GSEB Class 7 Maths Solutions Chapter 4 Simple Equations Exercise 4.3

Question 1. Solve the following equations.
(a) \( 2y + \frac { 5 }{ 2 } = \frac { 37 }{ 2 } \)
(b) \( 5t + 28 = 10 \)
(c) \( \frac { a }{ 5 } + 3 = 2 \)
(d) \( \frac { q }{ 4 } + 7 = 5 \)
(e) \( \frac { 5 }{ 2 }x = -10 \)
(f) \( \frac { 5 }{ 2 }x = \frac { 25 }{ 4 } \)
(g) \( 7m + \frac { 19 }{ 2 } = 13 \)
(h) \( 6z + 10 = -2 \)
(i) \( \frac { 3I }{ 2 } = \frac { 2 }{ 3 } \)
(j) \( \frac { 2b }{ 3 } - 5 = 3 \)
Answer:
(a) We have the equation: \( 2y + \frac { 5 }{ 2 } = \frac { 37 }{ 2 } \)
Move \( \frac { 5 }{ 2 } \) to the right side by subtracting it:
\( 2y = \frac { 37 }{ 2 } - \frac { 5 }{ 2 } \)
Simplify the right side:
\( 2y = \frac { 37-5 }{ 2 } \)
\( 2y = \frac { 32 }{ 2 } \)
\( 2y = 16 \)
Divide both sides by 2:
\( y = \frac { 16 }{ 2 } \)
\( y = 8 \)
So, the required solution is \( y = 8 \).

(b) We have the equation: \( 5t + 28 = 10 \)
Move 28 to the right side by subtracting it:
\( 5t = 10 - 28 \)
\( 5t = -18 \)
Divide both sides by 5:
\( t = \frac { -18 }{ 5 } \)
So, the required solution is \( t = - \frac { 18 }{ 5 } \).

(c) We have the equation: \( \frac { a }{ 5 } + 3 = 2 \)
Move 3 to the right side by subtracting it:
\( \frac { a }{ 5 } = 2 - 3 \)
\( \frac { a }{ 5 } = -1 \)
Multiply both sides by 5:
\( a = -1 \times 5 \)
\( a = -5 \)
So, the required solution is \( a = -5 \).

(d) We have the equation: \( \frac { q }{ 4 } + 7 = 5 \)
Move 7 to the right side by subtracting it:
\( \frac { q }{ 4 } = 5 - 7 \)
\( \frac { q }{ 4 } = -2 \)
Multiply both sides by 4:
\( q = -2 \times 4 \)
\( q = -8 \)
So, the required solution is \( q = -8 \).

(e) We have the equation: \( \frac { 5 }{ 2 }x = -10 \)
Multiply both sides by 2:
\( 5x = -10 \times 2 \)
\( 5x = -20 \)
Divide both sides by 5:
\( x = \frac { -20 }{ 5 } \)
\( x = -4 \)
So, the required solution is \( x = -4 \).

(f) We have the equation: \( \frac { 5 }{ 2 }x = \frac { 25 }{ 4 } \)
Multiply both sides by 2:
\( 5x = \frac { 25 }{ 4 } \times 2 \)
\( 5x = \frac { 25 }{ 2 } \)
Divide both sides by 5:
\( x = \frac { 25 }{ 2 } \times \frac { 1 }{ 5 } \)
\( x = \frac { 5 }{ 2 } \)
So, the required solution is \( x = \frac { 5 }{ 2 } \).

(g) We have the equation: \( 7m + \frac { 19 }{ 2 } = 13 \)
Move \( \frac { 19 }{ 2 } \) to the right side by subtracting it:
\( 7m = 13 - \frac { 19 }{ 2 } \)
Simplify the right side:
\( 7m = \frac { 26 - 19 }{ 2 } \)
\( 7m = \frac { 7 }{ 2 } \)
Divide both sides by 7:
\( m = \frac { 7 }{ 2 } \times \frac { 1 }{ 7 } \)
\( m = \frac { 1 }{ 2 } \)
So, the required solution is \( m = \frac { 1 }{ 2 } \).

(h) We have the equation: \( 6z + 10 = -2 \)
Move 10 to the right side by subtracting it:
\( 6z = -2 - 10 \)
\( 6z = -12 \)
Divide both sides by 6:
\( z = \frac { -12 }{ 6 } \)
\( z = -2 \)
So, the required solution is \( z = -2 \).

(i) We have the equation: \( \frac { 3I }{ 2 } = \frac { 2 }{ 3 } \)
Multiply both sides by 2:
\( 3I = \frac { 2 }{ 3 } \times 2 \)
\( 3I = \frac { 4 }{ 3 } \)
Divide both sides by 3:
\( I = \frac { 4 }{ 3 } \times \frac { 1 }{ 3 } \)
\( I = \frac { 4 }{ 9 } \)
So, the required solution is \( I = \frac { 4 }{ 9 } \).

(j) We have the equation: \( \frac { 2b }{ 3 } - 5 = 3 \)
Move -5 to the right side by adding it:
\( \frac { 2b }{ 3 } = 3 + 5 \)
\( \frac { 2b }{ 3 } = 8 \)
Multiply both sides by \( \frac { 3 }{ 2 } \):
\( b = 8 \times \frac { 3 }{ 2 } \)
\( b = 4 \times 3 \)
\( b = 12 \)
So, the required solution is \( b = 12 \).
In simple words: To solve these equations, your main goal is to get the unknown letter (like y, t, a, q, x, m, z, I, or b) all by itself on one side of the equals sign. You do this by moving numbers to the other side, using opposite operations: add to subtract, subtract to add, multiply to divide, and divide to multiply. Make sure to do the same thing to both sides of the equation to keep it balanced.

Exam Tip: Always perform operations to both sides of the equation to maintain balance. Double-check your arithmetic, especially with fractions and negative numbers, to avoid calculation errors.

Question 2. Solve the following equations.
(a) \( 2(x + 4) = 12 \)
(b) \( 3(n - 5) = 21 \)
(c) \( 3(n - 5) = -21 \)
(d) \( -4(2 + x) = 8 \)
(e) \( 4(2 - x) = 8 \)
Answer:
(a) We have the equation: \( 2(x + 4) = 12 \)
Divide both sides by 2:
\( x + 4 = \frac { 12 }{ 2 } \)
\( x + 4 = 6 \)
Move 4 to the right side by subtracting it:
\( x = 6 - 4 \)
\( x = 2 \)
So, the required solution is \( x = 2 \).

(b) We have the equation: \( 3(n - 5) = 21 \)
Divide both sides by 3:
\( n - 5 = \frac { 21 }{ 3 } \)
\( n - 5 = 7 \)
Move -5 to the right side by adding it:
\( n = 7 + 5 \)
\( n = 12 \)
So, the required solution is \( n = 12 \).

(c) We have the equation: \( 3(n - 5) = -21 \)
Divide both sides by 3:
\( n - 5 = \frac { -21 }{ 3 } \)
\( n - 5 = -7 \)
Move -5 to the right side by adding it:
\( n = -7 + 5 \)
\( n = -2 \)
So, the required solution is \( n = -2 \).

(d) We have the equation: \( -4(2 + x) = 8 \)
Divide both sides by -4:
\( 2 + x = \frac { 8 }{ -4 } \)
\( 2 + x = -2 \)
Move 2 to the right side by subtracting it:
\( x = -2 - 2 \)
\( x = -4 \)
So, the required solution is \( x = -4 \).

(e) We have the equation: \( 4(2 - x) = 8 \)
Divide both sides by 4:
\( 2 - x = \frac { 8 }{ 4 } \)
\( 2 - x = 2 \)
Move 2 to the right side by subtracting it:
\( -x = 2 - 2 \)
\( -x = 0 \)
Multiply by -1 to get x:
\( x = 0 \)
So, the required solution is \( x = 0 \).
In simple words: When a number is outside brackets, like \( 2(x + 4) \), you can either divide both sides by that number first, or multiply everything inside the brackets by that number. Both ways will get you to the same correct answer. Always simplify steps and use opposite operations to isolate the variable.

Exam Tip: When solving equations with brackets, dividing by the coefficient outside the bracket first often simplifies the problem. Remember to handle negative signs carefully when transposing or dividing.

Question 3. Solve the following equations:
(a) \( 4 = 5(p - 2) \)
(b) \( -4 = 5(p - 2) \)
(c) \( 16 = 4 + 3(t + 2) \)
(d) \( 4 + 5(p - 1) = 34 \)
(e) \( 0 = 16 + 4(m - 6) \)
Answer:
(a) We have the equation: \( 4 = 5(p - 2) \)
First, swap the sides to put the variable on the left:
\( 5(p - 2) = 4 \)
Divide both sides by 5:
\( p - 2 = \frac { 4 }{ 5 } \)
Move -2 to the right side by adding it:
\( p = \frac { 4 }{ 5 } + 2 \)
To add, find a common denominator:
\( p = \frac { 4 + 10 }{ 5 } \)
\( p = \frac { 14 }{ 5 } \)
So, the required solution is \( p = \frac { 14 }{ 5 } \).

(b) We have the equation: \( -4 = 5(p - 2) \)
Swap the sides to put the variable on the left:
\( 5(p - 2) = -4 \)
Divide both sides by 5:
\( p - 2 = - \frac { 4 }{ 5 } \)
Move -2 to the right side by adding it:
\( p = - \frac { 4 }{ 5 } + 2 \)
To add, find a common denominator:
\( p = \frac { -4 + 10 }{ 5 } \)
\( p = \frac { 6 }{ 5 } \)
So, the required solution is \( p = \frac { 6 }{ 5 } \).

(c) We have the equation: \( 16 = 4 + 3(t + 2) \)
Swap the sides to put the variable on the left:
\( 4 + 3(t + 2) = 16 \)
Move 4 to the right side by subtracting it:
\( 3(t + 2) = 16 - 4 \)
\( 3(t + 2) = 12 \)
Divide both sides by 3:
\( t + 2 = \frac { 12 }{ 3 } \)
\( t + 2 = 4 \)
Move 2 to the right side by subtracting it:
\( t = 4 - 2 \)
\( t = 2 \)
So, the required solution is \( t = 2 \).

(d) We have the equation: \( 4 + 5(p - 1) = 34 \)
Move 4 to the right side by subtracting it:
\( 5(p - 1) = 34 - 4 \)
\( 5(p - 1) = 30 \)
Divide both sides by 5:
\( p - 1 = \frac { 30 }{ 5 } \)
\( p - 1 = 6 \)
Move -1 to the right side by adding it:
\( p = 6 + 1 \)
\( p = 7 \)
So, the required solution is \( p = 7 \).

(e) We have the equation: \( 0 = 16 + 4(m - 6) \)
Swap the sides to put the variable on the left:
\( 16 + 4(m - 6) = 0 \)
Move 16 to the right side by subtracting it:
\( 4(m - 6) = 0 - 16 \)
\( 4(m - 6) = -16 \)
Divide both sides by 4:
\( m - 6 = \frac { -16 }{ 4 } \)
\( m - 6 = -4 \)
Move -6 to the right side by adding it:
\( m = -4 + 6 \)
\( m = 2 \)
So, the required solution is \( m = 2 \).
In simple words: These equations involve numbers outside brackets and also numbers added or subtracted. Start by moving any added or subtracted numbers away from the bracketed term. Then, handle the number outside the bracket by dividing. Finally, solve for the variable inside the bracket. Remember to use opposite operations and keep the equation balanced at each step.

Exam Tip: For equations like these, it's generally best to isolate the term with the variable (the bracketed part) before dealing with the multiplier outside the bracket. This simplifies the process and reduces errors.

Question 4.
(a) Construct 3 equations starting with \( x = 2 \).
(b) Construct 3 equations starting with \( x = -2 \).
Answer:
(a) Starting with \( x = 2 \):
I. First equation:
Given: \( x = 2 \)
Multiply both sides by 5:
\( 5 \times x = 5 \times 2 \)
\( 5x = 10 \)
Subtract 3 from both sides:
\( 5x - 3 = 10 - 3 \)
\( 5x - 3 = 7 \)
So, the first equation is \( 5x - 3 = 7 \).

II. Second equation:
Given: \( x = 2 \)
Multiply both sides by 7:
\( 7 \times x = 7 \times 2 \)
\( 7x = 14 \)
Add 5 to both sides:
\( 7x + 5 = 14 + 5 \)
\( 7x + 5 = 19 \)
So, the second equation is \( 7x + 5 = 19 \).

III. Third equation:
Given: \( x = 2 \)
Divide both sides by 3:
\( \frac { x }{ 3 } = \frac { 2 }{ 3 } \)
Subtract 4 from both sides:
\( \frac { x }{ 3 } - 4 = \frac { 2 }{ 3 } - 4 \)
\( \frac { x }{ 3 } - 4 = \frac { 2 - 12 }{ 3 } \)
\( \frac { x }{ 3 } - 4 = \frac { -10 }{ 3 } \)
So, the third equation is \( \frac { x }{ 3 } - 4 = \frac { -10 }{ 3 } \).

(b) Starting with \( x = -2 \):
I. First equation:
Given: \( x = -2 \)
Add 8 to both sides:
\( x + 8 = -2 + 8 \)
\( x + 8 = 6 \)
So, the first equation is \( x + 8 = 6 \).

II. Second equation:
Given: \( x = -2 \)
Subtract 10 from both sides:
\( x - 10 = -2 - 10 \)
\( x - 10 = -12 \)
So, the second equation is \( x - 10 = -12 \).

III. Third equation:
Given: \( x = -2 \)
Multiply both sides by 8:
\( 8 \times x = (-2) \times 8 \)
\( 8x = -16 \)
Subtract 2 from both sides:
\( 8x - 2 = -16 - 2 \)
\( 8x - 2 = -18 \)
So, the third equation is \( 8x - 2 = -18 \).
In simple words: To create new equations, you can start with a simple value for x, like \( x = 2 \). Then, you can do any math operation (add, subtract, multiply, divide) to both sides of the equation. Each time you do an operation, you get a new, balanced equation that still holds true for your starting x value. Do this a few times to build different equations.

Exam Tip: When constructing equations, remember that any operation performed on one side of the equation must also be performed on the other side to keep the equation valid. Practice with different operations to create a variety of equations.