GSEB Class 7 Maths Solutions Chapter 4 Simple Equations Exercise 4.2

Question 1. Give 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) \( y + 4 = -4 \)
Answer:
(a) \( x - 1 = 0 \)
The first step involves adding 1 to both sides of the equation.
Adding 1 to both sides, we get: \( x - 1 + 1 = 0 + 1 \)
\( x = 1 \) To check, substitute \( x = 1 \) into the left-hand side (L.H.S.): L.H.S. \( = x - 1 = 1 - 1 = 0 \) This equals the right-hand side (R.H.S.). Therefore, \( x = 1 \) is the correct solution.
(b) \( x + 1 = 0 \)
The first step involves subtracting 1 from both sides of the equation.
Subtracting 1 from both sides, we get: \( x + 1 - 1 = 0 - 1 \)
\( x = -1 \) To check, substitute \( x = -1 \) into the L.H.S.: L.H.S. \( = x + 1 = (-1) + 1 = 0 \) This equals the R.H.S. Therefore, the solution \( x = -1 \) is correct.
(c) \( x - 1 = 5 \)
The first step involves adding 1 to both sides of the equation.
Adding 1 to both sides, we get: \( x - 1 + 1 = 5 + 1 \)
\( x = 6 \) To check, substitute \( x = 6 \) into the L.H.S.: L.H.S. \( = x - 1 = 6 - 1 = 5 \) This equals the R.H.S. Therefore, the solution \( x = 6 \) is correct.
(d) \( x + 6 = 2 \)
The first step involves subtracting 6 from both sides of the equation.
Subtracting 6 from both sides, we get: \( x + 6 - 6 = 2 - 6 \)
\( x = -4 \) To check, substitute \( x = -4 \) into the L.H.S.: L.H.S. \( = x + 6 = -4 + 6 = 2 \) This equals the R.H.S. Therefore, the solution \( x = -4 \) is correct.
(e) \( y - 4 = - 7 \)
The first step involves adding 4 to both sides of the equation.
Adding 4 to both sides, we get: \( y - 4 + 4 = -7 + 4 \)
\( y = -3 \) To check, substitute \( y = -3 \) into the L.H.S.: L.H.S. \( = y - 4 = -3 - 4 = -7 \) This equals the R.H.S. Therefore, the solution \( y = -3 \) is correct.
(f) \( y + 4 = 4 \)
The first step involves subtracting 4 from both sides of the equation.
Subtracting 4 from both sides, we get: \( y + 4 - 4 = 4 - 4 \)
\( y = 0 \) To check, substitute \( y = 0 \) into the L.H.S.: L.H.S. \( = y + 4 = 0 + 4 = 4 \) This equals the R.H.S. Therefore, the solution \( y = 0 \) is correct.
(g) \( y + 4 = 4 \)
The first step involves subtracting 4 from both sides of the equation.
Subtracting 4 from both sides, we get: \( y + 4 - 4 = 4 - 4 \)
\( y = 0 \) To check, substitute \( y = 0 \) into the L.H.S.: L.H.S. \( = y + 4 = 0 + 4 = 4 \) This equals the R.H.S. Therefore, the solution \( y = 0 \) is correct.
(h) \( y + 4 = - 4 \)
The first step involves subtracting 4 from both sides of the equation.
Subtracting 4 from both sides, we get: \( y + 4 - 4 = -4 - 4 \)
\( y = -8 \) To check, substitute \( y = -8 \) into the L.H.S.: L.H.S. \( = y + 4 = -8 + 4 = -4 \) This equals the R.H.S. Therefore, the solution \( y = -8 \) is correct.In simple words: To solve these simple equations, we always do the opposite operation on both sides to isolate the variable. For example, if you have a minus, you add; if you have a plus, you subtract.

Exam Tip: Always remember the golden rule of equations: whatever operation you perform on one side, you must perform the exact same operation on the other side to keep the equation balanced.

Question 2. Give first the step you will use to separate the variable and then solve the equation:
(a) \( 3l = 42 \)
(b) \( \frac {b}{ 2 } = 6 \)
(c) \( \frac { p }{ 7 } = 4 \)
(d) \( 4x = 25 \)
(e) \( 8y = 36 \)
(f) \( \frac { z }{ 3 } = \frac { 5 }{ 4 } \)
(g) \( \frac { a }{ 5 } = \frac { 7 }{ 15 } \)
(h) \( 20t = -10 \)
Answer:
(a) \( 3l = 42 \)
The first step involves dividing both sides by 3.
Dividing both sides by 3, we get: \( \frac { 3l }{ 3 } = \frac { 42 }{ 3 } \)
\( l = 14 \) To check, substitute \( l = 14 \) into the L.H.S.: L.H.S. \( = 3l = 3 \times 14 = 42 \) This equals the R.H.S. Therefore, \( l = 14 \) is the correct solution.
(b) \( \frac {b}{ 2 } = 6 \)
The first step involves multiplying both sides by 2.
Multiplying both sides by 2, we get: \( \frac { b }{ 2 } \times 2 = 6 \times 2 \)
\( b = 12 \) To check, substitute \( b = 12 \) into the L.H.S.: L.H.S. \( = \frac { b }{ 2 } = \frac { 12 }{ 2 } = 6 \) This equals the R.H.S. Therefore, \( b = 12 \) is the correct solution.
(c) \( \frac { p }{ 7 } = 4 \)
The first step involves multiplying both sides by 7.
Multiplying both sides by 7, we get: \( \frac { p }{ 7 } \times 7 = 4 \times 7 \)
\( p = 28 \) To check, substitute \( p = 28 \) into the L.H.S.: L.H.S. \( = \frac { p }{ 7 } = \frac { 28 }{ 7 } = 4 \) This equals the R.H.S. Therefore, \( p = 28 \) is the correct solution.
(d) \( 4x = 25 \)
The first step involves dividing both sides by 4.
Dividing both sides by 4, we get: \( \frac { 4x }{ 4 } = \frac { 25 }{ 4 } \)
\( x = \frac { 25 }{ 4 } \) To check, substitute \( x = \frac { 25 }{ 4 } \) into the L.H.S.: L.H.S. \( = 4x = 4 \times \frac { 25 }{ 4 } = 25 \) This equals the R.H.S. Therefore, \( x = \frac { 25 }{ 4 } \) is the correct solution.
(e) \( 8y = 36 \)
The first step involves dividing both sides by 8.
Dividing both sides by 8, we get: \( \frac { 8y }{ 8 } = \frac { 36 }{ 8 } \)
\( y = \frac { 9 }{ 2 } \) To check, substitute \( y = \frac { 9 }{ 2 } \) into the L.H.S.: L.H.S. \( = 8y = 8 \times \frac { 9 }{ 2 } = 36 \) This equals the R.H.S. Therefore, \( y = \frac { 9 }{ 2 } \) is the correct solution.
(f) \( \frac { z }{ 3 } = \frac { 5 }{ 4 } \)
The first step involves multiplying both sides by 3.
Multiplying both sides by 3, we get: \( \frac { z }{ 3 } \times 3 = \frac { 5 }{ 4 } \times 3 \)
\( z = \frac { 15 }{ 4 } \) To check, substitute \( z = \frac { 15 }{ 4 } \) into the L.H.S.: L.H.S. \( = \frac { z }{ 3 } = \frac { 15 }{ 4 } \times \frac { 1 }{ 3 } = \frac { 5 }{ 4 } \) This equals the R.H.S. Therefore, \( z = \frac { 15 }{ 4 } \) is the correct solution.
(g) \( \frac { a }{ 5 } = \frac { 7 }{ 15 } \)
The first step involves multiplying both sides by 5.
Multiplying both sides by 5, we get: \( \frac { a }{ 5 } \times 5 = \frac { 7 }{ 15 } \times 5 \)
\( a = \frac { 7 }{ 3 } \) To check, substitute \( a = \frac { 7 }{ 3 } \) into the L.H.S.: L.H.S. \( = \frac { a }{ 5 } = \frac { 7 }{ 3 } \times \frac { 1 }{ 5 } = \frac { 7 }{ 15 } \) This equals the R.H.S. Therefore, \( a = \frac { 7 }{ 3 } \) is the correct solution.
(h) \( 20t = -10 \)
The first step involves dividing both sides by 20.
Dividing both sides by 20, we get: \( \frac { 20t }{ 20 } = \frac { -10 }{ 20 } \)
\( t = - \frac { 1 }{ 2 } \) To check, substitute \( t = - \frac { 1 }{ 2 } \) into the L.H.S.: L.H.S. \( = 20t = 20 \times (- \frac { 1 }{ 2 }) = -10 \) This equals the R.H.S. Therefore, \( t = - \frac { 1 }{ 2 } \) is the correct solution.In simple words: When a variable is multiplied by a number, we divide both sides by that number. When it's divided by a number, we multiply both sides by that number to find its value.

Exam Tip: For equations involving multiplication or division, perform the inverse operation on both sides. This means multiplying if the variable is divided, or dividing if the variable is multiplied.

Question 3. Give the steps you will use to separate the variable and then solve the equation:
(a) \( 3n - 2 = 46 \)
(b) \( 5m + 7 = 17 \)
(c) \( \frac { 20p }{ 3 } = 40 \)
(d) \( \frac { 3p }{ 10 } = 6 \)
Answer:
(a) \( 3n - 2 = 46 \)
Step I: The first step involves adding 2 to both sides.
Adding 2 to both sides, we get: \( 3n - 2 + 2 = 46 + 2 \)
\( 3n = 48 \) Step II: The next step involves dividing both sides by 3.
Dividing both sides by 3, we get: \( \frac { 3n }{ 3 } = \frac { 48 }{ 3 } \)
\( n = 16 \) Therefore, \( n = 16 \) is the required solution.
(b) \( 5m + 7 = 17 \)
Step I: The first step involves subtracting 7 from both sides.
Subtracting 7 from both sides, we get: \( 5m + 7 - 7 = 17 - 7 \)
\( 5m = 10 \) Step II: The next step involves dividing both sides by 5.
Dividing both sides by 5, we get: \( \frac { 5m }{ 5 } = \frac { 10 }{ 5 } \)
\( m = 2 \) Therefore, \( m = 2 \) is the required solution.
(c) \( \frac { 20p }{ 3 } = 40 \)
Step I: The first step involves multiplying both sides by 3.
Multiplying both sides by 3, we get: \( \frac { 20p }{ 3 } \times 3 = 40 \times 3 \)
\( 20p = 120 \) Step II: The next step involves dividing both sides by 20.
Dividing both sides by 20, we get: \( \frac { 20p }{ 20 } = \frac { 120 }{ 20 } \)
\( p = 6 \) Therefore, \( p = 6 \) is the required solution.
(d) \( \frac { 3p }{ 10 } = 6 \)
Step I: The first step involves multiplying both sides by 10.
Multiplying both sides by 10, we get: \( \frac { 3p }{ 10 } \times 10 = 6 \times 10 \)
\( 3p = 60 \) Step II: The next step involves dividing both sides by 3.
Dividing both sides by 3, we get: \( \frac { 3p }{ 3 } = \frac { 60 }{ 3 } \)
\( p = 20 \) Therefore, \( p = 20 \) is the required solution.In simple words: For multi-step equations, first undo any addition or subtraction by performing the opposite operation on both sides. Then, undo any multiplication or division to isolate the variable.

Exam Tip: When solving two-step equations, always deal with addition/subtraction before multiplication/division to simplify the process and avoid common errors.

Question 4. Solve the following equations:
(a) \( 10p = 100 \)
(b) \( 10p + 10 = 100 \)
(c) \( \frac { p }{ 4 } = 5 \)
(d) \( \frac { -p }{ 3 } = 5 \)
(e) \( \frac { 3p }{ 4 } = 6 \)
(f) \( \frac { -p }{ 3 } = 5 \)
(g) \( 3s + 12 = 0 \)
(h) \( 3s = 0 \)
(i) \( 2q = 6 \)
(j) \( 2q - 6 = 0 \)
(k) \( 2q + 6 = 0 \)
(l) \( 2q + 6 = 12 \)
Answer:
(a) \( 10p = 100 \)
Divide both sides by 10: \( \frac { 10p }{ 10 } = \frac { 100 }{ 10 } \)
\( p = 10 \) Therefore, \( p = 10 \) is the required solution.
(b) \( 10p + 10 = 100 \)
Subtract 10 from both sides: \( 10p + 10 - 10 = 100 - 10 \)
\( 10p = 90 \) Divide both sides by 10: \( \frac { 10p }{ 10 } = \frac { 90 }{ 10 } \)
\( p = 9 \) Therefore, \( p = 9 \) is the required solution.
(c) \( \frac { p }{ 4 } = 5 \)
Multiply both sides by 4: \( \frac { p }{ 4 } \times 4 = 5 \times 4 \)
\( p = 20 \) Therefore, \( p = 20 \) is the required solution.
(d) \( \frac { -p }{ 3 } = 5 \)
Multiply both sides by 3: \( \frac { -p }{ 3 } \times 3 = 5 \times 3 \)
\( -p = 15 \) Multiply both sides by (-1): \( -p \times (-1) = 15 \times (-1) \)
\( p = -15 \) Therefore, \( p = -15 \) is the required solution.
(e) \( \frac { 3p }{ 4 } = 6 \)
Multiply both sides by 4: \( \frac { 3p }{ 4 } \times 4 = 6 \times 4 \)
\( 3p = 24 \) Divide both sides by 3: \( \frac { 3p }{ 3 } = \frac { 24 }{ 3 } \)
\( p = 8 \) Therefore, \( p = 8 \) is the required solution.
(f) \( \frac { -p }{ 3 } = 5 \)
Multiply both sides by 3: \( \frac { -p }{ 3 } \times 3 = 5 \times 3 \)
\( -p = 15 \) Multiply both sides by (-1): \( -p \times (-1) = 15 \times (-1) \)
\( p = -15 \) Therefore, \( p = -15 \) is the required solution.
(g) \( 3s + 12 = 0 \)
Subtract 12 from both sides: \( 3s + 12 - 12 = 0 - 12 \)
\( 3s = -12 \) Divide both sides by 3: \( \frac { 3s }{ 3 } = \frac { -12 }{ 3 } \)
\( s = -4 \) Therefore, \( s = -4 \) is the required solution.
(h) \( 3s = 0 \)
Divide both sides by 3: \( \frac { 3s }{ 3 } = \frac { 0 }{ 3 } \)
\( s = 0 \) Therefore, \( s = 0 \) is the required solution.
(i) \( 2q = 6 \)
Divide both sides by 2: \( \frac { 2q }{ 2 } = \frac { 6 }{ 2 } \)
\( q = 3 \) Therefore, \( q = 3 \) is the required solution.
(j) \( 2q - 6 = 0 \)
Add 6 to both sides: \( 2q - 6 + 6 = 0 + 6 \)
\( 2q = 6 \) Divide both sides by 2: \( \frac { 2q }{ 2 } = \frac { 6 }{ 2 } \)
\( q = 3 \) Therefore, \( q = 3 \) is the required solution.
(k) \( 2q + 6 = 0 \)
Subtract 6 from both sides: \( 2q + 6 - 6 = 0 - 6 \)
\( 2q = -6 \) Divide both sides by 2: \( \frac { 2q }{ 2 } = \frac { -6 }{ 2 } \)
\( q = -3 \) Therefore, \( q = -3 \) is the required solution.
(l) \( 2q + 6 = 12 \)
Subtract 6 from both sides: \( 2q + 6 - 6 = 12 - 6 \)
\( 2q = 6 \) Divide both sides by 2: \( \frac { 2q }{ 2 } = \frac { 6 }{ 2 } \)
\( q = 3 \) Therefore, \( q = 3 \) is the required solution.In simple words: To solve these equations, we use inverse operations to get the variable by itself. This means adding what's subtracted, subtracting what's added, multiplying what's divided, and dividing what's multiplied, always doing the same thing to both sides.

Exam Tip: Be careful with negative numbers and signs when performing operations on both sides of the equation, as a small error can lead to a completely wrong answer.