**Simple Equations**

## Exercise 4.1

1). Complete the last column of the table.

Answer:

S No. | Equation | Value | Say, whether the Equation is satisfied (Yes/No) |

i | x + 3 = 0 |
x = 3 |
No |

ii | x + 3 = 0 |
x = 0 |
No |

iii | x + 3 = 0 |
x = -3 |
Yes |

iv | x – 7 = 1 |
x = 7 |
No |

v | x – 7 = 1 |
x = 8 |
Yes |

vi | 5x = 25 |
x = 0 |
No |

vii | 5x = 25 |
x = 5 |
Yes |

viii | 5x = 25 |
x = -5 |
No |

ix | /^{m}_{3} = 2 |
m = -6 |
No |

x | /^{m}_{3} = 2 |
m = 0 |
No |

xi | /^{m}_{3} = 2 |
m = 6 |
Yes |

i). *x *+ 3 = 0

put *x* = 3 in L.H.S

L.H.S = 3 + 3 = 6

L.H.S ≠ R.H.S

*x* = 3 is not the solution.

ii). *x *+ 3 = 0

put *x* = 0 in L.H.S

L.H.S = 0 + 3 = 3

L.H.S ≠ R.H.S

*x* = 0 is not the solution.

iii). *x *+ 3 = 0

put *x* = -3 in L.H.S

L.H.S = -3 + 3 = 0

L.H.S = R.H.S

*x* = -3 is the solution.

iv). *x *– 7 = 1

put *x* = 7 in L.H.S

L.H.S = 7 – 7 = 0

L.H.S ≠ R.H.S

*x* = 7 is not the solution.

v). *x *– 7 = 1

put *x* = 8 in L.H.S

L.H.S = 8 – 7 = 1

L.H.S = R.H.S

*x* = 8 is the solution.

vi). 5*x * = 25

put *x* = 0 in L.H.S

L.H.S = 5 X 0 = 0

L.H.S ≠ R.H.S

*x* = 0 is not the solution.

vii). 5*x * = 25

put *x* = 5 in L.H.S

L.H.S = 5 X 5 = 25

L.H.S = R.H.S

*x* = 5 is the solution.

viii). 5*x * = 25

put *x* = -5 in L.H.S

L.H.S = 5 X -5 = -25

L.H.S ≠ R.H.S

*x* = -5 is not the solution.

ix). * ^{m}*/

_{3}= 2

put *m* = -6 in L.H.S

L.H.S = ^{-6}/_{3} = -2

L.H.S ≠ R.H.S

*m* = -6 is not the solution.

x). * ^{m}*/

_{3}= 2

put *m* = 0 in L.H.S

L.H.S = ^{0}/_{3} = 0

L.H.S ≠ R.H.S

*m* = 0 is not the solution.

xi). * ^{m}*/

_{3}= 2

put *m* = 6 in L.H.S

L.H.S = ^{6}/_{3} = 2

L.H.S = R.H.S

*m* = 6 is the solution.

2). Check whether the value given in the brackets is a solution to the given equation or not:

(a) *n *+ 5 = 19 (*n *= 1) (b) 7*n *+ 5 = 19 (*n *= – 2)

(c) 7*n *+ 5 = 19 (*n *= 2) (d) 4*p *– 3 = 13 (*p *= 1)

(e) 4*p *– 3 = 13 (*p *= – 4) (f) 4*p *– 3 = 13 (*p *= 0)

Solutions:

(a) *n *+ 5 = 19 (*n *= 1)

L.H.S = *n* + 5 = 1 + 5 = 6

R.H.S = 19

L.H.S ≠ R.H.S

Therefore, *n *=1 is not the solution of the given equation.

(b) 7*n *+ 5 = 19 (*n *= – 2)

L.H.S = 7*n *+ 5 = 7(-2) + 5

= -14 + 5

= -9

R.H.S = 19

L.H.S ≠ R.H.S

Therefore, *n *= -2 is not the solution of the given equation.

(c) 7*n *+ 5 = 19 (*n *= 2)

L.H.S = 7*n* + 5 = 7(2) + 5

= 14 + 5

= 19

R.H.S = 19

L.H.S = R.H.S

Therefore, *n *= 2 is the solution of the given equation.

(d) 4*p *– 3 = 13 (*p *= 1)

L.H.S = 4*p *– 3

= 4(1) – 3

= 4 – 3

= 1

R.H.S = 13

L.H.S ≠ R.H.S

Therefore, *p *= 1 is not the solution of the given equation.

(e) 4*p *– 3 = 13 (*p *= – 4)

L.H.S = 4*p *– 3

= 4(-4) -3

= -16- 3

= -19

R.H.S = 13

L.H.S ≠ R.H.S

Therefore, *p *= -4 is not the solution of the given equation.

(f) 4*p *– 3 = 13 (*p *= 0)

L.H.S = 4*p *– 3

= 4(0) -3

= 0- 3

= -3

R.H.S = 13

L.H.S ≠ R.H.S

Therefore, *p *= 0 is not the solution of the given equation.

3). Solve the following equations by trial and error method:

(i) 5*p *+ 2 = 17 (ii) 3*m *– 14 = 4

Solution:

(i) 5*p *+ 2 = 17

Let *p* = 1

L.H.S = 5*p *+ 2

= 5(1) + 2

= 5 + 2

= 7

R.H.S = 17

L.H.S ≠ R.H.S

Let *p* = 2

L.H.S = 5*p *+ 2

= 5(2) + 2

= 10 + 2

= 12

R.H.S = 17

L.H.S ≠ R.H.S

Let *p* = 3

L.H.S = 5*p *+ 2

= 5(3) + 2

= 15 + 2

= 17

R.H.S = 17

L.H.S = R.H.S

*p* = 3 is the solution of the given equation.

(ii) 3*m *– 14 = 4

Let *m* = 1

L.H.S = 3*m* – 14

= 3(1) – 14

= 3 – 14

= -11

R.H.S = 4

L.H.S ≠ R.H.S

Let *m* = 2

L.H.S = 3*m* – 14

= 3(2) – 14

= 6 – 14

= -8

R.H.S = 4

L.H.S ≠ R.H.S

Let *m* = 3

L.H.S = 3*m* – 14

= 3(3) – 14

= 9 – 14

= -5

R.H.S = 4

L.H.S ≠ R.H.S

Let *m* = 4

L.H.S = 3*m* – 14

= 3(4) – 14

= 12 – 14

= -2

R.H.S = 4

L.H.S ≠ R.H.S

Let *m* = 5

L.H.S = 3*m* – 14

= 3(5) – 14

= 15 – 14

= 1

R.H.S = 4

L.H.S ≠ R.H.S

Let *m* = 6

L.H.S = 3*m* – 14

= 3(6) – 14

= 18 – 14

= 4

R.H.S = 4

L.H.S = R.H.S

*m* = 6 is the solution of the given equation.

4). Write equations for the following statements:

(i) The sum of numbers *x *and 4 is 9.

*x* + 4 = 9

(ii) 2 subtracted from *y *is 8.

*y* – 2 = 8

(iii) Ten times *a *is 70.

10*a* = 70

(iv) The number *b *divided by 5 gives 6.

* ^{b}*/

_{5}= 6

(v) Three-fourth of *t *is 15.

¾ X *t* = 15

(vi) Seven times *m *plus 7 gets you 77.

7*m* + 7 = 77

(vii) One-fourth of a number *x *minus 4 gives 4.

¼ X *x* -4 = 4

(viii) If you take away 6 from 6 times *y*, you get 60.

6*y* – 6 = 60

(ix) If you add 3 to one-third of *z*, you get 30.

^{1}/_{3} X *z* + 3 = 30

5). Write the following equations in statement forms:

(i) *p *+ 4 = 15 (ii) *m *– 7 = 3 (iii) 2*m *= 7 (iv)^{m}*/*_{5} = 3

(v)^{3}^{m}*/*_{5}= 6 (vi) 3*p *+ 4 = 25 (vii) 4*p *– 2 = 18

(viii)^{p}*/*_{2 }+ 2 = 8

Answer:

(i) *p *+ 4 = 15

The Sum of a number *p* and 4 is 15.

(ii) *m *– 7 = 3

7 subtracted from a number *m* gives 3.

(iii) 2*m *= 7

Two times *m* is 7.

(iv)^{m}*/*_{5} = 3

*m* divided by 5 is 3.

(v)^{3}^{m}*/*_{5}= 6

3 times of *m* divided by 5 gives 6.

(vi) 3*p *+ 4 = 25

The sum of 3 times of a number *p* and 4 is 25.

(vii) 4*p *– 2 = 18

The difference of 4 times of a *p* and 2 is 18.

(viii)^{p}*/*_{2 }+ 2 = 8

The sum of half of *p* and 2 gives 8.

6). Set up an equation in the following cases:

(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. (Take *m *to be the number of Parmit’s marbles.)

Number of marbles with Irfan = 37

Let the number of marbles with Parmit be *m*

The required equation is

5*m* + 7 = 37

(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be *y *years.)

Let the age of Laxmi be *y* years

Age of Laxmi’s father = 49

3*y* + 4 = 49

(iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be *l*.)

Let the lowest score be l

Highest score = 87

2*l* + 7 = 87

(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be *b *in degrees. Remember that the sum of angles of a triangle is 180 degrees).

Let the base angle be b in degrees

Base angles of isosceles triangle are congruent.

Vertex angle = 2b

Sum of the three angles of a triangle is 180^{0}.

required equation is 2b + b + b = 180

4b = 180.

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