Kaliyuga Ekalavya

NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.2 Question 1(ii) Question: Solve the following pair of linear equations by the substitution method: s – t = 3 and s 3 + t 2 = 6 Given: The given equations are: s – t = 3 ---(1) s 3 + t 2 = 6 ---(2) To Find: We need to find the values of s and t using the substitution method. Formula: Substitution method involves solving one equation for one variable and substituting it into the other equation. Solution: From equation (1), we have s = t + 3. Substitute s = t + 3 into equation (2): t+3 3 + t 2 = 6 Multiplying both sides by 6 to eliminate fractions: 2(t+3) + 3t = 36 ⇒ 2t + 6 + 3t = 36 ⇒ 5t = 30 ⇒ t = 6 Substitute t = 6 into s = t + 3: s = 6 + 3 = 9 Result: Therefore, the solution is s = 9 and t = 6. Next question solution: NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.2 Question 1(iii) Explore more in Pair of L...

NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.2 Question 1(iii) Question: Solve the following pair of linear equations by the substitution method: 3x – y = 3 9x – 3y = 9 Given: The given pair of linear equations is: 3x – y = 3 ---(1) 9x – 3y = 9 ---(2) To Find: We need to find the values of x and y that satisfy both equations using the substitution method. Formula: Substitution method involves solving one equation for one variable and substituting it into the other equation. Solution: From equation (1), we can express y in terms of x: y = 3x – 3 Substitute this expression for y into equation (2): 9x – 3(3x – 3) = 9 Simplify and solve for x: 9x – 9x + 9 = 9 9 = 9 This equation is always true, indicating that the two equations are linearly dependent (represent the same line). Therefore, there are infinitely many solutions. Any point (x, y) that satisfies equation (1) also satisfies equation (2). Result: Th...

NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.2 Question 1(i) Question: Solve the following pair of linear equations by the substitution method. x + y = 14 x – y = 4 Given: x + y = 14 ...(1) x – y = 4 ...(2) To Find: The values of x and y that satisfy both equations. Formula: Substitution method: Solve one equation for one variable, substitute into the other equation. Solution: From equation (2): x = y + 4 Substitute x = y + 4 into equation (1): (y + 4) + y = 14 ⇒ 2y + 4 = 14 ⇒ 2y = 10 ⇒ y = 5 Substitute y = 5 into x = y + 4: x = 5 + 4 ⇒ x = 9 Result: Therefore, x = 9 and y = 5 Next question solution: NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.2 Question 1(ii) Explore more in Pair of Linear Equations: Click this link to explore more NCERT Class X Chapter 3 Pair of Linear Equations Explore more: Click this link to explore more NCERT Class X chapter solut...

NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.2 Question 1(iv) Question: Solve the following pair of linear equations by the substitution method: 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3 Given: The given equations are: 0.2x + 0.3y = 1.3 ---(1) 0.4x + 0.5y = 2.3 ---(2) To Find: We need to find the values of x and y that satisfy both equations. Formula: Substitution method: Solve one equation for one variable and substitute it into the other equation. Solution: From equation (1), we can express x in terms of y: 0.2x = 1.3 - 0.3y ⇒ x = 1.3 - 0.3y 0.2 Substitute this value of x into equation (2): 0.4( 1.3 - 0.3y 0.2 ) + 0.5y = 2.3 Simplify and solve for y: 2(1.3 - 0.3y) + 0.5y = 2.3 ⇒ 2.6 - 0.6y + 0.5y = 2.3 ⇒ -0.1y = -0.3 ⇒ y = 3 Substitute y = 3 back into the expression for x: x = 1.3 - 0.3(3) 0.2 = 1.3 - 0.9 0.2 = 0.4 0.2 = 2 Result: Therefore, the solution to the pair of linear equations is x = 2 and y...

NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.2 Question 1(v) Question: Solve the following pair of linear equations by the substitution method. √2x + √3y = 0 √3x - √8y = 0 Given: √2x + √3y = 0 √3x - √8y = 0 To Find: The values of x and y that satisfy both equations. Formula: Substitution method for solving linear equations. Solution: From the first equation, √2x + √3y = 0 ⇒ √2x = -√3y ⇒ x = -√3y √2 Substitute this value of x into the second equation: √3( -√3y √2 ) - √8y = 0 Simplifying, we get: -3y √2 - √8y = 0 ⇒ -3y √2 = 2√2y -3y = 4y ⇒ 7y = 0 ⇒ y = 0 Substitute y = 0 into x = -√3y √2 ⇒ x = 0 Result: x = 0, y = 0 Next question solution: NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.2 Question 1(vi) Explore more in Pair of Linear Equations: Click this link to explore more NCERT Class X Chapter 3 Pair of Linear Equations Explore more: Click this link to exp...

NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.2 Question 1(vi) Question: Solve the following pair of linear equations by the substitution method. 3x 2 - 5y 3 = -2 x 3 + y 2 = 13 6 Given: The given pair of linear equations are: 3x 2 - 5y 3 = -2 ...(1) x 3 + y 2 = 13 6 ...(2) To Find: We need to solve the given pair of linear equations by the substitution method. Formula: Substitution method involves solving one equation for one variable and substituting it into the other equation. Solution: From equation (2): x 3 + y 2 = 13 6 ⇒ x + 3y 2 = 13 2 ⇒ 2x + 3y = 13 ...(3) From equation (1): 3x 2 - 5y 3 = -2 ⇒ 9x - 10y = -12 ...(4) From (3), x = 13-3y 2 Substitute this value of x in (4): 9( 13-3y 2 ) - 10y = -12 ⇒ 9(13 - 3y) - 20y = -24 ⇒ 117 - 27y - 20y = -24 ⇒ 117 - 47y =...

NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.2 Question 2 Question: Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘ m ’ for which y = mx + 3. using substitution method Given: 2x + 3y = 11 ---(1) 2x – 4y = – 24 ---(2) y = mx + 3 ---(3) To Find: The values of x and y, and the value of m. Formula: Substitution method Solution: From equation (1), 2x = 11 - 3y ⇒ x = 11 - 3y 2 Substitute this value of x in equation (2): 2( 11 - 3y 2 ) – 4y = –24 11 - 3y -4y = -24 11 - 7y = -24 -7y = -35 y = 5 Substitute y = 5 in equation (1): 2x + 3(5) = 11 2x + 15 = 11 2x = -4 x = -2 Substitute x = -2 and y = 5 in equation (3): 5 = m(-2) + 3 5 = -2m + 3 2 = -2m m = -1 Result: x = -2, y = 5, m = -1 Next question solution: NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.2 Question 3(i) Explore more in Pair of Linear Equations: Click this link to exp...

NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.2 Question 3(i) Question: Form the pair of linear equations for the following problems and find their solution by substitution method. The difference between two numbers is 26 and one number is three times the other. Find them. Given The difference between two numbers is 26. One number is three times the other. To Find: The two numbers. Formula: Substitution method for solving linear equations. Solution: Let the two numbers be x and y. According to the problem, we have: x - y = 26 ...(1) x = 3y ...(2) Substitute equation (2) into equation (1): 3y - y = 26 2y = 26 ⇒ y = 26 2 = 13 Substitute y = 13 into equation (2): x = 3(13) = 39 Result: Therefore, the two numbers are 39 and 13. Next question solution: NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.2 Question 3(ii) Explore more in Pair of Linear Equations: ...

NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.2 Question 3(ii) Question: Form the pair of linear equations for the following problems and find their solution by substitution method. The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them. Given: Let the two supplementary angles be x and y, where x is the larger angle and y is the smaller angle. The larger angle exceeds the smaller angle by 18 degrees. To Find: We need to find the values of the two supplementary angles x and y. Formula: Supplementary angles add up to 180 degrees. Therefore, x + y = 180 Solution: We have two equations: x + y = 180 ---(1) x - y = 18 ---(2) From equation (2), we can express x in terms of y: x = y + 18 ---(3) Substitute equation (3) into equation (1): (y + 18) + y = 180 Simplify and solve for y: 2y + 18 = 180 2y = 180 - 18 2y = 162 y = 162 2 = 81 Substitute the value of y (81) back into eq...

NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.2 Question 3(iv) Question: Form the pair of linear equations for the following problems and find their solution by substitution method. 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 km? How much does a person have to pay for travelling a distance of 25 km? Given: For a distance of 10 km, the charge paid is Rs 105. For a journey of 15 km, the charge paid is Rs 155. To Find: The fixed charges and the charge per km. Also, the charge for travelling a distance of 25 km. Formula: Let the fixed charge be Rs x and the charge per km be Rs y. Then the equations are: x + 10y = 105 x + 15y = 155 Solution: From the first equation, x = 105 - 10y Substitute this value of x in the seco...

NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.2 Question 3(iii) Question: Form the pair of linear equations for the following problems and find their solution by substitution method. 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. Given: Cost of 7 bats and 6 balls = Rs 3800 Cost of 3 bats and 5 balls = Rs 1750 To Find: Cost of each bat and each ball Formula: Substitution method Solution: Let the cost of each bat be x and the cost of each ball be y. From the given information, we can form the following equations: 7x + 6y = 3800 ---(1) 3x + 5y = 1750 ---(2) From equation (2), we can express x in terms of y: 3x = 1750 - 5y ⇒ x = 1750 - 5y 3 Substitute this value of x into equation (1): 7( 1750 - 5y 3 ) + 6y = 3800 ...

NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.2 Question 3(v) Question: Form the pair of linear equations for the following problems and find their solution by substitution method. A fraction becomes 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 5 6 . Find the fraction Given: A fraction becomes 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 5 6 . To Find: The fraction. Formula: Let the fraction be x y . Then, x+2 y+2 = 9 11 and x+3 y+3 = 5 6 Solution: 11(x+2) = 9(y+2) ⇒ 11x + 22 = 9y + 18 ⇒ 11x - 9y = -4 ...(1) 6(x+3) = 5(y+3) ⇒ 6x + 18 = 5y + 15 ⇒ 6x - 5y = -3 ...(2) From (2), 6x = 5y -3 ⇒ x = 5y-3 6 Substitute in (1): 11( 5y-3 6 ) - 9y = -4 ⇒ 55y - 33 - 54y = -24 ⇒ y = 9 Substitute y = 9 in x = 5y-3 6 ⇒ x = 5(9)-3 6 = 42 6 = 7 Result: ...

NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.2 Question 3(vi) Question: Form the pair of linear equations for the following problems and find their solution by substitution method. 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 ages? Given: Let Jacob's present age be J years and his son's present age be S years. Five years hence, Jacob's age will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. To Find: Their present ages (J and S). Formula: Formulate two linear equations based on the given information and solve them using the substitution method. Solution: Five years hence, Jacob's age will be three times that of his son: J + 5 = 3(S + 5) ⇒ J + 5 = 3S + 15 ⇒ J = 3S + 10 ...(1) Five years ago, Jacob’s age was seven times that of his son: J - 5 = 7(S -...