Simple equations are mathematical statements that equate two expressions. They involve variables (letters that represent unknown values), constants (fixed numerical values), and mathematical operations (addition, subtraction, multiplication, division).
Key Concepts:
- Equation: A mathematical statement with an equal sign (=) showing that two expressions are equal.
- Variable: A letter used to represent an unknown value.
- Constant: A fixed numerical value.
- Solution: The value of the variable that makes the equation true.
Solving Simple Equations:
To solve an equation, the goal is to isolate the variable on one side of the equation. This is achieved by performing the same operation on both sides of the equation.
Common operations:
- Addition
- Subtraction
- Multiplication
- Division
Real-life Applications:
Simple equations are used to solve various real-life problems, such as:
- Finding unknown quantities
- Solving age-related problems
- Determining costs
- Calculating distances and speeds
By understanding simple equations, you can develop problem-solving skills and apply them to various mathematical and real-world scenarios.
Exercise 4.1
1. Complete the given column of the table:
Ans :
| 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 |
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) 7n + 5 = 19; in – -2)
(c) 7n + 5 = 19; (n = 2)
(d) 4p – 3 = 13; (p = 1)
(e) 4p – 3 = 13; (p = -4)
(f) 4p-3 = 13; (p = 0)
Ans :
(a) n + 5 = 19; (n = 1)
- Substituting n = 1, we get: 1 + 5 = 19
- This is false (6 ≠ 19), so n = 1 is not a solution.
(b) 7n + 5 = 19; (n = -2)
- Substituting n = -2, we get: 7(-2) + 5 = 19
- This simplifies to -14 + 5 = 19, which is false (-9 ≠ 19), so n = -2 is not a solution.
(c) 7n + 5 = 19; (n = 2)
- Substituting n = 2, we get: 7(2) + 5 = 19
- This simplifies to 14 + 5 = 19, which is true (19 = 19), so n = 2 is a solution.
(d) 4p – 3 = 13; (p = 1)
- Substituting p = 1, we get: 4(1) – 3 = 13
- This simplifies to 4 – 3 = 13, which is false (1 ≠ 13), so p = 1 is not a solution.
(e) 4p – 3 = 13; (p = -4)
- Substituting p = -4, we get: 4(-4) – 3 = 13
- This simplifies to -16 – 3 = 13, which is false (-19 ≠ 13), so p = -4 is not a solution.
(f) 4p – 3 = 13; (p = 0)
- Substituting p = 0, we get: 4(0) – 3 = 13
- This simplifies to -3 = 13, which is false, so p = 0 is not a solution.
3. Solve the following equations by trial and error method:
(i) 5p + 2 = 17
(ii) 3m – 14 = 4
Ans :
(i) 5p + 2 = 17
Let’s try different values for p:
- If p = 1, then 5(1) + 2 = 7 ≠ 17
- If p = 2, then 5(2) + 2 = 12 ≠ 17
- If p = 3, then 5(3) + 2 = 17
So, p = 3 is the solution.
(ii) 3m – 14 = 4
Let’s try different values for m:
- If m = 3, then 3(3) – 14 = -5 ≠ 4
- If m = 5, then 3(5) – 14 = 1 ≠ 4
- If m = 6, then 3(6) – 14 = 4
So, m = 6 is the solution.
4. Write equations for the following statements:
(i) The sum of numbers x and 4 is 9.
(ii) 2 subtracted from y is 8.
(iii) Ten times a is 70.
(iv) The number b divided by 5 gives 6.
(v) Three-fourth of t is 15.
(vi) Seven times m plus 7 gets you 77.
(vii) One-fourth of a number x minus 4 gives 4.
(viii) If you take away 6 from 6 times y, you get 60.
(ix) If you add 3 to one-third of z, you get 30.
Ans :
(i) The sum of numbers x and 4 is 9.
- Equation: x + 4 = 9
(ii) 2 subtracted from y is 8.
- Equation: y – 2 = 8
(iii) Ten times a is 70.
- Equation: 10a = 70
(iv) The number b divided by 5 gives 6.
- Equation: b / 5 = 6
(v) Three-fourth of t is 15.
- Equation: (3/4)t = 15
(vi) Seven times m plus 7 gets you 77.
- Equation: 7m + 7 = 77
(vii) One-fourth of a number x minus 4 gives 4.
- Equation: (1/4)x – 4 = 4
(viii) If you take away 6 from 6 times y, you get 60.
- Equation: 6y – 6 = 60
(ix) If you add 3 to one-third of z, you get 30.
- Equation: (1/3)z + 3 = 30
5. Write the following equations in statement forms.
Ans :
i) The sum of p and 4 is 15.
(ii) When 7 is subtracted from m, the result is 3.
(iii) Twice the value of m is equal to 7.
(iv) When m is divided by 5, the result is 3.
(v) Three times m, divided by 5, equals 6.
(vi) Three times p, plus 4, equals 25.
(vii) Four times p, minus 2, equals 18.
(viii) When p is divided by 2 and then added to 2, the result is 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)
(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)
(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 1)
(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).
Ans :
(i) Equation: 5m + 7 = 37
(ii) Equation: 3y + 4 = 49
(iii) Equation: 2l + 7 = 87
(iv) Equation: b + b + 2b = 180
Exercise 4.2
1. Given first the step you will use to separate the variable and then solve the equation:
(a) x – 1 = 0
(b) x + 1 = 0
(c) x – 1 = 5
(d) x + 6 = 2
(e) y – 4 = -7
(f) y -4 = 4
(g) y + 4 = 4
(h) y + 4 = -4
Ans :
(a) x – 1 = 0
- Step: Add 1 to both sides to isolate x.
- Solution: x – 1 + 1 = 0 + 1 => x = 1
(b) x + 1 = 0
- Step: Subtract 1 from both sides to isolate x.
- Solution: x + 1 – 1 = 0 – 1 => x = -1
(c) x – 1 = 5
- Step: Add 1 to both sides to isolate x.
- Solution: x – 1 + 1 = 5 + 1 => x = 6
(d) x + 6 = 2
- Step: Subtract 6 from both sides to isolate x.
- Solution: x + 6 – 6 = 2 – 6 => x = -4
(e) y – 4 = -7
- Step: Add 4 to both sides to isolate y.
- Solution: y – 4 + 4 = -7 + 4 => y = -3
(f) y – 4 = 4
- Step: Add 4 to both sides to isolate y.
- Solution: y – 4 + 4 = 4 + 4 => y = 8
(g) y + 4 = 4
- Step: Subtract 4 from both sides to isolate y.
- Solution: y + 4 – 4 = 4 – 4 => y = 0
(h) y + 4 = -4
- Step: Subtract 4 from both sides to isolate y.
- Solution: y + 4 – 4 = -4 – 4 => y = -8
2. Give first the step you will use to separate the variable and then solve the following equation:
Ans :
(a) 3l = 42
- Step: Divide both sides by 3 to isolate l.
- Solution: 3l / 3 = 42 / 3 => l = 14
(b) b/2 = 6
- Step: Multiply both sides by 2 to isolate b.
- Solution: (b/2) * 2 = 6 * 2 => b = 12
(c) p/7 = 4
- Step: Multiply both sides by 7 to isolate p.
- Solution: (p/7) * 7 = 4 * 7 => p = 28
(d) 4x = 25
- Step: Divide both sides by 4 to isolate x.
- Solution: 4x / 4 = 25 / 4 => x = 6.25
(e) 8y = 36
- Step: Divide both sides by 8 to isolate y.
- Solution: 8y / 8 = 36 / 8 => y = 4.5
(f) z/3 = 5/4
- Step: Multiply both sides by 3 to isolate z.
- Solution: (z/3) * 3 = (5/4) * 3 => z = 15/4
(g) a/5 = 7/15
- Step: Multiply both sides by 5 to isolate a.
- Solution: (a/5) * 5 = (7/15) * 5 => a = 7/3
(h) 20t = -10
- Step: Divide both sides by 20 to isolate t.
- Solution: 20t / 20 = -10 / 20 => t = -0.5
3. Give the steps you will use to separate the variables and then solve the equation:
(a) 3n – 2 = 46
(b) 5m + 7 = 17
(c) 20p/3=40
(d) 3p/10=6
Ans :
(a) 3n – 2 = 46
- Step 1: Add 2 to both sides to isolate 3n.
- Step 2: Divide both sides by 3 to isolate n.
Solution:
- 3n – 2 + 2 = 46 + 2
- 3n = 48
- n = 48 / 3
- n = 16
(b) 5m + 7 = 17
- Step 1: Subtract 7 from both sides to isolate 5m.
- Step 2: Divide both sides by 5 to isolate m.
Solution:
- 5m + 7 – 7 = 17 – 7
- 5m = 10
- m = 10 / 5
- m = 2
(c) 20p/3 = 40
- Step 1: Multiply both sides by 3 to isolate 20p.
- Step 2: Divide both sides by 20 to isolate p.
Solution:
- (20p/3) * 3 = 40 * 3
- 20p = 120
- p = 120 / 20
- p = 6
(d) 3p/10 = 6
- Step 1: Multiply both sides by 10 to isolate 3p.
- Step 2: Divide both sides by 3 to isolate p.
Solution:
- (3p/10) * 10 = 6 * 10
- 3p = 60
- p = 60 / 3
- p = 20
4. Solve the following equations:
(a) 10p = 100
(b) 10p + 10 = 100
(c) p/4=5
(d) −p/3=5
(e) 3p/4=6
(f) 3s = -9
(g) 3s + 12 = 0
(h) 3s = 0
(i) 2q = 6
(j) 2q – 6 = 0
(k) 2q + 6 = 0
(l) 2q + 6 = 12
Ans :
(a) 10p = 100
Step: Divide both sides by 10 to isolate p.
Solution: 10p / 10 = 100 / 10 => p = 10
(b) 10p + 10 = 100
Step 1: Subtract 10 from both sides to isolate 10p.
Step 2: Divide both sides by 10 to isolate p.
Solution: 10p + 10 – 10 = 100 – 10 => 10p = 90 => p = 90 / 10 => p = 9
(c) p/4 = 5
Step: Multiply both sides by 4 to isolate p.
Solution: (p/4) * 4 = 5 * 4 => p = 20
(d) -p/3 = 5
Step 1: Multiply both sides by -3 to isolate p.
Solution: (-p/3) * (-3) = 5 * (-3) => p = -15
(e) 3p/4 = 6
Step 1: Multiply both sides by 4 to isolate 3p.
Step 2: Divide both sides by 3 to isolate p.
Solution: (3p/4) * 4 = 6 * 4 => 3p = 24 => p = 24 / 3 => p = 8
(f) 3s = -9
Step: Divide both sides by 3 to isolate s.
Solution: 3s / 3 = -9 / 3 => s = -3
(g) 3s + 12 = 0
Step 1: Subtract 12 from both sides to isolate 3s.
Step 2: Divide both sides by 3 to isolate s.
Solution: 3s + 12 – 12 = 0 – 12 => 3s = -12 => s = -12 / 3 => s = -4
(h) 3s = 0
Step: Divide both sides by 3 to isolate s.
Solution: 3s / 3 = 0 / 3 => s = 0
(i) 2q = 6
Step: Divide both sides by 2 to isolate q.
Solution: 2q / 2 = 6 / 2 => q = 3
(j) 2q – 6 = 0
Step 1: Add 6 to both sides to isolate 2q.
Step 2: Divide both sides by 2 to isolate q.
Solution: 2q – 6 + 6 = 0 + 6 => 2q = 6 => q = 6 / 2 => q = 3
(k) 2q + 6 = 0
Step 1: Subtract 6 from both sides to isolate 2q.
Step 2: Divide both sides by 2 to isolate q.
Solution: 2q + 6 – 6 = 0 – 6 => 2q = -6 => q = -6 / 2 => q = -3
(l) 2q + 6 = 12
Step 1: Subtract 6 from both sides to isolate 2q.
Step 2: Divide both sides by 2 to isolate q.
Solution: 2q + 6 – 6 = 12 – 6 => 2q = 6 => q = 6 / 2 => q = 3
Exercise 4.3
1. Set up equations and solve them to find the unknown numbers in the following cases:
(a) Add 4 to eight times a number; you get 60.
(b) One-fifth of a number minus 4 gives 3.
(c) If I take three-fourths of a number and add 3 to it, I get 21.
(d) When I subtracted 11 from twice a number, the result was 15.
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
(g) Anwar thinks of a number. If he takes away 7 from 5/2 of the numbers, the result is 23.
Ans :
(a) Add 4 to eight times a number; you get 60.
Let the number be x. Equation: 8x + 4 = 60
- Solving: 8x = 60 – 4 8x = 56 x = 7
(b) One-fifth of a number minus 4 gives 3.
Let the number be y. Equation: (y/5) – 4 = 3
- Solving: (y/5) = 3 + 4 y/5 = 7 y = 35
(c) If I take three-fourths of a number and add 3 to it, I get 21.
Let the number be z. Equation: (3/4)z + 3 = 21
- Solving: (3/4)z = 21 – 3 (3/4)z = 18 z = 18 * (4/3) z = 24
(d) When I subtracted 11 from twice a number, the result was 15.
Let the number be p. Equation: 2p – 11 = 15
- Solving: 2p = 15 + 11 2p = 26 p = 13
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
Let the number of notebooks be n. Equation: 50 – 3n = 8
- Solving: -3n = 8 – 50 -3n = -42 n = 14
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
Let the number be m. Equation: (m + 19) / 5 = 8
- Solving: m + 19 = 8 * 5 m + 19 = 40 m = 40 – 19 m = 21
(g) Anwar thinks of a number. If he takes away 7 from 5/2 of the numbers, the result is 23.
Let the number be x. Equation: (5/2)x – 7 = 23
- Solving: (5/2)x = 23 + 7 (5/2)x = 30 x = 30 * (2/5) x = 12
2. Solve the following:
(a) 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. What is the lowest score?
(b) In an isosceles triangle, the base angle are equal. The vertex angle is 40°. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°?)
(c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?
Ans :
(a) 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. What is the lowest score?
Let the lowest score be x.
According to the problem, the highest score is 2x + 7.
Given, the highest score is 87.
So, 2x + 7 = 87
2x = 87 – 7
2x = 80
x = 40
Therefore, the lowest score is 40.
(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle? (Remember, the sum of angles of a triangle is 180°?)
Let the base angle be y.
Since it’s an isosceles triangle, both base angles are equal.
So, the sum of all angles is y + y + 40° = 180°
2y + 40° = 180°
2y = 180° – 40°
2y = 140°
y = 70°
Therefore, each base angle is 70°.
(c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?
Let Rahul’s score be x.
Sachin’s score is 2x.
Together, they scored x + 2x = 3x runs.
Given, they scored 2 runs less than a double century (200).
So, 3x = 200 – 2
3x = 198 x
= 66
Therefore, Rahul scored 66 runs and Sachin scored 2 * 66 = 132 runs.
3. Solve the following:
(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?
(iii) People of Sundargram planted trees in a village garden. Some of the trees were fruit trees. The number of non-fruit trees were 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?
Ans :
(i)
Let the number of marbles Parmit has be ‘x’.
So, Irfan has 5x + 7 marbles.
Given, Irfan has 37 marbles.
Therefore, 5x + 7 = 37
5x = 37 – 7
5x = 30
x = 30 / 5
x = 6
So, Parmit has 6 marbles.
(ii)
Let Laxmi’s age be ‘y’ years.
Her father’s age is 3y + 4 years.
Given, her father’s age is 49 years.
Therefore, 3y + 4 = 49
3y = 49 – 4
3y = 45
y = 45 / 3
y = 15
So, Laxmi’s age is 15 years.
(iii)
Let the number of fruit trees be ‘z’.
Number of non-fruit trees = 3z + 2
Given, number of non-fruit trees = 77
Therefore, 3z + 2 = 77
3z = 77 – 2
3z = 75
z = 75 / 3
z = 25
So, the number of fruit trees planted was 25.
4. Solve the following riddle:
I am a number,
Tell my identity!
Take me seven times over
And add a fifty!
To reach a triple century
You still need forty!
Ans :
Understanding the Riddle:
- Let’s assume the number is ‘x’.
- Seven times the number is 7x.
- Adding 50 to it gives 7x + 50.
- This value is 40 less than a triple century (300).
Setting up the equation: 7x + 50 = 300 – 40
Solving the equation: 7x + 50 = 260 7x = 260 – 50 7x = 210 x = 210 / 7 x = 30
So, the number is 30.
