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Chapter 14

Simple Linear Equations — Exercise 14(A)

Class - 7 Concise Mathematics Selina



Exercise 14(A)

Question 1

Solve the following equation :

xx + 5 = 10

Answer

xx + 5 = 10

Transposing 5 to R.H.S.,

xx = 10 − 5

xx = 5

Hence, x\bm{x} = 5.

Question 2

Solve the following equation :

2 + yy = 7

Answer

⇒ 2 + yy = 7

Transposing 2 to R.H.S.,

yy = 7 − 2

yy = 5

Hence, y\bm{y} = 5.

Question 3

Solve the following equation :

aa − 2 = 6

Answer

aa − 2 = 6

Transposing −2 to R.H.S.,

aa = 6 + 2

aa = 8

Hence, a\bm{a} = 8.

Question 4

Solve the following equation :

xx − 5 = 8

Answer

xx − 5 = 8

Transposing −5 to R.H.S.,

xx = 8 + 5

xx = 13

Hence, x\bm{x} = 13.

Question 5

Solve the following equation :

5 − dd = 12

Answer

⇒ 5 − dd = 12

Transposing 5 to R.H.S.,

⇒ −dd = 12 − 5

⇒ −dd = 7

Multiplying both sides by −1,

⇒ (−dd) × (−1) = 7 × (−1)

dd = −7

Hence, d\bm{d} = −7.

Question 6

Solve the following equation :

3p3p = 12

Answer

3p3p = 12

Dividing both sides by 3,

3p3=123\dfrac{3p}{3} = \dfrac{12}{3}

pp = 4

Hence, p\bm{p} = 4.

Question 7

Solve the following equation :

14 = 7m7m

Answer

⇒ 14 = 7m7m

Dividing both sides by 7,

147=7m7\dfrac{14}{7} = \dfrac{7m}{7}

⇒ 2 = mm

Hence, m\bm{m} = 2.

Question 8

Solve the following equation :

2x2x = 0

Answer

2x2x = 0

Dividing both sides by 2,

2x2=02\dfrac{2x}{2} = \dfrac{0}{2}

xx = 0

Hence, x\bm{x} = 0.

Question 9

Solve the following equation :

x9=2\dfrac{x}{9} = 2

Answer

x9=2\dfrac{x}{9} = 2

Multiplying both sides by 9,

x9×9\dfrac{x}{9} \times 9 = 2 × 9

xx = 18

Hence, x\bm{x} = 18.

Question 10

Solve the following equation :

y12=4\dfrac{y}{-12} = -4

Answer

y12=4\dfrac{y}{-12} = -4

Multiplying both sides by −12,

y12×(12)\dfrac{y}{-12} \times (-12) = (−4) × (−12)

yy = 48

Hence, y\bm{y} = 48.

Question 11

Solve the following equation :

8x8x − 2 = 38

Answer

8x8x − 2 = 38

Transposing −2 to R.H.S.,

8x8x = 38 + 2

8x8x = 40

Dividing both sides by 8,

8x8=408\dfrac{8x}{8} = \dfrac{40}{8}

xx = 5

Hence, x\bm{x} = 5.

Question 12

Solve the following equation :

2x2x + 5 = 5

Answer

2x2x + 5 = 5

Transposing 5 to R.H.S.,

2x2x = 5 − 5

2x2x = 0

Dividing both sides by 2,

2x2=02\dfrac{2x}{2} = \dfrac{0}{2}

xx = 0

Hence, x\bm{x} = 0.

Question 13

Solve the following equation :

5x5x − 1 = 74

Answer

5x5x − 1 = 74

Transposing −1 to R.H.S.,

5x5x = 74 + 1

5x5x = 75

Dividing both sides by 5,

5x5=755\dfrac{5x}{5} = \dfrac{75}{5}

xx = 15

Hence, x\bm{x} = 15.

Question 14

Solve the following equation :

14 = 27 − xx

Answer

⇒ 14 = 27 − xx

Transposing −xx to L.H.S. and 14 to R.H.S.,

xx = 27 − 14

xx = 13

Hence, x\bm{x} = 13.

Question 15

Solve the following equation :

10 + 6a6a = 40

Answer

⇒ 10 + 6a6a = 40

Transposing 10 to R.H.S.,

6a6a = 40 − 10

6a6a = 30

Dividing both sides by 6,

6a6=306\dfrac{6a}{6} = \dfrac{30}{6}

aa = 5

Hence, a\bm{a} = 5.

Question 16

Solve the following equation :

c12=13c - \dfrac{1}{2} = \dfrac{1}{3}

Answer

c12=13c - \dfrac{1}{2} = \dfrac{1}{3}

Transposing 12-\dfrac{1}{2} to R.H.S.,

c=13+12c = \dfrac{1}{3} + \dfrac{1}{2}

c=2+36c = \dfrac{2 + 3}{6}

c=56c = \dfrac{5}{6}

cc = 56\dfrac{5}{6}

Hence, c\bm{c} = 56\dfrac{5}{6}.

Question 17

Solve the following equation :

a152=0\dfrac{a}{15} - 2 = 0

Answer

a152=0\dfrac{a}{15} - 2 = 0

Transposing −2 to R.H.S.,

a15=2\dfrac{a}{15} = 2

Multiplying both sides by 15,

a15×15\dfrac{a}{15} \times 15 = 2 × 15

aa = 30

Hence, a\bm{a} = 30.

Question 18

Solve the following equation :

12 = cc − 2

Answer

⇒ 12 = cc − 2

Transposing −2 to L.H.S.,

⇒ 12 + 2 = cc

⇒ 14 = cc

Hence, c\bm{c} = 14.

Question 19

Solve the following equation :

4 = xx − 2.5

Answer

⇒ 4 = xx − 2.5

Transposing −2.5 to L.H.S.,

⇒ 4 + 2.5 = xx

⇒ 6.5 = xx

Hence, x\bm{x} = 6.5.

Question 20

Solve the following equation :

y+5=814y + 5 = 8\dfrac{1}{4}

Answer

y+5=814y + 5 = 8\dfrac{1}{4}

y+5=334y + 5 = \dfrac{33}{4}

Transposing 5 to R.H.S.,

y=3345y = \dfrac{33}{4} - 5

y=33204y = \dfrac{33 - 20}{4}

y=134=314y = \dfrac{13}{4} = 3\dfrac{1}{4}

Hence, y\bm{y} = 3143\dfrac{1}{4}.

Question 21

Solve the following equation :

x+14=38x + \dfrac{1}{4} = -\dfrac{3}{8}

Answer

x+14=38x + \dfrac{1}{4} = -\dfrac{3}{8}

Transposing 14\dfrac{1}{4} to R.H.S.,

x=3814x = -\dfrac{3}{8} - \dfrac{1}{4}

x=328x = \dfrac{-3 - 2}{8}

x=58x = -\dfrac{5}{8}

Hence, x\bm{x} = 58-\dfrac{5}{8}.

Question 22

Solve the following equation :

pp + 0.02 = 0.08

Answer

pp + 0.02 = 0.08

Transposing 0.02 to R.H.S.,

pp = 0.08 − 0.02

pp = 0.06

Hence, p\bm{p} = 0.06.

Question 23

Solve the following equation :

p12=223p - 12 = 2\dfrac{2}{3}

Answer

p12=223p - 12 = 2\dfrac{2}{3}

p12=83p - 12 = \dfrac{8}{3}

Transposing −12 to R.H.S.,

p=83+12p = \dfrac{8}{3} + 12

p=8+363p = \dfrac{8 + 36}{3}

p=443=1423p = \dfrac{44}{3} = 14\dfrac{2}{3}

Hence, p\bm{p} = 142314\dfrac{2}{3}.

Question 24

Solve the following equation :

3x3x = 15

Answer

⇒ −3x3x = 15

Dividing both sides by −3,

3x3=153\dfrac{-3x}{-3} = \dfrac{15}{-3}

xx = −5

Hence, x\bm{x} = −5.

Question 25

Solve the following equation :

1.3b1.3b = 39

Answer

1.3b1.3b = 39

Dividing both sides by 1.3,

b=391.3=39013b = \dfrac{39}{1.3} = \dfrac{390}{13}

bb = 30

Hence, b\bm{b} = 30.

Question 26

Solve the following equation :

58n=20\dfrac{5}{8}n = 20

Answer

58n=20\dfrac{5}{8}n = 20

Multiplying both sides by 85\dfrac{8}{5},

58n×85=20×85\dfrac{5}{8}n \times \dfrac{8}{5} = 20 \times \dfrac{8}{5}

nn = 4 × 8 = 32

Hence, n\bm{n} = 32.

Question 27

Solve the following equation :

316m=21\dfrac{3}{16}m = 21

Answer

316m=21\dfrac{3}{16}m = 21

Multiplying both sides by 163\dfrac{16}{3},

316m×163=21×163\dfrac{3}{16}m \times \dfrac{16}{3} = 21 \times \dfrac{16}{3}

mm = 7 × 16 = 112

Hence, m\bm{m} = 112.

Question 28

Solve the following equation :

2a2a − 3 = 5

Answer

2a2a − 3 = 5

Transposing −3 to R.H.S.,

2a2a = 5 + 3

2a2a = 8

Dividing both sides by 2,

2a2=82\dfrac{2a}{2} = \dfrac{8}{2}

aa = 4

Hence, a\bm{a} = 4.

Question 29

Solve the following equation :

3p3p − 1 = 8

Answer

3p3p − 1 = 8

Transposing −1 to R.H.S.,

3p3p = 8 + 1

3p3p = 9

Dividing both sides by 3,

3p3=93\dfrac{3p}{3} = \dfrac{9}{3}

pp = 3

Hence, p\bm{p} = 3.

Question 30

Solve the following equation :

9y9y − 7 = 20

Answer

9y9y − 7 = 20

Transposing −7 to R.H.S.,

9y9y = 20 + 7

9y9y = 27

Dividing both sides by 9,

9y9=279\dfrac{9y}{9} = \dfrac{27}{9}

yy = 3

Hence, y\bm{y} = 3.

Question 31

Solve the following equation :

2b2b − 14 = 8

Answer

2b2b − 14 = 8

Transposing −14 to R.H.S.,

2b2b = 8 + 14

2b2b = 22

Dividing both sides by 2,

2b2=222\dfrac{2b}{2} = \dfrac{22}{2}

bb = 11

Hence, b\bm{b} = 11.

Question 32

Solve the following equation :

710x+6=41\dfrac{7}{10}x + 6 = 41

Answer

710x+6=41\dfrac{7}{10}x + 6 = 41

Transposing 6 to R.H.S.,

710x=416\dfrac{7}{10}x = 41 - 6

710x=35\dfrac{7}{10}x = 35

Multiplying both sides by 107\dfrac{10}{7},

710x×107=35×107\dfrac{7}{10}x \times \dfrac{10}{7} = 35 \times \dfrac{10}{7}

xx = 5 × 10 = 50

Hence, x\bm{x} = 50.

Question 33

Solve the following equation :

512m12=48\dfrac{5}{12}m - 12 = 48

Answer

512m12=48\dfrac{5}{12}m - 12 = 48

Transposing −12 to R.H.S.,

512m=48+12\dfrac{5}{12}m = 48 + 12

512m=60\dfrac{5}{12}m = 60

Multiplying both sides by 125\dfrac{12}{5},

512m×125=60×125\dfrac{5}{12}m \times \dfrac{12}{5} = 60 \times \dfrac{12}{5}

mm = 12 × 12 = 144

Hence, m\bm{m} = 144.

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