GSEB Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.4

Question 1. Solve the following pair of linear equations by the elimination method and the substitution method:
1. \( x + y = 5 \) and \( 2x - 3y = 4 \)
2. \( 3x + 4y = 10 \) and \( 2x - 2y = 2 \)
3. \( 3x - 5y - 4 = 0 \) and \( 9x = 2y + 7 \)
4. \( \frac{x}{2}+\frac{2 y}{3} = -1 \) and \( x - \frac{y}{3} = 3 \)
Answer:

1. \( x + y = 5 \) and \( 2x - 3y = 4 \)

I. By Elimination Method:
The provided set of equations is:
\( x + y = 5 \) ...(1)
\( 2x - 3y = 4 \) ...(2)
To make the \( y \) coefficients equal, we multiply equation (1) by 3:
\( 3(x + y) = 3(5) \)
\( \implies 3x + 3y = 15 \) ...(3)
Now, we add equation (2) and equation (3):
\( (2x - 3y) + (3x + 3y) = 4 + 15 \)
\( \implies 5x = 19 \)
\( \implies x = \frac{19}{5} \)
Next, we substitute this value of \( x \) into equation (1):
\( \frac{19}{5} + y = 5 \)
\( \implies y = 5 - \frac{19}{5} \)
\( \implies y = \frac{25 - 19}{5} \)
\( \implies y = \frac{6}{5} \)
Thus, the solution for the given system of equations is \( x = \frac{19}{5} \) and \( y = \frac{6}{5} \).
In simple words: We used the elimination method to solve the equations. First, we multiplied the first equation by 3 to make the 'y' terms match. Then, we added the two equations to remove 'y' and found the value of 'x'. Finally, we put 'x' back into an original equation to find 'y'.

Exam Tip: In the elimination method, aim to make the coefficients of one variable opposite numbers so they cancel out when you add or subtract the equations.

II. By Substitution Method:
The given system of equations is:
\( x + y = 5 \) ...(1)
\( 2x - 3y = 4 \) ...(2)
From equation (1), we can express \( y \) in terms of \( x \):
\( y = 5 - x \) ...(3)
Now, we substitute this expression for \( y \) into equation (2):
\( 2x - 3(5 - x) = 4 \)
\( \implies 2x - 15 + 3x = 4 \)
\( \implies 5x - 15 = 4 \)
\( \implies 5x = 15 + 4 \)
\( \implies 5x = 19 \)
\( \implies x = \frac{19}{5} \)
Next, we substitute this value of \( x \) back into equation (3):
\( y = 5 - \frac{19}{5} \)
\( \implies y = \frac{25 - 19}{5} \)
\( \implies y = \frac{6}{5} \)
So, the solution for the given system of equations is \( x = \frac{19}{5} \) and \( y = \frac{6}{5} \).
Verification:
Substituting \( x = \frac{19}{5} \) and \( y = \frac{6}{5} \) into both original equations shows they are satisfied:
For equation (1): \( x + y = \frac{19}{5} + \frac{6}{5} = \frac{19+6}{5} = \frac{25}{5} = 5 \)
For equation (2): \( 2x - 3y = 2(\frac{19}{5}) - 3(\frac{6}{5}) = \frac{38}{5} - \frac{18}{5} = \frac{38-18}{5} = \frac{20}{5} = 4 \)
Both equations hold true, hence the solution is correct.
In simple words: In the substitution method, we picked one equation to find 'y' in terms of 'x' (or 'x' in terms of 'y'). Then, we put that expression into the other equation to solve for one variable. After that, we put the found value back into the expression for the second variable to find its value. Both methods give the same answer.

Exam Tip: Remember to always verify your solutions by plugging the values of \( x \) and \( y \) back into both original equations. This step confirms the correctness of your answer.

2. \( 3x + 4y = 10 \) and \( 2x - 2y = 2 \)

I. By Elimination Method:
The given system of equations is:
\( 3x + 4y = 10 \) ...(1)
\( 2x - 2y = 2 \) ...(2)
To make the \( y \) coefficients equal and opposite, we multiply equation (2) by 2:
\( 2(2x - 2y) = 2(2) \)
\( \implies 4x - 4y = 4 \) ...(3)
Now, we add equation (1) and equation (3):
\( (3x + 4y) + (4x - 4y) = 10 + 4 \)
\( \implies 7x = 14 \)
\( \implies x = \frac{14}{7} \)
\( \implies x = 2 \)
Next, we substitute this value of \( x \) into equation (2):
\( 2(2) - 2y = 2 \)
\( \implies 4 - 2y = 2 \)
\( \implies 2y = 4 - 2 \)
\( \implies 2y = 2 \)
\( \implies y = \frac{2}{2} \)
\( \implies y = 1 \)
Thus, the solution for the given system of equations is \( x = 2 \) and \( y = 1 \).
In simple words: We used elimination. We multiplied the second equation by 2 so the 'y' terms would cancel out when added. After adding, we found 'x', and then we put 'x' back into an equation to find 'y'.

Exam Tip: When using elimination, choose the variable that requires the simplest multiplication to match or cancel out its coefficients in both equations.

II. By Substitution Method:
The given system of equations is:
\( 3x + 4y = 10 \) ...(1)
\( 2x - 2y = 2 \) ...(2)
From equation (2), we can express \( y \) in terms of \( x \):
\( 2y = 2x - 2 \)
\( \implies y = \frac{2x - 2}{2} \)
\( \implies y = x - 1 \) ...(3)
Now, we substitute this expression for \( y \) into equation (1):
\( 3x + 4(x - 1) = 10 \)
\( \implies 3x + 4x - 4 = 10 \)
\( \implies 7x - 4 = 10 \)
\( \implies 7x = 10 + 4 \)
\( \implies 7x = 14 \)
\( \implies x = \frac{14}{7} \)
\( \implies x = 2 \)
Next, we substitute this value of \( x \) back into equation (3):
\( y = 2 - 1 \)
\( \implies y = 1 \)
So, the solution for the given system of equations is \( x = 2 \) and \( y = 1 \).
Verification:
Substituting \( x = 2 \) and \( y = 1 \) into both original equations shows they are satisfied:
For equation (1): \( 3x + 4y = 3(2) + 4(1) = 6 + 4 = 10 \)
For equation (2): \( 2x - 2y = 2(2) - 2(1) = 4 - 2 = 2 \)
Both equations hold true, hence the solution is correct.
In simple words: For substitution, we isolated 'y' from the second equation. Then we put that 'y' expression into the first equation to solve for 'x'. After finding 'x', we used it to find 'y'. We checked our work to ensure both equations were correct.

Exam Tip: Simplify equations before applying methods if possible. For example, \( 2x - 2y = 2 \) can be simplified to \( x - y = 1 \), which might make substitution easier.

3. \( 3x - 5y - 4 = 0 \) and \( 9x = 2y + 7 \)

I. By Elimination Method:
The provided set of equations is:
\( 3x - 5y - 4 = 0 \) ...(1)
\( 9x = 2y + 7 \)
We rearrange the second equation to the standard form \( Ax + By + C = 0 \):
\( 9x - 2y - 7 = 0 \) ...(2)
To make the \( x \) coefficients equal, we multiply equation (1) by 3:
\( 3(3x - 5y - 4) = 3(0) \)
\( \implies 9x - 15y - 12 = 0 \) ...(3)
Now, we subtract equation (3) from equation (2):
\( (9x - 2y - 7) - (9x - 15y - 12) = 0 - 0 \)
\( \implies 9x - 2y - 7 - 9x + 15y + 12 = 0 \)
\( \implies 13y + 5 = 0 \)
\( \implies 13y = -5 \)
\( \implies y = -\frac{5}{13} \)
Next, we substitute this value of \( y \) into equation (1):
\( 3x - 5(-\frac{5}{13}) - 4 = 0 \)
\( \implies 3x + \frac{25}{13} - 4 = 0 \)
\( \implies 3x + \frac{25 - 52}{13} = 0 \)
\( \implies 3x - \frac{27}{13} = 0 \)
\( \implies 3x = \frac{27}{13} \)
\( \implies x = \frac{27}{13 \times 3} \)
\( \implies x = \frac{9}{13} \)
Thus, the solution for the given system of equations is \( x = \frac{9}{13} \) and \( y = -\frac{5}{13} \).
In simple words: First, we put both equations into a standard form. Then, we multiplied the first equation by 3 to match the 'x' terms. Subtracting the equations helped us find 'y'. Finally, we plugged 'y' back into an equation to find 'x'.

Exam Tip: Always rearrange equations into the standard form \( Ax + By + C = 0 \) or \( Ax + By = C \) before starting to apply elimination or substitution methods to avoid errors.

II. By Substitution Method:
The given system of equations is:
\( 3x - 5y - 4 = 0 \) ...(1)
\( 9x = 2y + 7 \)
We rearrange the second equation:
\( 9x - 2y - 7 = 0 \)
From this second equation, we can express \( x \) in terms of \( y \):
\( 9x = 2y + 7 \)
\( \implies x = \frac{2y + 7}{9} \) ...(3)
Now, we substitute this expression for \( x \) into equation (1):
\( 3(\frac{2y + 7}{9}) - 5y - 4 = 0 \)
\( \implies \frac{2y + 7}{3} - 5y - 4 = 0 \)
To remove the fraction, multiply the entire equation by 3:
\( 3(\frac{2y + 7}{3}) - 3(5y) - 3(4) = 3(0) \)
\( \implies 2y + 7 - 15y - 12 = 0 \)
\( \implies -13y - 5 = 0 \)
\( \implies 13y = -5 \)
\( \implies y = -\frac{5}{13} \)
Next, we substitute this value of \( y \) back into equation (3):
\( x = \frac{2(-\frac{5}{13}) + 7}{9} \)
\( \implies x = \frac{-\frac{10}{13} + \frac{7 \times 13}{13}}{9} \)
\( \implies x = \frac{\frac{-10 + 91}{13}}{9} \)
\( \implies x = \frac{\frac{81}{13}}{9} \)
\( \implies x = \frac{81}{13 \times 9} \)
\( \implies x = \frac{9}{13} \)
So, the solution for the given system of equations is \( x = \frac{9}{13} \) and \( y = -\frac{5}{13} \).
Verification:
Substituting \( x = \frac{9}{13} \) and \( y = -\frac{5}{13} \) into both original equations shows they are satisfied:
For equation (1): \( 3x - 5y - 4 = 3(\frac{9}{13}) - 5(-\frac{5}{13}) - 4 = \frac{27}{13} + \frac{25}{13} - 4 = \frac{52}{13} - 4 = 4 - 4 = 0 \)
For equation (2): \( 9x = 2y + 7 \implies 9(\frac{9}{13}) = 2(-\frac{5}{13}) + 7 \implies \frac{81}{13} = -\frac{10}{13} + \frac{91}{13} \implies \frac{81}{13} = \frac{81}{13} \)
Both equations hold true, hence the solution is correct.
In simple words: We used the substitution approach. We rearranged the second equation to get 'x' by itself. Then, we put that expression for 'x' into the first equation and solved for 'y'. After finding 'y', we used it to calculate 'x'. The verification showed both equations were correct with these values.

Exam Tip: When dealing with fractions in equations, it is often helpful to multiply the entire equation by the least common multiple of the denominators to clear the fractions and simplify calculations.

4. \( \frac{x}{2}+\frac{2 y}{3} = -1 \) and \( x - \frac{y}{3} = 3 \)

I. By Elimination Method:
The provided set of equations is:
\( \frac{x}{2}+\frac{2 y}{3} = -1 \)
To clear fractions, multiply by 6: \( 3x + 4y = -6 \) ...(1)
\( x - \frac{y}{3} = 3 \)
To clear fractions, multiply by 3: \( 3x - y = 9 \) ...(2)
To eliminate \( x \), we can subtract equation (2) from equation (1):
\( (3x + 4y) - (3x - y) = -6 - 9 \)
\( \implies 3x + 4y - 3x + y = -15 \)
\( \implies 5y = -15 \)
\( \implies y = \frac{-15}{5} \)
\( \implies y = -3 \)
Next, we substitute this value of \( y \) into equation (2):
\( 3x - (-3) = 9 \)
\( \implies 3x + 3 = 9 \)
\( \implies 3x = 9 - 3 \)
\( \implies 3x = 6 \)
\( \implies x = \frac{6}{3} \)
\( \implies x = 2 \)
Thus, the solution for the given system of equations is \( x = 2 \) and \( y = -3 \).
In simple words: First, we cleared all the fractions from both equations. Then, we subtracted the second equation from the first to eliminate 'x' and found 'y'. Finally, we put 'y' back into one of the simpler equations to find 'x'.

Exam Tip: It is usually best to eliminate fractions by multiplying by the least common multiple (LCM) of the denominators before proceeding with either elimination or substitution methods. This helps to prevent calculation errors.

II. By Substitution Method:
The given system of equations is:
\( \frac{x}{2}+\frac{2 y}{3} = -1 \)
To clear fractions, multiply by 6: \( 3x + 4y = -6 \) ...(1)
\( x - \frac{y}{3} = 3 \)
To clear fractions, multiply by 3: \( 3x - y = 9 \) ...(2)
From equation (2), we can express \( y \) in terms of \( x \):
\( y = 3x - 9 \) ...(3)
Now, we substitute this expression for \( y \) into equation (1):
\( 3x + 4(3x - 9) = -6 \)
\( \implies 3x + 12x - 36 = -6 \)
\( \implies 15x - 36 = -6 \)
\( \implies 15x = -6 + 36 \)
\( \implies 15x = 30 \)
\( \implies x = \frac{30}{15} \)
\( \implies x = 2 \)
Next, we substitute this value of \( x \) back into equation (3):
\( y = 3(2) - 9 \)
\( \implies y = 6 - 9 \)
\( \implies y = -3 \)
So, the solution for the given system of equations is \( x = 2 \) and \( y = -3 \).
Verification:
Substituting \( x = 2 \) and \( y = -3 \) into the original equations shows they are satisfied:
For equation (1): \( \frac{x}{2}+\frac{2 y}{3} = \frac{2}{2}+\frac{2(-3)}{3} = 1 - 2 = -1 \)
For equation (2): \( x - \frac{y}{3} = 2 - \frac{(-3)}{3} = 2 + 1 = 3 \)
Both equations hold true, hence the solution is correct. The substitution method was most efficient in this case.
In simple words: We first got rid of all the fractions. Then, we rearranged the second equation to get 'y' by itself. We put this 'y' expression into the first equation to solve for 'x'. After finding 'x', we used it to get 'y'. We checked the answer, and it was right.

Exam Tip: When equations involve fractions, converting them into integer coefficients first often makes the substitution method much simpler and less prone to errors.

Question 2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
1. If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \( \frac{1}{2} \) if we only add 1 to the denominator. What is the fraction?
2. Five years ago, Nuri was thrice as old as Sonu. Ten years later. Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
3. The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the number. Find the number.
4. Meena went to a bank to withdraw Rs. 2000. She asked the cashier to give her Rs. 50 and Rs. 100 notes only. Meena got 25 notes in all. Find how many notes of Rs. 50 and Rs. 100 she received.
5. A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs. 27 for a book kept for seven days, while Susy paid Rs. 21 for the book she kept five days. Find the fixed charge and the charge for each extra day.
Answer:

1. What is the fraction?
Let the fraction be \( \frac{x}{y} \).
According to the first condition, if we add 1 to the numerator and subtract 1 from the denominator, the fraction becomes 1:
\( \frac{x+1}{y-1} = 1 \)
\( \implies x + 1 = y - 1 \)
\( \implies x - y = -2 \) ...(1)
According to the second condition, if we only add 1 to the denominator, the fraction becomes \( \frac{1}{2} \):
\( \frac{x}{y+1} = \frac{1}{2} \)
\( \implies 2x = y + 1 \)
\( \implies 2x - y = 1 \) ...(2)
Now, we solve equations (1) and (2) by the elimination method.
Subtract equation (1) from equation (2):
\( (2x - y) - (x - y) = 1 - (-2) \)
\( \implies 2x - y - x + y = 1 + 2 \)
\( \implies x = 3 \)
Next, we substitute this value of \( x \) into equation (1):
\( 3 - y = -2 \)
\( \implies y = 3 + 2 \)
\( \implies y = 5 \)
Therefore, the required fraction is \( \frac{3}{5} \).
Verification:
Substituting \( x = 3 \) and \( y = 5 \) into the conditions:
1. \( \frac{3+1}{5-1} = \frac{4}{4} = 1 \)
2. \( \frac{3}{5+1} = \frac{3}{6} = \frac{1}{2} \)
Both conditions are satisfied, so the solution is correct.
In simple words: We set up two equations based on the information given about the fraction. By subtracting the first equation from the second, we found 'x'. Then, we used 'x' to find 'y'. The fraction is \( \frac{3}{5} \).

Exam Tip: For problems involving fractions, always define the numerator and denominator as separate variables, then form equations based on the given conditions. This simplifies the process of setting up and solving.

2. How old are Nuri and Sonu?
Let Nuri's current age be \( x \) years and Sonu's current age be \( y \) years.
According to the first condition (five years ago):
Nuri's age 5 years ago \( = x - 5 \)
Sonu's age 5 years ago \( = y - 5 \)
Nuri was thrice as old as Sonu:
\( x - 5 = 3(y - 5) \)
\( \implies x - 5 = 3y - 15 \)
\( \implies x - 3y = -10 \) ...(1)
According to the second condition (ten years later):
Nuri's age 10 years later \( = x + 10 \)
Sonu's age 10 years later \( = y + 10 \)
Nuri will be twice as old as Sonu:
\( x + 10 = 2(y + 10) \)
\( \implies x + 10 = 2y + 20 \)
\( \implies x - 2y = 10 \) ...(2)
Now, we solve equations (1) and (2) by the elimination method.
Subtract equation (2) from equation (1):
\( (x - 3y) - (x - 2y) = -10 - 10 \)
\( \implies x - 3y - x + 2y = -20 \)
\( \implies -y = -20 \)
\( \implies y = 20 \)
Next, we substitute this value of \( y \) into equation (2):
\( x - 2(20) = 10 \)
\( \implies x - 40 = 10 \)
\( \implies x = 40 + 10 \)
\( \implies x = 50 \)
Therefore, Nuri is currently 50 years old and Sonu is currently 20 years old.
Verification:
Substituting \( x = 50 \) and \( y = 20 \) into the conditions:
1. Five years ago: Nuri was \( 50 - 5 = 45 \), Sonu was \( 20 - 5 = 15 \). \( 45 = 3 \times 15 \) (True)
2. Ten years later: Nuri will be \( 50 + 10 = 60 \), Sonu will be \( 20 + 10 = 30 \). \( 60 = 2 \times 30 \) (True)
Both conditions are satisfied, so the solution is correct.
In simple words: We set up equations for Nuri and Sonu's ages based on the past and future information. By subtracting the second equation from the first, we found Sonu's age. Then, we used Sonu's age to find Nuri's age. Nuri is 50 and Sonu is 20.

Exam Tip: For age-related problems, always define variables for the current ages, then adjust them for past or future scenarios to form your equations correctly.

3. Find the number.
Let the unit's digit be \( x \) and the ten's digit be \( y \).
Then, the original two-digit number is \( 10y + x \).
The number obtained by reversing the order of the digits is \( 10x + y \).
According to the first condition, the sum of the digits is 9:
\( x + y = 9 \) ...(1)
According to the second condition, nine times the original number is twice the number obtained by reversing the digits:
\( 9(10y + x) = 2(10x + y) \)
\( \implies 90y + 9x = 20x + 2y \)
\( \implies 90y - 2y = 20x - 9x \)
\( \implies 88y = 11x \)
\( \implies 11x - 88y = 0 \)
Dividing by 11:
\( \implies x - 8y = 0 \) ...(2)
Now, we solve equations (1) and (2) by the elimination method.
Subtract equation (2) from equation (1):
\( (x + y) - (x - 8y) = 9 - 0 \)
\( \implies x + y - x + 8y = 9 \)
\( \implies 9y = 9 \)
\( \implies y = \frac{9}{9} \)
\( \implies y = 1 \)
Next, we substitute this value of \( y \) into equation (1):
\( x + 1 = 9 \)
\( \implies x = 9 - 1 \)
\( \implies x = 8 \)
The unit's digit is 8 and the ten's digit is 1. Therefore, the required number is \( 10(1) + 8 = 18 \).
Verification:
Substituting \( x = 8 \) and \( y = 1 \) into the conditions:
1. Sum of digits: \( 8 + 1 = 9 \) (True)
2. Nine times the number is twice the reversed number: Original number = 18, Reversed number = 81. \( 9 \times 18 = 162 \), \( 2 \times 81 = 162 \). \( 162 = 162 \) (True)
Both conditions are satisfied, so the solution is correct.
In simple words: We set up equations based on the sum of the digits and the relationship between the original number and its reversed form. We eliminated 'x' to find 'y', then used 'y' to find 'x'. The number is 18.

Exam Tip: When dealing with two-digit number problems, remember that a number with a ten's digit \( y \) and a unit's digit \( x \) is represented as \( 10y + x \), not \( yx \).

4. Find how many notes of Rs. 50 and Rs. 100 she received.
Let Meena receive \( x \) notes of Rs. 50 and \( y \) notes of Rs. 100.
According to the first condition, Meena received 25 notes in all:
\( x + y = 25 \) ...(1)
According to the second condition, the total amount withdrawn was Rs. 2000:
\( 50x + 100y = 2000 \)
To simplify, we can divide the entire equation by 50:
\( \frac{50x}{50} + \frac{100y}{50} = \frac{2000}{50} \)
\( \implies x + 2y = 40 \) ...(2)
Now, we solve equations (1) and (2) by the elimination method.
Subtract equation (1) from equation (2):
\( (x + 2y) - (x + y) = 40 - 25 \)
\( \implies x + 2y - x - y = 15 \)
\( \implies y = 15 \)
Next, we substitute this value of \( y \) into equation (1):
\( x + 15 = 25 \)
\( \implies x = 25 - 15 \)
\( \implies x = 10 \)
Therefore, Meena received 10 notes of Rs. 50 and 15 notes of Rs. 100.
Verification:
Substituting \( x = 10 \) and \( y = 15 \) into the conditions:
1. Total number of notes: \( 10 + 15 = 25 \) (True)
2. Total amount: \( 50(10) + 100(15) = 500 + 1500 = 2000 \) (True)
Both conditions are satisfied, so the solution is correct.
In simple words: We made two equations: one for the total number of notes and one for the total amount of money. By subtracting the first equation from the second, we found the number of Rs. 100 notes. Then, we used that to find the number of Rs. 50 notes.

Exam Tip: In word problems involving quantities and values, create one equation for the total number of items and another for the total value to successfully set up the system of linear equations.

5. Find the fixed charge and the charge for each extra day.
Let the fixed charge for the first three days be Rs. \( a \).
Let the additional charge for each extra day be Rs. \( b \).
According to the first condition, Saritha paid Rs. 27 for a book kept for seven days.
Extra days \( = 7 - 3 = 4 \)
So, \( a + 4b = 27 \) ...(1)
According to the second condition, Susy paid Rs. 21 for the book she kept five days.
Extra days \( = 5 - 3 = 2 \)
So, \( a + 2b = 21 \) ...(2)
Now, we solve equations (1) and (2) by the elimination method.
Subtract equation (2) from equation (1):
\( (a + 4b) - (a + 2b) = 27 - 21 \)
\( \implies a + 4b - a - 2b = 6 \)
\( \implies 2b = 6 \)
\( \implies b = \frac{6}{2} \)
\( \implies b = 3 \)
Next, we substitute this value of \( b \) into equation (2):
\( a + 2(3) = 21 \)
\( \implies a + 6 = 21 \)
\( \implies a = 21 - 6 \)
\( \implies a = 15 \)
Therefore, the fixed charge is Rs. 15 and the charge for each extra day is Rs. 3.
Verification:
Substituting \( a = 15 \) and \( b = 3 \) into the conditions:
1. Saritha's payment: \( a + 4b = 15 + 4(3) = 15 + 12 = 27 \) (True)
2. Susy's payment: \( a + 2b = 15 + 2(3) = 15 + 6 = 21 \) (True)
Both conditions are satisfied, so the solution is correct.
In simple words: We set up two equations for the library charges: one for Saritha and one for Susy. By subtracting the second equation from the first, we found the cost for each extra day. Then, we used that value to find the fixed charge.

Exam Tip: In problems involving fixed charges and additional per-unit charges, carefully identify the 'base' amount (fixed charge) and the 'variable' amount (additional charge per unit) to build your equations accurately.