GSEB Class 7 Maths Solutions Chapter 4 Simple Equations Exercise 4.1

Question 1. Complete the last column of the table:

Exam Tip: To complete such tables, always substitute the given variable value into the equation's Left Hand Side (LHS) and then compare it with the Right Hand Side (RHS) to check if they are equal.

Question 2. Check whether the value given in the brackets is a solution to the given equation or not:
Answer:
(a) \( n + 5 = 19 \) (\( n = 1 \))
When we put \( n = 1 \) into \( n + 5 \), we get \( 1 + 5 = 6 \). This result, 6, is not equal to 19, which is the Right Hand Side (RHS) of the equation. So, \( n = 1 \) does not solve the equation \( n + 5 = 19 \).
(b) \( 7n + 5 = 19 \) (\( n = -2 \))
If we substitute \( n = -2 \) into the expression \( 7n + 5 \), we get \( 7 \times (-2) + 5 \). This calculates to \( -14 + 5 \), which equals \( -9 \). Since \( -9 \) is not equal to 19 (the RHS), \( n = -2 \) is not a valid solution for the equation \( 7n + 5 = 19 \).
(c) \( 7n + 5 = 19 \) (\( n = 2 \))
By substituting \( n = 2 \) into \( 7n + 5 \), we compute \( 7 \times 2 + 5 \). This simplifies to \( 14 + 5 \), which gives 19. This value is exactly equal to the Right Hand Side (RHS) of the equation. Therefore, \( n = 2 \) is a correct solution for \( 7n + 5 = 19 \).
(d) \( 4p - 3 = 13 \) (\( p = 1 \))
Let's put \( p = 1 \) into the expression \( 4p - 3 \). We calculate \( 4 \times 1 - 3 \), which results in \( 4 - 3 = 1 \). This value, 1, is not equal to 13, which is the Right Hand Side (RHS). Thus, \( p = 1 \) is not a solution for the equation \( 4p - 3 = 13 \).
(e) \( 4p - 3 = 13 \) (\( p = -4 \))
If we substitute \( p = -4 \) into \( 4p - 3 \), we perform \( 4 \times (-4) - 3 \). This calculation yields \( -16 - 3 \), which amounts to \( -19 \). Since \( -19 \) does not match 13 (the RHS), \( p = -4 \) is not a valid solution for the equation \( 4p - 3 = 13 \).
(f) \( 4p - 3 = 13 \) (\( p = 0 \))
When we insert \( p = 0 \) into \( 4p - 3 \), the expression becomes \( (4 \times 0) - 3 \). This simplifies to \( 0 - 3 \), which equals \( -3 \). Because \( -3 \) is not equivalent to 13 (the RHS), \( p = 0 \) is not a solution for \( 4p - 3 = 13 \).
In simple words: To check if a number is a solution, put that number into the equation where the letter is. Then, work out both sides. If they are the same, it's a solution. If they are different, it's not.

Exam Tip: To check if a value is a solution, substitute it into the equation and confirm if both sides are equal. Remember to apply integer multiplication rules correctly when substituting negative values.

Question 3. Solve the following equations by trial and error method:
(i) \( 5p + 2 = 17 \)
(ii) \( 3m - 14 = 4 \)
Answer:
(i) \( 5p + 2 = 17 \)
We will try different values for \( p \).
When \( p = 0 \), the expression \( 5p + 2 \) becomes \( (5 \times 0) + 2 = 0 + 2 = 2 \). This value, 2, is not equal to 17.
When \( p = 1 \), the expression \( 5p + 2 \) becomes \( (5 \times 1) + 2 = 5 + 2 = 7 \). This value, 7, is not equal to 17.
When \( p = 2 \), the expression \( 5p + 2 \) becomes \( (5 \times 2) + 2 = 10 + 2 = 12 \). This value, 12, is not equal to 17.
When \( p = 3 \), the expression \( 5p + 2 \) becomes \( (5 \times 3) + 2 = 15 + 2 = 17 \). This value, 17, is exactly equal to the Right Hand Side (RHS). Therefore, \( p = 3 \) is the correct solution for the equation \( 5p + 2 = 17 \).
(ii) \( 3m - 14 = 4 \)
We test different values for \( m \) using the trial and error approach.
If \( m = 0 \), then the Left Hand Side (LHS) is \( 3m - 14 = 3(0) - 14 = 0 - 14 = -14 \). This is not equal to 4.
If \( m = 1 \), then the LHS is \( 3m - 14 = (3 \times 1) - 14 = 3 - 14 = -11 \). This is not equal to 4.
If \( m = 2 \), then the LHS is \( 3m - 14 = (3 \times 2) - 14 = 6 - 14 = -8 \). This is not equal to 4.
If \( m = 3 \), then the LHS is \( 3m - 14 = (3 \times 3) - 14 = 9 - 14 = -5 \). This is not equal to 4.
If \( m = 4 \), then the LHS is \( 3m - 14 = (3 \times 4) - 14 = 12 - 14 = -2 \). This is not equal to 4.
If \( m = 5 \), then the LHS is \( 3m - 14 = (3 \times 5) - 14 = 15 - 14 = 1 \). This is not equal to 4.
If \( m = 6 \), then the LHS is \( 3m - 14 = (3 \times 6) - 14 = 18 - 14 = 4 \). This value, 4, is equal to the Right Hand Side (RHS).
Therefore, \( m = 6 \) is the solution to the equation \( 3m - 14 = 4 \).
In simple words: To solve by trial and error, try different numbers one by one in place of the letter. Keep trying until the equation makes sense (Left Side = Right Side).

Exam Tip: In the trial and error method, start with small integer values (0, 1, 2, ...) and methodically substitute them until the equation is satisfied. Be thorough with your trials and maintain accuracy in all arithmetic operations.

Question 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.
Answer:
(i) \( x + 4 = 9 \)
(ii) \( y - 2 = 8 \)
(iii) \( 10a = 70 \)
(iv) \( \frac{b}{5} = 6 \)
(v) \( \frac{3t}{4} = 15 \)
(vi) \( 7m + 7 = 77 \)
(vii) \( \frac{1}{4}x - 4 = 4 \)
(viii) \( 6y - 6 = 60 \)
(ix) \( \frac{z}{3} + 3 = 30 \)
In simple words: Read each statement carefully and turn the words into numbers and symbols to build an algebraic equation. 'Is' usually means equals.

Exam Tip: Pay close attention to keywords: "sum" means addition, "subtracted from" means subtraction (order matters), "times" means multiplication, and "divided by" means division.

Question 5. Write the following equations in statement
(i) \( p + 4 = 15 \)
(ii) \( m - 7 = 3 \)
(iii) \( 2m = 7 \)
(iv) \( \frac{m}{5} = 3 \)
(v) \( \frac{3m}{5} = 6 \)
(vi) \( 3p + 4 = 25 \)
(vii) \( 4p - 2 = 18 \)
(viii) \( \frac{p}{2} + 2 = 8 \)
Answer:
(i) The total of \( p \) and 4 equals 15.
(ii) If 7 is taken away from \( m \), the result is 3.
(iii) Twice the number \( m \) is 7.
(iv) One-fifth of a number \( m \) is 3.
(v) Three-fifth of a number \( m \) is 6.
(vi) Adding 4 to 3 times a number \( p \) results in 25.
(vii) Subtracting 2 from four times a number \( p \) yields 18.
(viii) Two added to half of a number \( p \) is 8.
In simple words: Convert the math symbols and numbers back into simple sentences. Each part of the equation becomes a phrase in the sentence.

Exam Tip: Practice recognizing common phrases like "sum of", "difference of", "product of", and "quotient of" to easily convert between equations and statements.

Question 6. Set up an equation in each of the following cases:
(i) Irfan says that he has 7 marbles more than five times the marbles Permit 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 1. The highest score is 87. (Take the lowest score to be l.)
(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.)
Answer:
(i) Let \( m \) represent the count of marbles that Parmit possesses. Five times Parmit's marbles would be \( 5m \). Irfan states that he has 7 marbles more than five times Parmit's marbles, so Irfan's marbles can be expressed as \( 5m + 7 \). We are told that Irfan actually has 37 marbles. Therefore, we can set up the equation as \( 5m + 7 = 37 \).
(ii) Let Laxmi's age be denoted by \( y \) years. Three times Laxmi's age would be \( 3y \). Laxmi's father is 4 years older than this, so his age can be written as \( 3y + 4 \). We know that Laxmi's father is 49 years old. Thus, the equation can be formed as \( 3y + 4 = 49 \).
(iii) Let \( l \) represent the lowest score obtained. Twice the lowest marks would be \( 2l \). The highest marks are 7 more than twice the lowest marks, so they can be written as \( 2l + 7 \). We are given that the highest score is 87. Thus, the equation to represent this situation is \( 2l + 7 = 87 \).
(iv) Let \( b \) represent the measure of each base angle in degrees. In an isosceles triangle, the two base angles are always equal, so both base angles are \( b \) degrees. The vertex angle is specified as twice either base angle, which means the vertex angle is \( 2b \) degrees. We know that the total sum of all three angles inside any triangle is 180 degrees. Therefore, the equation for this situation is \( b + b + 2b = 180 \), which simplifies to \( 4b = 180 \).
In simple words: For each story, figure out what number is unknown and call it a letter. Then, write down how the numbers and letters are connected to make an equation.

Exam Tip: Break down word problems into smaller parts. Identify the unknown, define a variable for it, and then translate each phrase into an algebraic expression. Recall properties of geometric figures, like the sum of angles in a triangle, and combine them with given relationships to form the equation.