GSEB Class 7 Maths Solutions Chapter 4 સાદા સમીકરણ Exercise 4.3

1. Solve the following equations:

Question (a). \( 2y + \frac { 5 }{ 2 } = \frac { 37 }{ 2 } \)
Answer:
We are given the equation: \( 2y + \frac { 5 }{ 2 } = \frac { 37 }{ 2 } \)
Move \( \frac { 5 }{ 2 } \) to the right side:
\( 2y = \frac { 37 }{ 2 } - \frac { 5 }{ 2 } \)
Subtract the fractions:
\( 2y = \frac { 37 - 5 }{ 2 } \)
\( 2y = \frac { 32 }{ 2 } \)
\( 2y = 16 \)
Divide both sides by 2:
\( \frac { 2y }{ 2 } = \frac { 16 }{ 2 } \)
\( \implies y = 8 \)
The solution is \( y = 8 \).
In simple words: To solve this, first move the fraction from the left side to the right. Then, subtract the fractions. Finally, divide both sides by 2 to find the value of y.

Exam Tip: Remember to combine like terms and isolate the variable by performing the opposite operations on both sides of the equation.

Question (b). \( 5t + 28 = 10 \)
Answer:
We are given the equation: \( 5t + 28 = 10 \)
Move 28 to the right side by subtracting it:
\( 5t = 10 - 28 \)
\( 5t = -18 \)
Divide both sides by 5:
\( \frac { 5t }{ 5 } = \frac { -18 }{ 5 } \)
\( \implies t = - \frac { 18 }{ 5 } \)
The solution is \( t = - \frac { 18 }{ 5 } \).
In simple words: First, move the constant number (28) to the other side by changing its sign. Then, divide both sides by the number next to 't' (which is 5) to find the value of 't'.

Exam Tip: Always pay close attention to signs, especially when moving terms across the equals sign. A common mistake is forgetting to change a positive to a negative, or vice versa.

Question (c). \( \frac { a }{ 5 } + 3 = 2 \)
Answer:
We are given 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:
\( \frac { a }{ 5 } \times 5 = -1 \times 5 \)
\( \implies a = -5 \)
The solution is \( a = -5 \).
In simple words: To solve, first subtract 3 from both sides. Then, multiply both sides by 5 to get the value of 'a'.

Exam Tip: When dealing with fractions, multiplying by the denominator is a good way to clear the fraction and simplify the equation.

Question (d). \( \frac { q }{ 4 } + 7 = 5 \)
Answer:
We are given 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:
\( \frac { q }{ 4 } \times 4 = -2 \times 4 \)
\( \implies q = -8 \)
The solution is \( q = -8 \).
In simple words: First, take the number 7 to the right side and subtract it. Then, multiply both sides by 4 to find the value of 'q'.

Exam Tip: Be careful with integer arithmetic when subtracting numbers, especially when dealing with negative results.

Question (e). \( \frac { 5 }{ 2 }x = -10 \)
Answer:
We are given the equation: \( \frac { 5 }{ 2 }x = -10 \)
To isolate x, multiply both sides by the reciprocal of \( \frac { 5 }{ 2 } \), which is \( \frac { 2 }{ 5 } \).
\( \frac { 5 }{ 2 }x \times \frac { 2 }{ 5 } = -10 \times \frac { 2 }{ 5 } \)
\( x = - \frac { 20 }{ 5 } \)
\( \implies x = -4 \)
The solution is \( x = -4 \).
In simple words: To get 'x' by itself, multiply both sides of the equation by the flipped version of the fraction that's with 'x'. For \( \frac { 5 }{ 2 } \), use \( \frac { 2 }{ 5 } \).

Exam Tip: When a variable is multiplied by a fraction, you can isolate it by multiplying both sides by the reciprocal of that fraction.

Question (f). \( \frac { 5 }{ 2 }x = \frac { 25 }{ 4 } \)
Answer:
We are given the equation: \( \frac { 5 }{ 2 }x = \frac { 25 }{ 4 } \)
To isolate x, multiply both sides by the reciprocal of \( \frac { 5 }{ 2 } \), which is \( \frac { 2 }{ 5 } \).
\( \frac { 5 }{ 2 }x \times \frac { 2 }{ 5 } = \frac { 25 }{ 4 } \times \frac { 2 }{ 5 } \)
\( x = \frac { 50 }{ 20 } \)
\( \implies x = \frac { 5 }{ 2 } \)
The solution is \( x = \frac { 5 }{ 2 } \).
In simple words: To find 'x', multiply both sides of the equation by the inverse fraction of \( \frac { 5 }{ 2 } \), which is \( \frac { 2 }{ 5 } \). Then simplify the result.

Exam Tip: Always simplify fractions to their lowest terms at the end of the calculation for full marks.

Question (g). \( 7m + \frac { 19 }{ 2 } = 13 \)
Answer:
We are given the equation: \( 7m + \frac { 19 }{ 2 } = 13 \)
Move \( \frac { 19 }{ 2 } \) to the right side by subtracting it:
\( 7m = 13 - \frac { 19 }{ 2 } \)
To subtract, find a common denominator for 13:
\( 7m = \frac { 26 }{ 2 } - \frac { 19 }{ 2 } \)
\( 7m = \frac { 26 - 19 }{ 2 } \)
\( 7m = \frac { 7 }{ 2 } \)
Divide both sides by 7:
\( \frac { 7m }{ 7 } = \frac { 7 }{ 2 } \times \frac { 1 }{ 7 } \)
\( \implies m = \frac { 1 }{ 2 } \)
The solution is \( m = \frac { 1 }{ 2 } \).
In simple words: First, move the fraction to the other side. Then, subtract the numbers by getting a common bottom number. Finally, divide by 7 to solve for 'm'.

Exam Tip: When subtracting a fraction from a whole number, express the whole number as a fraction with the same denominator to easily combine them.

Question (h). \( 6z + 10 = -2 \)
Answer:
We are given the equation: \( 6z + 10 = -2 \)
Move 10 to the right side by subtracting it:
\( 6z = -2 - 10 \)
\( 6z = -12 \)
Divide both sides by 6:
\( \frac { 6z }{ 6 } = \frac { -12 }{ 6 } \)
\( \implies z = -2 \)
The solution is \( z = -2 \).
In simple words: Start by moving the 10 to the other side of the equation, changing its sign. Then, divide both sides by 6 to get the value of 'z'.

Exam Tip: Be careful when combining negative numbers; remember that subtracting a positive number from a negative number makes it even more negative.

Question (i). \( \frac { 3l }{ 2 } = \frac { 2 }{ 3 } \)
Answer:
We are given the equation: \( \frac { 3l }{ 2 } = \frac { 2 }{ 3 } \)
To isolate \( l \), multiply both sides by \( \frac { 2 }{ 3 } \):
\( \frac { 3l }{ 2 } \times \frac { 2 }{ 3 } = \frac { 2 }{ 3 } \times \frac { 2 }{ 3 } \)
\( l = \frac { 4 }{ 9 } \)
The solution is \( l = \frac { 4 }{ 9 } \).
In simple words: To get 'l' by itself, multiply both sides of the equation by the reciprocal of the fraction \( \frac { 3 }{ 2 } \), which is \( \frac { 2 }{ 3 } \). This will cancel out the fraction next to 'l'.

Exam Tip: When isolating a variable that has a fraction coefficient, multiplying by the reciprocal is a more direct approach than multiplying by the denominator first and then dividing by the numerator.

Question (j). \( \frac { 2b }{ 3 } - 5 = 3 \)
Answer:
We are given 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 3:
\( \frac { 2b }{ 3 } \times 3 = 8 \times 3 \)
\( 2b = 24 \)
Divide both sides by 2:
\( \frac { 2b }{ 2 } = \frac { 24 }{ 2 } \)
\( \implies b = 12 \)
The solution is \( b = 12 \).
In simple words: First, add 5 to both sides. Then, multiply by 3 to remove the fraction. Finally, divide by 2 to find 'b'.

Exam Tip: Always perform operations to isolate the variable in the reverse order of operations (PEMDAS/BODMAS): usually, addition/subtraction first, then multiplication/division.

2. Solve the following equations:

Question (a). \( 2(x + 4) = 12 \)
Answer:
We are given the equation: \( 2(x + 4) = 12 \)
Divide both sides by 2:
\( \frac { 2(x + 4) }{ 2 } = \frac { 12 }{ 2 } \)
\( x + 4 = 6 \)
Move 4 to the right side by subtracting it:
\( x = 6 - 4 \)
\( \implies x = 2 \)
The solution is \( x = 2 \).
In simple words: First, divide both sides by 2 to get rid of the number outside the bracket. Then, subtract 4 from both sides to find 'x'.

Exam Tip: When a number is multiplying an entire bracket, dividing both sides by that number is often the easiest first step to simplify.

Question (b). \( 3(n - 5) = 21 \)
Answer:
We are given the equation: \( 3(n - 5) = 21 \)
Divide both sides by 3:
\( \frac { 3(n - 5) }{ 3 } = \frac { 21 }{ 3 } \)
\( n - 5 = 7 \)
Move -5 to the right side by adding it:
\( n = 7 + 5 \)
\( \implies n = 12 \)
The solution is \( n = 12 \).
In simple words: Divide both sides by 3 to remove the multiplier. Then, add 5 to both sides to get the value of 'n'.

Exam Tip: Be careful with the signs when moving terms. If a term is negative on one side, it becomes positive when moved to the other side.

Question (c). \( 3(n - 5) = -21 \)
Answer:
We are given the equation: \( 3(n - 5) = -21 \)
Divide both sides by 3:
\( \frac { 3(n - 5) }{ 3 } = \frac { -21 }{ 3 } \)
\( n - 5 = -7 \)
Move -5 to the right side by adding it:
\( n = -7 + 5 \)
\( \implies n = -2 \)
The solution is \( n = -2 \).
In simple words: Divide by 3 first. Then, add 5 to both sides to solve for 'n'. Watch out for the negative numbers.

Exam Tip: Dividing a negative number by a positive number results in a negative number. This is a fundamental rule for arithmetic operations with signed numbers.

Question (d). \( -4(2 + x) = 8 \)
Answer:
We are given the equation: \( -4(2 + x) = 8 \)
Divide both sides by -4:
\( \frac { -4(2 + x) }{ -4 } = \frac { 8 }{ -4 } \)
\( 2 + x = -2 \)
Move 2 to the right side by subtracting it:
\( x = -2 - 2 \)
\( \implies x = -4 \)
The solution is \( x = -4 \).
In simple words: Start by dividing both sides by -4. Then, subtract 2 from both sides to get 'x'.

Exam Tip: Dividing by a negative number reverses the sign of the result. For example, positive 8 divided by negative 4 gives negative 2.

Question (e). \( 4(2 - x) = 8 \)
Answer:
We are given the equation: \( 4(2 - x) = 8 \)
Divide both sides by 4:
\( \frac { 4(2 - x) }{ 4 } = \frac { 8 }{ 4 } \)
\( 2 - x = 2 \)
Move 2 to the right side by subtracting it:
\( -x = 2 - 2 \)
\( -x = 0 \)
Multiply by -1 to solve for x:
\( x = 0 \)
The solution is \( x = 0 \).
In simple words: First, divide both sides by 4. Then, move 2 to the other side by subtracting it. If you get -x = 0, then x is also 0.

Exam Tip: If an equation simplifies to \( -x = 0 \), then \( x \) must also be \( 0 \), as \( -0 \) is the same as \( 0 \).

3. Solve the following equations:

Question (a). \( 4 = 5(p - 2) \)
Answer:
We are given the equation: \( 4 = 5(p - 2) \)
Swap the sides to place the variable on the left:
\( 5(p - 2) = 4 \)
Divide both sides by 5:
\( \frac { 5(p - 2) }{ 5 } = \frac { 4 }{ 5 } \)
\( p - 2 = \frac { 4 }{ 5 } \)
Move -2 to the right side by adding it:
\( p = \frac { 4 }{ 5 } + 2 \)
To add, find a common denominator for 2:
\( p = \frac { 4 }{ 5 } + \frac { 10 }{ 5 } \)
\( p = \frac { 4 + 10 }{ 5 } \)
\( \implies p = \frac { 14 }{ 5 } \)
The solution is \( p = \frac { 14 }{ 5 } \).
In simple words: First, divide both sides by 5. Then, add 2 to both sides. To add the fraction and whole number, make the whole number a fraction with the same bottom number.

Exam Tip: It's often helpful to rearrange equations so that the variable term is on the left side, which can make the solving process feel more natural.

Question (b). \( -4 = 5(p - 2) \)
Answer:
We are given the equation: \( -4 = 5(p - 2) \)
Swap the sides to place the variable on the left:
\( 5(p - 2) = -4 \)
Divide both sides by 5:
\( \frac { 5(p - 2) }{ 5 } = \frac { -4 }{ 5 } \)
\( p - 2 = - \frac { 4 }{ 5 } \)
Move -2 to the right side by adding it:
\( p = - \frac { 4 }{ 5 } + 2 \)
To add, find a common denominator for 2:
\( p = - \frac { 4 }{ 5 } + \frac { 10 }{ 5 } \)
\( p = \frac { -4 + 10 }{ 5 } \)
\( \implies p = \frac { 6 }{ 5 } \)
The solution is \( p = \frac { 6 }{ 5 } \).
In simple words: Swap the equation so 'p' is on the left. Divide by 5. Then, add 2 to both sides. Convert the whole number to a fraction with the same denominator to combine them.

Exam Tip: When combining a negative fraction with a positive whole number, remember the rules for adding integers with different signs: subtract their absolute values and use the sign of the larger absolute value.

Question (c). \( 16 = 4 + 3(t + 2) \)
Answer:
We are given the equation: \( 16 = 4 + 3(t + 2) \)
Swap the sides to place 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:
\( \frac { 3(t + 2) }{ 3 } = \frac { 12 }{ 3 } \)
\( t + 2 = 4 \)
Move 2 to the right side by subtracting it:
\( t = 4 - 2 \)
\( \implies t = 2 \)
The solution is \( t = 2 \).
In simple words: First, switch the sides so 't' is on the left. Subtract 4 from both sides. Then, divide by 3. Finally, subtract 2 to find 't'.

Exam Tip: Remember to perform operations in the correct order: first isolate the term with the variable (the \( 3(t+2) \) part), then simplify the bracket, and finally isolate the variable itself.

Question (d). \( 4 + 5(p - 1) = 34 \)
Answer:
We are given 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:
\( \frac { 5(p - 1) }{ 5 } = \frac { 30 }{ 5 } \)
\( p - 1 = 6 \)
Move -1 to the right side by adding it:
\( p = 6 + 1 \)
\( \implies p = 7 \)
The solution is \( p = 7 \).
In simple words: First, subtract 4 from both sides. Then, divide both sides by 5. Finally, add 1 to both sides to find 'p'.

Exam Tip: Isolate the term containing the parenthesis first before attempting to divide or distribute. This simplifies the equation gradually.

Question (e). \( 0 = 16 + 4(m - 6) \)
Answer:
We are given the equation: \( 0 = 16 + 4(m - 6) \)
Swap the sides to place the variable on the left:
\( 16 + 4(m - 6) = 0 \)
Move 16 to the right side by subtracting it:
\( 4(m - 6) = -16 \)
Divide both sides by 4:
\( \frac { 4(m - 6) }{ 4 } = \frac { -16 }{ 4 } \)
\( m - 6 = -4 \)
Move -6 to the right side by adding it:
\( m = -4 + 6 \)
\( \implies m = 2 \)
The solution is \( m = 2 \).
In simple words: Switch the sides of the equation. Subtract 16 from both sides. Divide by 4. Then, add 6 to find 'm'.

Exam Tip: When adding or subtracting a positive and a negative number, find the difference between their absolute values and use the sign of the number with the larger absolute value.

4.

Question (a). Create 3 equations starting with \( x = 2 \).
Answer:
Starting with \( x = 2 \):
**Equation 1:**
Multiply both sides by 3:
\( 3x = 2 \times 3 \)
\( 3x = 6 \)
Subtract 1 from both sides:
\( 3x - 1 = 6 - 1 \)
\( \implies 3x - 1 = 5 \)
**Equation 2:**
Divide both sides by 3:
\( \frac { x }{ 3 } = \frac { 2 }{ 3 } \)
Add 5 to both sides:
\( \frac { x }{ 3 } + 5 = \frac { 2 }{ 3 } + 5 \)
\( \frac { x }{ 3 } + 5 = \frac { 2 + 15 }{ 3 } \)
\( \implies \frac { x }{ 3 } + 5 = \frac { 17 }{ 3 } \)
**Equation 3:**
Multiply both sides by 4:
\( 4x = 2 \times 4 \)
\( 4x = 8 \)
Subtract 3 from both sides:
\( 4x - 3 = 8 - 3 \)
\( \implies 4x - 3 = 5 \)
In simple words: You start with \( x = 2 \). To make new equations, do the same math step to both sides. You can multiply, divide, add, or subtract any number to both sides, and the equation will still be true.

Exam Tip: When constructing equations, always perform the exact same operation on both sides to maintain equality. You can use any arithmetic operation.

Question (b). Create 3 equations starting with \( x = -2 \).
Answer:
Starting with \( x = -2 \):
**Equation 1:**
Multiply both sides by 7:
\( 7x = -2 \times 7 \)
\( 7x = -14 \)
Subtract 1 from both sides:
\( 7x - 1 = -14 - 1 \)
\( \implies 7x - 1 = -15 \)
**Equation 2:**
Divide both sides by 5:
\( \frac { x }{ 5 } = \frac { -2 }{ 5 } \)
Add 2 to both sides:
\( \frac { x }{ 5 } + 2 = \frac { -2 }{ 5 } + 2 \)
\( \frac { x }{ 5 } + 2 = \frac { -2 + 10 }{ 5 } \)
\( \implies \frac { x }{ 5 } + 2 = \frac { 8 }{ 5 } \)
**Equation 3:**
Subtract 3 from both sides:
\( x - 3 = -2 - 3 \)
\( x - 3 = -5 \)
Multiply both sides by 4:
\( 4(x - 3) = -5 \times 4 \)
\( \implies 4(x - 3) = -20 \)
In simple words: Begin with \( x = -2 \). For each new equation, apply the same mathematical operation (like multiplying, dividing, adding, or subtracting) to both sides of the equation. This ensures the equation remains balanced and correct.

Exam Tip: When dealing with negative numbers, be extra careful with arithmetic operations. A common error is incorrectly adding or subtracting negative numbers.