GSEB Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3

Question 1. Solve the following pair of equations by substitution method.
1. \( x + y = 14 \), \( x - y = 4 \)
2. \( s - t = 3 \), \( \frac{s}{3}+\frac{t}{2}=6 \)
3. \( 3x - y = 3 \), \( 9x - 3y = 9 \)
4. \( 0.2x + 0.3y = 1.3 \), \( 0.4x + 0.5y = 2.3 \)
5. \( \sqrt{2} x+\sqrt{3} y=0 \), \( \sqrt{3} x-\sqrt{8} y=0 \)
6. \( \frac{3 x}{2}-\frac{5 y}{3}=-2 \), \( \frac{x}{3}+\frac{y}{2}=\frac{13}{6} \)
Answer:
1. The given pair of linear equations is:
\( x + y = 14 \) .....(1)
\( x - y = 4 \) .....(2)
From equation (1), we can find y:
\( y = 14 - x \) ...(3)
Substitute this value of y into equation (2):
\( x - (14 - x) = 4 \)
\( \implies x - 14 + x = 4 \)
\( \implies 2x - 14 = 4 \)
\( \implies 2x = 4 + 14 \)
\( \implies 2x = 18 \)
\( \implies x = \frac{18}{2} \)
\( \implies x = 9 \)
Now, substitute this value of x back into equation (3):
\( y = 14 - 9 \)
\( \implies y = 5 \)
Therefore, the solution is \( x = 9, y = 5 \).
Verification: By substituting \( x = 9 \) and \( y = 5 \), we observe that both equations (1) and (2) are satisfied:
\( x + y = 9 + 5 = 14 \)
\( x - y = 9 - 5 = 4 \)
This confirms the Solution.
In simple words: First, we change one equation to express one variable using the other. Then, we put that expression into the second equation to find the value of the first variable. Finally, we use that value to find the second variable. We check our answers by putting both numbers back into the original equations.
2. The given pair of linear equations is:
\( s - t = 3 \) .....(1)
\( \frac{s}{3}+\frac{t}{2}=6 \) .....(2)
From equation (1), we can find s:
\( s = t + 3 \) .....(3)
Substitute this value of s into equation (2):
\( \frac{t+3}{3}+\frac{t}{2}=6 \)
To remove the denominators, multiply the entire equation by the least common multiple of 3 and 2, which is 6:
\( \implies 2(t+3) + 3t = 36 \)
\( \implies 2t + 6 + 3t = 36 \)
\( \implies 5t + 6 = 36 \)
\( \implies 5t = 36 - 6 \)
\( \implies 5t = 30 \)
\( \implies t = \frac{30}{5} \)
\( \implies t = 6 \)
Now, substitute this value of t back into equation (3):
\( s = 6 + 3 \)
\( \implies s = 9 \)
Therefore, the solution is \( s = 9, t = 6 \).
Verification: By substituting \( s = 9 \) and \( t = 6 \), we observe that both equations (1) and (2) are satisfied:
\( s - t = 9 - 6 = 3 \)
\( \frac{s}{3}+\frac{t}{2}=\frac{9}{3}+\frac{6}{2} = 3 + 3 = 6 \)
This confirms the solution.
In simple words: Similar to the previous problem, we isolate one variable from the first equation and substitute it into the second equation. After simplifying and solving, we get the values for both variables, and then we check if they work in the original equations.
3. The given pair of linear equations is:
\( 3x - y = 3 \) .....(1)
\( 9x - 3y = 9 \) .....(2)
From equation (1), we can find y:
\( y = 3x - 3 \) .....(3)
Substitute this value of y into equation (2):
\( 9x - 3(3x - 3) = 9 \)
\( \implies 9x - 9x + 9 = 9 \)
\( \implies 9 = 9 \)
This statement is true. Therefore, equations (1) and (2) have infinitely many solutions, meaning they represent the same line.
In simple words: When you try to solve these equations, you end up with a true statement, like 9 equals 9. This shows that both equations are essentially the same, so there are many, many answers that work for them.
4. The given pair of linear equations is:
\( 0.2x + 0.3y = 1.3 \) .....(1)
\( 0.4x + 0.5y = 2.3 \) .....(2)
From equation (1), we can find y:
\( 0.3y = 1.3 - 0.2x \)
\( \implies y = \frac{1.3-0.2 x}{0.3} \) .....(3)
Substitute this value of y into equation (2):
\( 0.4x + 0.5\left(\frac{1.3-0.2 x}{0.3}\right) = 2.3 \)
To remove the decimal from the denominator, multiply the term by 10/10. To remove all other decimals, multiply the entire equation by 10 or 100 as needed. Let's first simplify the fraction inside the parentheses:
\( 0.4x + \frac{0.5 \times (1.3-0.2x)}{0.3} = 2.3 \)
\( 0.4x + \frac{0.65 - 0.1x}{0.3} = 2.3 \)
Multiply the whole equation by 0.3 to get rid of the denominator:
\( \implies 0.3(0.4x) + (0.65 - 0.1x) = 0.3(2.3) \)
\( \implies 0.12x + 0.65 - 0.1x = 0.69 \)
\( \implies 0.12x - 0.1x = 0.69 - 0.65 \)
\( \implies 0.02x = 0.04 \)
\( \implies x = \frac{0.04}{0.02} \)
\( \implies x = 2 \)
Now, substitute this value of x back into equation (3):
\( y = \frac{1.3-0.2(2)}{0.3} \)
\( \implies y = \frac{1.3-0.4}{0.3} \)
\( \implies y = \frac{0.9}{0.3} \)
\( \implies y = 3 \)
Therefore, the solution is \( x = 2, y = 3 \).
Verification: By substituting \( x = 2 \) and \( y = 3 \), we observe that both equations (1) and (2) are satisfied:
\( 0.2x + 0.3y = (0.2)(2) + (0.3)(3) = 0.4 + 0.9 = 1.3 \)
\( 0.4x + 0.5y = (0.4)(2) + (0.5)(3) = 0.8 + 1.5 = 2.3 \)
This confirms the solution.
In simple words: We deal with equations that have decimals by using substitution. We solve for one variable, place it into the other equation, and then simplify the decimals by multiplying to get whole numbers. After finding both values, we double-check them with the original equations.
5. The given pair of linear equations is:
\( \sqrt{2} x+\sqrt{3} y=0 \) .....(1)
\( \sqrt{3} x-\sqrt{8} y=0 \) .....(2)
From equation (2), we can find x:
\( \sqrt{3} x = \sqrt{8} y \)
\( \implies x = \frac{\sqrt{8}}{\sqrt{3}} y \) .....(3)
Substitute this value of x into equation (1):
\( \sqrt{2} \left(\frac{\sqrt{8}}{\sqrt{3}} y\right) + \sqrt{3} y = 0 \)
\( \implies \frac{\sqrt{16}}{\sqrt{3}} y + \sqrt{3} y = 0 \)
\( \implies \frac{4}{\sqrt{3}} y + \sqrt{3} y = 0 \)
Factor out y:
\( \implies \left(\frac{4}{\sqrt{3}} + \sqrt{3}\right) y = 0 \)
Since \( \left(\frac{4}{\sqrt{3}} + \sqrt{3}\right) \ne 0 \) (because it's a sum of positive numbers), for the product to be zero, y must be zero.
\( \implies y = 0 \)
Now, substitute this value of y back into equation (3):
\( x = \frac{\sqrt{8}}{\sqrt{3}}(0) \)
\( \implies x = 0 \)
Therefore, the solution is \( x = 0, y = 0 \).
Verification: By substituting \( x = 0 \) and \( y = 0 \), we observe that both equations (1) and (2) are satisfied:
\( \sqrt{2}x + \sqrt{3}y = \sqrt{2}(0) + \sqrt{3}(0) = 0 \)
\( \sqrt{3}x - \sqrt{8}y = \sqrt{3}(0) - \sqrt{8}(0) = 0 \)
This confirms the solution.
In simple words: When dealing with square roots, we isolate one variable from an equation and then substitute it into the other. If the coefficients of the variable that remains don't add up to zero, then the variable itself must be zero. This gives us zero for both variables, which we then verify.
6. The given system of linear equations is:
\( \frac{3 x}{2}-\frac{5 y}{3}=-2 \) .....(1)
\( \frac{x}{3}+\frac{y}{2}=\frac{13}{6} \) .....(2)
To clear fractions, multiply equation (1) by 6 and equation (2) by 6:
For equation (1): \( 6 \left(\frac{3x}{2} - \frac{5y}{3}\right) = 6(-2) \)
\( \implies 9x - 10y = -12 \) .....(3)
For equation (2): \( 6 \left(\frac{x}{3} + \frac{y}{2}\right) = 6\left(\frac{13}{6}\right) \)
\( \implies 2x + 3y = 13 \) .....(4)
From equation (3), we can find x:
\( 9x = 10y - 12 \)
\( \implies x = \frac{10y-12}{9} \) .....(5)
Substitute the value of x from equation (5) into equation (4):
\( 2\left(\frac{10y-12}{9}\right) + 3y = 13 \)
Multiply the entire equation by 9 to eliminate the denominator:
\( \implies 2(10y-12) + 9(3y) = 9(13) \)
\( \implies 20y - 24 + 27y = 117 \)
\( \implies 47y - 24 = 117 \)
\( \implies 47y = 117 + 24 \)
\( \implies 47y = 141 \)
\( \implies y = \frac{141}{47} \)
\( \implies y = 3 \)
Now, substitute the value of y back into equation (5):
\( x = \frac{10(3)-12}{9} \)
\( \implies x = \frac{30-12}{9} \)
\( \implies x = \frac{18}{9} \)
\( \implies x = 2 \)
Therefore, the solution is \( x = 2, y = 3 \).
Verification: By substituting \( x = 2 \) and \( y = 3 \), we observe that both original equations are satisfied:
For equation (1): \( \frac{3x}{2} - \frac{5y}{3} = \frac{3(2)}{2} - \frac{5(3)}{3} = \frac{6}{2} - \frac{15}{3} = 3 - 5 = -2 \)
For equation (2): \( \frac{x}{3} + \frac{y}{2} = \frac{2}{3} + \frac{3}{2} = \frac{4}{6} + \frac{9}{6} = \frac{13}{6} \)
This confirms the solution.
In simple words: First, we clear the fractions by multiplying each equation by its least common denominator. Then, we use the substitution method as before: solve for one variable, put that into the other equation, find the values, and then verify them.

Exam Tip: Always clear fractions in linear equations by multiplying by the LCM of the denominators before attempting substitution; it greatly simplifies the process.

Question 2. Solve \( 2x + 3y = 11 \) and \( 2x - 4y = -24 \) and hence find the value of 'm' for which \( y = mx + 3 \).
Answer: The given pair of linear equations is:
\( 2x + 3y = 11 \) .....(1)
\( 2x - 4y = -24 \) .....(2)
From equation (1), we can express 2x:
\( 2x = 11 - 3y \)
Substitute this expression for 2x into equation (2):
\( (11 - 3y) - 4y = -24 \)
\( \implies 11 - 7y = -24 \)
\( \implies -7y = -24 - 11 \)
\( \implies -7y = -35 \)
\( \implies y = \frac{-35}{-7} \)
\( \implies y = 5 \)
Now, substitute this value of y back into the expression for 2x from equation (1):
\( 2x = 11 - 3(5) \)
\( \implies 2x = 11 - 15 \)
\( \implies 2x = -4 \)
\( \implies x = \frac{-4}{2} \)
\( \implies x = -2 \)
Verification: By substituting \( x = -2 \) and \( y = 5 \), we find that both equations (1) and (2) are satisfied:
\( 2x + 3y = 2(-2) + 3(5) = -4 + 15 = 11 \)
\( 2x - 4y = 2(-2) - 4(5) = -4 - 20 = -24 \)
This verifies the solution.
Now, to find the value of 'm' using \( y = mx + 3 \), substitute the values \( x = -2 \) and \( y = 5 \):
\( 5 = m(-2) + 3 \)
\( \implies 5 = -2m + 3 \)
\( \implies -2m = 5 - 3 \)
\( \implies -2m = 2 \)
\( \implies m = \frac{2}{-2} \)
\( \implies m = -1 \)
In simple words: First, solve the two given equations for 'x' and 'y' using the substitution method. After finding those values, plug them into the third equation, \( y = mx + 3 \), to calculate the value of 'm'.

Exam Tip: Remember to always verify the solution of the system of equations before using the values in subsequent calculations; it helps catch errors early.

Question 3. Form the pair of linear equations in the following problems and find their solution by substitution method:
1. The difference between two numbers is 26 and one number is three times the other. Find them.
2. The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
3. The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.
4. The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per kilometre? How much does a person have to pay for travelling a distance of 25 km?
5. A fraction becomes \( \frac{9}{11} \), if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes \( \frac{5}{6} \), find the fraction. [CBSE]
6. Five years hence, the age of Jacob will be three times that of his son, five years ago, Jacob's age was seven times that of his son. What are their present age?
Answer:
1. Let the two numbers be \( x \) and \( y \), where \( x > y \). According to the question, the pair of linear equations formed is:
\( x - y = 26 \) .....(1)
\( x = 3y \) .....(2)
Substitute the value of x from equation (2) into equation (1):
\( 3y - y = 26 \)
\( \implies 2y = 26 \)
\( \implies y = \frac{26}{2} \)
\( \implies y = 13 \)
Now, substitute this value of y back into equation (2):
\( x = 3(13) \)
\( \implies x = 39 \)
Hence, the required numbers are 39 and 13.
Verification: By substituting \( x = 39 \) and \( y = 13 \), we find that both equations (1) and (2) are satisfied:
\( x - y = 39 - 13 = 26 \)
\( 3y = 3(13) = 39 = x \)
This verifies the solution.
In simple words: We set up two equations based on the problem's details: one for the difference between the numbers and one showing that one number is three times the other. By replacing one variable with its expression from the other equation, we solve for both numbers and then check our answer.
2. Let the larger of the two supplementary angles be \( x^\circ \) and the smaller be \( y^\circ \).
According to the question, the pair of linear equations formed is:
\( x^\circ = y^\circ + 18^\circ \) .....(1)
\( x^\circ + y^\circ = 180^\circ \) (Since the two angles are supplementary) .....(2)
Substitute the value of \( x^\circ \) from equation (1) into equation (2):
\( (y^\circ + 18^\circ) + y^\circ = 180^\circ \)
\( \implies 2y^\circ + 18^\circ = 180^\circ \)
\( \implies 2y^\circ = 180^\circ - 18^\circ \)
\( \implies 2y^\circ = 162^\circ \)
\( \implies y^\circ = \frac{162^\circ}{2} \)
\( \implies y^\circ = 81^\circ \)
Now, substitute this value of \( y^\circ \) back into equation (1):
\( x^\circ = 81^\circ + 18^\circ \)
\( \implies x^\circ = 99^\circ \)
Hence, the larger and the smaller of the two supplementary angles are \( 99^\circ \) and \( 81^\circ \) respectively.
Verification: By substituting \( x^\circ = 99^\circ \) and \( y^\circ = 81^\circ \), we find that both equations (1) and (2) are satisfied:
\( x^\circ = 81^\circ + 18^\circ = 99^\circ \)
\( x^\circ + y^\circ = 99^\circ + 81^\circ = 180^\circ \)
This verifies the solution.
In simple words: We know supplementary angles add up to 180 degrees. We also know one angle is 18 degrees more than the other. We use these two facts to create equations, solve for both angles, and then make sure they meet the original conditions.
3. Let the cost of each bat be Rs \( x \) and the cost of each ball be Rs \( y \).
According to the question, the pair of linear equations formed is:
\( 7x + 6y = 3800 \) .....(1)
\( 3x + 5y = 1750 \) .....(2)
From equation (2), we can find y:
\( 5y = 1750 - 3x \)
\( \implies y = \frac{1750-3x}{5} \) .....(3)
Substitute this value of y into equation (1):
\( 7x + 6\left(\frac{1750-3x}{5}\right) = 3800 \)
Multiply the entire equation by 5 to eliminate the denominator:
\( \implies 5(7x) + 6(1750-3x) = 5(3800) \)
\( \implies 35x + 10500 - 18x = 19000 \)
\( \implies 17x + 10500 = 19000 \)
\( \implies 17x = 19000 - 10500 \)
\( \implies 17x = 8500 \)
\( \implies x = \frac{8500}{17} \)
\( \implies x = 500 \)
Now, substitute this value of x back into equation (3):
\( y = \frac{1750-3(500)}{5} \)
\( \implies y = \frac{1750-1500}{5} \)
\( \implies y = \frac{250}{5} \)
\( \implies y = 50 \)
Hence, the cost of each bat is Rs 500 and the cost of each ball is Rs 50.
Verification: By substituting \( x = 500 \) and \( y = 50 \), we find that both equations (1) and (2) are satisfied:
\( 7x + 6y = 7(500) + 6(50) = 3500 + 300 = 3800 \)
\( 3x + 5y = 3(500) + 5(50) = 1500 + 250 = 1750 \)
This verifies the solution.
In simple words: We set up equations for the cost of bats and balls for two different purchases. By expressing one variable (like ball cost) in terms of the other (bat cost) and plugging it into the second equation, we find the individual costs of bats and balls. Then, we check if these costs match the original purchase amounts.
4. Let the fixed charges be Rs \( x \) and the charge per kilometre be Rs \( y \).
According to the question, the pair of linear equations formed is:
For 10 km journey: \( x + 10y = 105 \) .....(1)
For 15 km journey: \( x + 15y = 155 \) .....(2)
From equation (1), we can find x:
\( x = 105 - 10y \) .....(3)
Substitute this value of x into equation (2):
\( (105 - 10y) + 15y = 155 \)
\( \implies 105 + 5y = 155 \)
\( \implies 5y = 155 - 105 \)
\( \implies 5y = 50 \)
\( \implies y = \frac{50}{5} \)
\( \implies y = 10 \)
Now, substitute this value of y back into equation (3):
\( x = 105 - 10(10) \)
\( \implies x = 105 - 100 \)
\( \implies x = 5 \)
Hence, the fixed charges are Rs 5 and the charge per kilometre is Rs 10.
Verification: By substituting \( x = 5 \) and \( y = 10 \), we find that both equations (1) and (2) are satisfied:
\( x + 10y = 5 + 10(10) = 5 + 100 = 105 \)
\( x + 15y = 5 + 15(10) = 5 + 150 = 155 \)
This verifies the solution.
Now, for travelling a distance of 25 km, a person will have to pay:
Total charge = Fixed charge + (Charge per km \( \times \) Distance)
Total charge = \( x + 25y \)
Total charge = \( 5 + 25(10) \)
Total charge = \( 5 + 250 \)
Total charge = Rs 255
In simple words: We set up equations for taxi fares, where a fixed fee is added to a per-kilometer charge. By solving these, we find the fixed charge and the per-kilometer rate. Then, we use these rates to figure out the total cost for a 25 km trip.
5. Let the fraction be \( \frac{x}{y} \).
According to the question, the pair of linear equations formed is:
When 2 is added to both numerator and denominator:
\( \frac{x+2}{y+2}=\frac{9}{11} \)
\( \implies 11(x + 2) = 9(y + 2) \)
\( \implies 11x + 22 = 9y + 18 \)
\( \implies 11x - 9y = 18 - 22 \)
\( \implies 11x - 9y = -4 \) .....(1)
When 3 is added to both numerator and denominator:
\( \frac{x+3}{y+3}=\frac{5}{6} \)
\( \implies 6(x + 3) = 5(y + 3) \)
\( \implies 6x + 18 = 5y + 15 \)
\( \implies 6x - 5y = 15 - 18 \)
\( \implies 6x - 5y = -3 \) .....(2)
From equation (1), we can find x:
\( 11x = 9y - 4 \)
\( \implies x = \frac{9y-4}{11} \) .....(3)
Substitute this value of x into equation (2):
\( 6\left(\frac{9y-4}{11}\right) - 5y = -3 \)
Multiply the entire equation by 11 to eliminate the denominator:
\( \implies 6(9y-4) - 11(5y) = 11(-3) \)
\( \implies 54y - 24 - 55y = -33 \)
\( \implies -y - 24 = -33 \)
\( \implies -y = -33 + 24 \)
\( \implies -y = -9 \)
\( \implies y = 9 \)
Now, substitute this value of y back into equation (3):
\( x = \frac{9(9)-4}{11} \)
\( \implies x = \frac{81-4}{11} \)
\( \implies x = \frac{77}{11} \)
\( \implies x = 7 \)
Hence, the required fraction is \( \frac{7}{9} \).
Verification: By substituting \( x = 7 \) and \( y = 9 \), we find that both initial conditions for the equations are satisfied:
Adding 2: \( \frac{x+2}{y+2} = \frac{7+2}{9+2} = \frac{9}{11} \)
Adding 3: \( \frac{x+3}{y+3} = \frac{7+3}{9+3} = \frac{10}{12} = \frac{5}{6} \)
This verifies the solution.
In simple words: We set up two equations based on how the fraction changes when you add different numbers to its top and bottom. We then solve these equations for the numerator and denominator using substitution to find the original fraction, and check it with the given rules.
6. Let the present age of Jacob be \( x \) years and his son's present age be \( y \) years.
According to the question, the pair of linear equations formed is:
Five years hence (in the future):
Jacob's age: \( x + 5 \)
Son's age: \( y + 5 \)
Condition: \( x + 5 = 3(y + 5) \)
\( \implies x + 5 = 3y + 15 \)
\( \implies x - 3y = 15 - 5 \)
\( \implies x - 3y = 10 \) .....(1)
Five years ago (in the past):
Jacob's age: \( x - 5 \)
Son's age: \( y - 5 \)
Condition: \( x - 5 = 7(y - 5) \)
\( \implies x - 5 = 7y - 35 \)
\( \implies x - 7y = -35 + 5 \)
\( \implies x - 7y = -30 \) .....(2)
From equation (1), we can find x:
\( x = 3y + 10 \) .....(3)
Substitute this value of x into equation (2):
\( (3y + 10) - 7y = -30 \)
\( \implies 10 - 4y = -30 \)
\( \implies -4y = -30 - 10 \)
\( \implies -4y = -40 \)
\( \implies y = \frac{-40}{-4} \)
\( \implies y = 10 \)
Now, substitute this value of y back into equation (3):
\( x = 3(10) + 10 \)
\( \implies x = 30 + 10 \)
\( \implies x = 40 \)
Hence, the present age of Jacob is 40 years and his son's present age is 10 years.
Verification: By substituting \( x = 40 \) and \( y = 10 \), we find that both equations (1) and (2) are satisfied:
Five years hence: \( x + 5 = 40 + 5 = 45 \). \( 3(y+5) = 3(10+5) = 3(15) = 45 \). (Matches)
Five years ago: \( x - 5 = 40 - 5 = 35 \). \( 7(y-5) = 7(10-5) = 7(5) = 35 \). (Matches)
This verifies the solution.
In simple words: We use information about Jacob's and his son's ages five years in the future and five years in the past to set up two equations. By solving these equations for 'x' and 'y', we find their current ages, and then we confirm that these ages fit the given conditions.

Exam Tip: For age problems, carefully translate "hence" (future) and "ago" (past) into mathematical expressions, always checking both conditions after finding the ages.