**Q 1) Say whether the equation is satisfied.**

**Equation: x + 3 = 0, Value: x = 3**

**Answer: **x + 3 = 0

Substituting x = 3 in above equation

L.H.S. = x + 3

= 3 + 3

= 6

R.H.S. = 0

L.H.S. ≠ R.H.S.

Therefore, equation is not satisfied.

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

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

**Answer: **n + 5 = 19

Substituting n = 1 in above equation

L.H.S. = n + 5

= 1 + 5

= 6

R.H.S. = 19

L.H.S. ≠ R.H.S.

Therefore, n = 1 is not a solution to the given equation.

**(ii) 7n + 5 = 19 (n = – 2)**

**Answer:** 7n + 5 = 19

Substituting n = -2 in above equation

L.H.S. = 7n + 5

= 7 x (-2) + 5

= (-14) + 5

= (-9)

R.H.S. = 19

L.H.S. ≠ R.H.S.

Therefore, n = -2 is not a solution to the given equation.

**Q 3) Solve the following equation by trial and error method.**

**5p + 2 = 17**

**Answer:**

5p + 2 = 17

Substituting p = 1 in above equation

L.H.S. = 5p + 2

= (5 x 1) + 2

= 5 + 2

= 7

R.H.S. = 17

L.H.S. ≠ R.H.S.

Therefore, p = 1 is not a solution to the given equation.

Substituting p = 2 in above equation

L.H.S. = 5p + 2

= (5 x 2) + 2

= 10 + 2

= 12

R.H.S. = 17

L.H.S. ≠ R.H.S.

Therefore, p = 2 is not a solution to the given equation.

Substituting p = 3 in above equation

L.H.S. = 5p + 2

= (5 x 3) + 2

= 15 + 2

= 17

R.H.S. = 17

L.H.S. = R.H.S.

Therefore, p = 3 is a solution to the given equation.

**Q 4) Write equations for the following statements.**

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

**Answer: **x + 4 = 9

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

**Answer:** b/5 = 6

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

**Answer:** 7m + 7 = 77

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

**Answer:** (x/4) – 4 = 4

**Q 5) Write the following equations in statement forms.**

**(i) p + 4 = 15**

**Answer: **The sum of numbers p and 4 is 15.

**(ii) 3p + 4 = 25**

**Answer:** Three times p plus 4 gives you 25.

**(iii) (p/2)** **+ 2 = 8**

**Answer:** If you add half of a number p to 2, you get 8.

**(iv) 4p – 2 = 18**

**Answer:** Four times p minus 2 gives you 18.

**Q 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).**

**Answer:**

__Given__

Irfan’s marbles = 37

__To Find__

Parmit’s marbles = ?

__Condition__

Irfan has 7 marbles more than five times Parmit’s marbles

__Solution__

Let Parmit’s marbles be m

Therefore, Irfan’s marbles = 5m + 7

But it is given that Irfan’s marbles = 37

Therefore, 5m + 7 = 37

**(ii) 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).**

**Answer:**

__Given__

Sum of angles of a triangle = 180^{o}

__To Find__

Measure of the base angles = ?

__Condition__

In an isosceles triangle, the vertex angle is twice either base angle

__Solution__

Let base angle be b degrees

Therefore, Vertex angle = 2b

As sum of angles of a triangle = 180^{o}

2b + b + b = 180^{o}

Therefore, 4b = 180^{o}

**Q 7) Give first the step you will use to separate the variable and then solve the equation.**

**(i) x – 1 = 0**

**Answer: **x – 1 = 0

Adding 1 to both sides of the equation

x – 1 + 1 = 0 + 1

x = 1

**(ii) y – 4 = -7**

**Answer: **y – 4 = -7

Adding 4 to both sides of the equation

y – 4 + 4 = -7 + 4

y = -3

**(iii) b/2**** ****= 6**

**Answer: ** = 6

Multiplying both sides of the equation by 2

(b/2) x 2 = 6 x 2

b = 12

**Q 8) Solve the following equations.**

**(i) 2q + 6 = 12**

**Answer: **2q + 6 = 12

Transpose 6 from LHS to RHS

2q = 12 – 6

2q = 6

Dividing both sides by 2

q = 6/2

q = 3

**(ii) 2q – 6 = 0**

**Answer: **2q – 6 = 0

Transpose (-6) from LHS to RHS

2q = 0 + 6

2q = 6

Dividing both sides by 2

q = 6/2

q = 3

**(iii) 4 + 5(p – 1) =34**

**Answer: **4 + 5(p – 1) =34

4 + (5 x p) – (5 x 1) = 34

4 + 5p – 5 = 34

5p – 1 = 34

Transpose (-1) from LHS to RHS

5p = 34 + 1

5p = 35

Dividing both sides by 5

p = 35/5

p = 7

**Q 9) Solve the following.**

**(i) The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus 7. The highest score is 87. What is the lowest score?**

**Answer:**

__Given__

Highest score in class = 87

__To Find__

Lowest score = ?

__Condition__

Highest score is twice the lowest score plus 7

__Solution__

Let lowest score be x

Therefore, highest score = 2x + 7

But it is given that highest score = 87

Therefore, 2x + 7 = 87

Transpose 7 from LHS to RHS

2x = 87 – 7

2x = 80

Dividing both sides by 2

x = 80/2

x = 40

Therefore, lowest score in class is 40 marks.

**(ii) People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees was two more than three times the number of fruit trees. What was the number of fruit trees planted if the number of non-fruit trees planted was 77?**

**Answer:**

__Given__

Non-fruit trees = 77

__To Find__

Fruit trees = ?

__Condition__

Number of non-fruit trees was two more than three times the number of fruit trees

__Solution__

Let fruit trees be x

Therefore, non-fruit trees = 3x + 2

But it is given that non-fruit trees = 77

Therefore, 3x + 2 = 77

Transpose 2 from LHS to RHS

3x = 77 – 2

3x = 75

Dividing both sides by 3

x = 75/3

x = 25

Therefore, number of fruit trees is 25.

**Practice Questions**

**If one-third of a number exceeds its one-fourth by 1, find the number.****The length of a rectangle is twice its breadth. If its perimeter is 60 cm, find the length and the breadth of the rectangle.****Seven times a number is 12 less than thirteen times the same number. Find the number.****The present age of a son is half the present age of his father. Ten years ago, the father was thrice as old as his son. What are their present age?****The sum of three consecutive multiples of 2 is 18. Find the three multiples.****Each of the 2 equal sides of an isosceles triangle is twice as large as the third side. If the perimeter of the triangle is 30 cm, find the length of each side of the triangle.**