x + 5(x + 1)}{10}[\text{ L.C.M. of 5 and 2 = 10 }]\\[1em] \Rightarrow \dfrac{2x + 5x + 5}{10}\\[1em] \Rightarrow \dfrac{7x + 5}{10}

Hence, x5+x+12=7x+510\bm{\dfrac{x}{5} + \dfrac{x + 1}{2} = \dfrac{7x + 5}{10}}.

Question 16

Simplify:

x+x+23x + \dfrac{x + 2}{3}

Answer

Simplifying,

⇒x+x+23⇒3x+(x+2)3[ L.C.M. of 1 and 3 = 3 ]⇒3x+x+23⇒4x+23\Rightarrow x + \dfrac{x + 2}{3}\\[1em] \Rightarrow \dfrac{3x + (x + 2)}{3}[\text{ L.C.M. of 1 and 3 = 3 }]\\[1em] \Rightarrow \dfrac{3x + x + 2}{3}\\[1em] \Rightarrow \dfrac{4x + 2}{3}

Hence, x+x+23=4x+23\bm{x + \dfrac{x + 2}{3} = \dfrac{4x + 2}{3}}.

Question 17

Simplify:

3y5−y+22\dfrac{3y}{5} - \dfrac{y + 2}{2}

Answer

Simplifying,

⇒3y5−y+22⇒6y−5(y+2)10[ L.C.M. of 5 and 2 = 10 ]⇒6y−5y−1010⇒y−1010\Rightarrow \dfrac{3y}{5} - \dfrac{y + 2}{2}\\[1em] \Rightarrow \dfrac{6y - 5(y + 2)}{10}[\text{ L.C.M. of 5 and 2 = 10 }]\\[1em] \Rightarrow \dfrac{6y - 5y - 10}{10}\\[1em] \Rightarrow \dfrac{y - 10}{10}

Hence, 3y5−y+22=y−1010\bm{\dfrac{3y}{5} - \dfrac{y + 2}{2} = \dfrac{y - 10}{10}}.

Question 18

Simplify:

2a+13+3a−12\dfrac{2a + 1}{3} + \dfrac{3a - 1}{2}

Answer

Simplifying,

⇒2a+13+3a−12⇒2(2a+1)+3(3a−1)6[ L.C.M. of 3 and 2 = 6 ]⇒4a+2+9a−36⇒13a−16\Rightarrow \dfrac{2a + 1}{3} + \dfrac{3a - 1}{2}\\[1em] \Rightarrow \dfrac{2(2a + 1) + 3(3a - 1)}{6}[\text{ L.C.M. of 3 and 2 = 6 }]\\[1em] \Rightarrow \dfrac{4a + 2 + 9a - 3}{6}\\[1em] \Rightarrow \dfrac{13a - 1}{6}

Hence, 2a+13+3a−12=13a−16\bm{\dfrac{2a + 1}{3} + \dfrac{3a - 1}{2} = \dfrac{13a - 1}{6}}.

Question 19

Simplify:

k+12+2k−13−k+34\dfrac{k + 1}{2} + \dfrac{2k - 1}{3} - \dfrac{k + 3}{4}

Answer

Simplifying,

⇒k+12+2k−13−k+34⇒6(k+1)+4(2k−1)−3(k+3)12[ L.C.M. of 2, 3 and 4 = 12 ]⇒6k+6+8k−4−3k−912⇒11k−712\Rightarrow \dfrac{k + 1}{2} + \dfrac{2k - 1}{3} - \dfrac{k + 3}{4}\\[1em] \Rightarrow \dfrac{6(k + 1) + 4(2k - 1) - 3(k + 3)}{12}[\text{ L.C.M. of 2, 3 and 4 = 12 }]\\[1em] \Rightarrow \dfrac{6k + 6 + 8k - 4 - 3k - 9}{12}\\[1em] \Rightarrow \dfrac{11k - 7}{12}

Hence, k+12+2k−13−k+34=11k−712\bm{\dfrac{k + 1}{2} + \dfrac{2k - 1}{3} - \dfrac{k + 3}{4} = \dfrac{11k - 7}{12}}.

Question 20

Simplify:

m5−m−23+m\dfrac{m}{5} - \dfrac{m - 2}{3} + m

Answer

Simplifying,

⇒m5−m−23+m⇒3m−5(m−2)+15m15[ L.C.M. of 5, 3 and 1 = 15 ]⇒3m−5m+10+15m15⇒13m+1015\Rightarrow \dfrac{m}{5} - \dfrac{m - 2}{3} + m\\[1em] \Rightarrow \dfrac{3m - 5(m - 2) + 15m}{15}[\text{ L.C.M. of 5, 3 and 1 = 15 }]\\[1em] \Rightarrow \dfrac{3m - 5m + 10 + 15m}{15}\\[1em] \Rightarrow \dfrac{13m + 10}{15}

Hence, m5−m−23+m=13m+1015\bm{\dfrac{m}{5} - \dfrac{m - 2}{3} + m = \dfrac{13m + 10}{15}}.

Question 21

Simplify:

5(x−4)3+2(5x−3)5+6(x−4)7\dfrac{5(x - 4)}{3} + \dfrac{2(5x - 3)}{5} + \dfrac{6(x - 4)}{7}

Answer

Simplifying,

⇒5(x−4)3+2(5x−3)5+6(x−4)7⇒175(x−4)+42(5x−3)+90(x−4)105[ L.C.M. of 3, 5 and 7 = 105 ]⇒175x−700+210x−126+90x−360105⇒475x−1186105\Rightarrow \dfrac{5(x - 4)}{3} + \dfrac{2(5x - 3)}{5} + \dfrac{6(x - 4)}{7}\\[1em] \Rightarrow \dfrac{175(x - 4) + 42(5x - 3) + 90(x - 4)}{105}[\text{ L.C.M. of 3, 5 and 7 = 105 }]\\[1em] \Rightarrow \dfrac{175x - 700 + 210x - 126 + 90x - 360}{105}\\[1em] \Rightarrow \dfrac{475x - 1186}{105}

Hence, 5(x−4)3+2(5x−3)5+6(x−4)7=475x−1186105\bm{\dfrac{5(x - 4)}{3} + \dfrac{2(5x - 3)}{5} + \dfrac{6(x - 4)}{7} = \dfrac{475x - 1186}{105}}.

Question 22

Simplify:

(p+p3)(2p+p2)(3p−2p3)\left(p + \dfrac{p}{3}\right)\left(2p + \dfrac{p}{2}\right)\left(3p - \dfrac{2p}{3}\right)

Answer

Simplifying,

⇒(p+p3)(2p+p2)(3p−2p3)⇒(3p+p3)(4p+p2)(9p−2p3)⇒4p3×5p2×7p3⇒140p318⇒70p39\Rightarrow \left(p + \dfrac{p}{3}\right)\left(2p + \dfrac{p}{2}\right)\left(3p - \dfrac{2p}{3}\right)\\[1em] \Rightarrow \left(\dfrac{3p + p}{3}\right)\left(\dfrac{4p + p}{2}\right)\left(\dfrac{9p - 2p}{3}\right)\\[1em] \Rightarrow \dfrac{4p}{3} \times \dfrac{5p}{2} \times \dfrac{7p}{3}\\[1em] \Rightarrow \dfrac{140p^3}{18}\\[1em] \Rightarrow \dfrac{70p^3}{9}

Hence, (p+p3)(2p+p2)(3p−2p3)=70p39\bm{\left(p + \dfrac{p}{3}\right)\left(2p + \dfrac{p}{2}\right)\left(3p - \dfrac{2p}{3}\right) = \dfrac{70p^3}{9}}.

Question 23

Simplify:

730of(p3+7p15)\dfrac{7}{30} of \left(\dfrac{p}{3} + \dfrac{7p}{15}\right)

Answer

Simplifying,

⇒730 of (p3+7p15)⇒730×(5p+7p15)[ L.C.M. of 3 and 15 = 15 ]⇒730×12p15⇒84p450⇒14p75\Rightarrow \dfrac{7}{30} \text{ of } \left(\dfrac{p}{3} + \dfrac{7p}{15}\right)\\[1em] \Rightarrow \dfrac{7}{30} \times \left(\dfrac{5p + 7p}{15}\right)[\text{ L.C.M. of 3 and 15 = 15 }]\\[1em] \Rightarrow \dfrac{7}{30} \times \dfrac{12p}{15}\\[1em] \Rightarrow \dfrac{84p}{450}\\[1em] \Rightarrow \dfrac{14p}{75}

Hence, 730 of (p3+7p15)=14p75\bm{\dfrac{7}{30} \text{ of } \left(\dfrac{p}{3} + \dfrac{7p}{15}\right) = \dfrac{14p}{75}}.

Question 24

Simplify:

(2p+p7)÷(9p10+4p)\left(2p + \dfrac{p}{7}\right) \div \left(\dfrac{9p}{10} + 4p\right)

Answer

Simplifying,

⇒(2p+p7)÷(9p10+4p)⇒(14p+p7)÷(9p+40p10)⇒15p7÷49p10⇒15p7×1049p⇒150343\Rightarrow \left(2p + \dfrac{p}{7}\right) \div \left(\dfrac{9p}{10} + 4p\right)\\[1em] \Rightarrow \left(\dfrac{14p + p}{7}\right) \div \left(\dfrac{9p + 40p}{10}\right)\\[1em] \Rightarrow \dfrac{15p}{7} \div \dfrac{49p}{10}\\[1em] \Rightarrow \dfrac{15p}{7} \times \dfrac{10}{49p}\\[1em] \Rightarrow \dfrac{150}{343}

Hence, (2p+p7)÷(9p10+4p)=150343\bm{\left(2p + \dfrac{p}{7}\right) \div \left(\dfrac{9p}{10} + 4p\right) = \dfrac{150}{343}}.

Question 25

Simplify:

(5k8−3k5)÷k4\left(\dfrac{5k}{8} - \dfrac{3k}{5}\right) \div \dfrac{k}{4}

Answer

Simplifying,

⇒(5k8−3k5)÷k4⇒(25k−24k40)÷k4[ L.C.M. of 8 and 5 = 40 ]⇒k40×4k⇒110\Rightarrow \left(\dfrac{5k}{8} - \dfrac{3k}{5}\right) \div \dfrac{k}{4}\\[1em] \Rightarrow \left(\dfrac{25k - 24k}{40}\right) \div \dfrac{k}{4}[\text{ L.C.M. of 8 and 5 = 40 }]\\[1em] \Rightarrow \dfrac{k}{40} \times \dfrac{4}{k}\\[1em] \Rightarrow \dfrac{1}{10}

Hence, (5k8−3k5)÷k4=110\bm{\left(\dfrac{5k}{8} - \dfrac{3k}{5}\right) \div \dfrac{k}{4} = \dfrac{1}{10}}.

Question 26

Simplify:

(y6+2y3)÷(y+2y−13)\left(\dfrac{y}{6} + \dfrac{2y}{3}\right) \div \left(y + \dfrac{2y - 1}{3}\right)

Answer

Simplifying,

⇒(y6+2y3)÷(y+2y−13)⇒(y+4y6)÷(3y+2y−13)⇒5y6÷5y−13⇒5y6×35y−1⇒5y2(5y−1)=5y10y−2\Rightarrow \left(\dfrac{y}{6} + \dfrac{2y}{3}\right) \div \left(y + \dfrac{2y - 1}{3}\right)\\[1em] \Rightarrow \left(\dfrac{y + 4y}{6}\right) \div \left(\dfrac{3y + 2y - 1}{3}\right)\\[1em] \Rightarrow \dfrac{5y}{6} \div \dfrac{5y - 1}{3}\\[1em] \Rightarrow \dfrac{5y}{6} \times \dfrac{3}{5y - 1}\\[1em] \Rightarrow \dfrac{5y}{2(5y - 1)} = \dfrac{5y}{10y - 2}

Hence, (y6+2y3)÷(y+2y−13)=5y2(5y−1)\bm{\left(\dfrac{y}{6} + \dfrac{2y}{3}\right) \div \left(y + \dfrac{2y - 1}{3}\right) = \dfrac{5y}{2(5y - 1)}}.

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