Yes, the ordered pair (4, 3) is a solution of the equation 5x - 6y = 2. To verify, substitute x = 4 and y = 3 into the left-hand side of the equation: LHS = 5(4) - 6(3) = 20 - 18 = 2. Since the calculated value equals the right-hand side (RHS = 2), the ordered pair satisfies the linear equation.Read more
Yes, the ordered pair (4, 3) is a solution of the equation 5x – 6y = 2. To verify, substitute x = 4 and y = 3 into the left-hand side of the equation: LHS = 5(4) – 6(3) = 20 – 18 = 2. Since the calculated value equals the right-hand side (RHS = 2), the ordered pair satisfies the linear equation.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
Two solutions for the equation 7x - 3y = 21 are (3, 0) and (0, -7). Setting y = 0 gives 7x - 3(0) = 21, which simplifies to 7x = 21, leading to x = 3. Setting x = 0 gives 7(0) - 3y = 21, which simplifies to -3y = 21, yielding y = -7. Both pairs satisfy the given equation. For more NCERT SolutRead more
Two solutions for the equation 7x – 3y = 21 are (3, 0) and (0, -7). Setting y = 0 gives 7x – 3(0) = 21, which simplifies to 7x = 21, leading to x = 3. Setting x = 0 gives 7(0) – 3y = 21, which simplifies to -3y = 21, yielding y = -7. Both pairs satisfy the given equation.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
Two solutions for the equation 2x + 3y = 5 are (1, 1) and (4, -1). When substituting x = 1, the equation becomes 2(1) + 3y = 5, giving 3y = 3, so y = 1. When substituting x = 4, the equation becomes 2(4) + 3y = 5, giving 8 + 3y = 5, which simplifies to 3y = -3, yielding y = -1. For more NCERTRead more
Two solutions for the equation 2x + 3y = 5 are (1, 1) and (4, -1). When substituting x = 1, the equation becomes 2(1) + 3y = 5, giving 3y = 3, so y = 1. When substituting x = 4, the equation becomes 2(4) + 3y = 5, giving 8 + 3y = 5, which simplifies to 3y = -3, yielding y = -1.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
Let the edge length of the cube be a. Since volume is given by the formula V = a³, we have a³ = 64 cm³, giving side a = 4 cm. The total surface area of a cube consists of six identical square faces, calculated using TSA = 6a². Substituting the side length gives TSA = 6 × (4 cm)² = 6 × 16 cm² = 96 cmRead more
Let the edge length of the cube be a. Since volume is given by the formula V = a³, we have a³ = 64 cm³, giving side a = 4 cm. The total surface area of a cube consists of six identical square faces, calculated using TSA = 6a². Substituting the side length gives TSA = 6 × (4 cm)² = 6 × 16 cm² = 96 cm².
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 14 Math of Space: Surface Area and Volume Question Answer (2026-27)
The cubical box has a side of 2 m, which equals 200 cm, giving a total volume of 200³ = 8,000,000 cm³. Each small cube has a side of 20 cm, yielding a volume of 20³ = 8,000 cm³. Dividing the volume of the box by the volume of one cube gives 8,000,000 / 8,000 = 1,000 small cubes packed tightly insideRead more
The cubical box has a side of 2 m, which equals 200 cm, giving a total volume of 200³ = 8,000,000 cm³. Each small cube has a side of 20 cm, yielding a volume of 20³ = 8,000 cm³. Dividing the volume of the box by the volume of one cube gives 8,000,000 / 8,000 = 1,000 small cubes packed tightly inside.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 14 Math of Space: Surface Area and Volume Question Answer (2026-27)
Verify if the ordered pair (4, 3) is a solution of 5x – 6y = 2. Explain your reasoning.
Yes, the ordered pair (4, 3) is a solution of the equation 5x - 6y = 2. To verify, substitute x = 4 and y = 3 into the left-hand side of the equation: LHS = 5(4) - 6(3) = 20 - 18 = 2. Since the calculated value equals the right-hand side (RHS = 2), the ordered pair satisfies the linear equation.Read more
Yes, the ordered pair (4, 3) is a solution of the equation 5x – 6y = 2. To verify, substitute x = 4 and y = 3 into the left-hand side of the equation: LHS = 5(4) – 6(3) = 20 – 18 = 2. Since the calculated value equals the right-hand side (RHS = 2), the ordered pair satisfies the linear equation.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-13/
See lessFind any two solutions for the equation 7x – 3y = 21.
Two solutions for the equation 7x - 3y = 21 are (3, 0) and (0, -7). Setting y = 0 gives 7x - 3(0) = 21, which simplifies to 7x = 21, leading to x = 3. Setting x = 0 gives 7(0) - 3y = 21, which simplifies to -3y = 21, yielding y = -7. Both pairs satisfy the given equation. For more NCERT SolutRead more
Two solutions for the equation 7x – 3y = 21 are (3, 0) and (0, -7). Setting y = 0 gives 7x – 3(0) = 21, which simplifies to 7x = 21, leading to x = 3. Setting x = 0 gives 7(0) – 3y = 21, which simplifies to -3y = 21, yielding y = -7. Both pairs satisfy the given equation.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-13/
See lessFind any two solutions for the equation 2x + 3y = 5.
Two solutions for the equation 2x + 3y = 5 are (1, 1) and (4, -1). When substituting x = 1, the equation becomes 2(1) + 3y = 5, giving 3y = 3, so y = 1. When substituting x = 4, the equation becomes 2(4) + 3y = 5, giving 8 + 3y = 5, which simplifies to 3y = -3, yielding y = -1. For more NCERTRead more
Two solutions for the equation 2x + 3y = 5 are (1, 1) and (4, -1). When substituting x = 1, the equation becomes 2(1) + 3y = 5, giving 3y = 3, so y = 1. When substituting x = 4, the equation becomes 2(4) + 3y = 5, giving 8 + 3y = 5, which simplifies to 3y = -3, yielding y = -1.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-13/
See lessThe volume of a cube is 64 cm³. What is its total surface area?
Let the edge length of the cube be a. Since volume is given by the formula V = a³, we have a³ = 64 cm³, giving side a = 4 cm. The total surface area of a cube consists of six identical square faces, calculated using TSA = 6a². Substituting the side length gives TSA = 6 × (4 cm)² = 6 × 16 cm² = 96 cmRead more
Let the edge length of the cube be a. Since volume is given by the formula V = a³, we have a³ = 64 cm³, giving side a = 4 cm. The total surface area of a cube consists of six identical square faces, calculated using TSA = 6a². Substituting the side length gives TSA = 6 × (4 cm)² = 6 × 16 cm² = 96 cm².
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 14 Math of Space: Surface Area and Volume Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-14/
See lessHow many small cubes with side 20 cm can be packed tight in a cubical box with side 2 m?
The cubical box has a side of 2 m, which equals 200 cm, giving a total volume of 200³ = 8,000,000 cm³. Each small cube has a side of 20 cm, yielding a volume of 20³ = 8,000 cm³. Dividing the volume of the box by the volume of one cube gives 8,000,000 / 8,000 = 1,000 small cubes packed tightly insideRead more
The cubical box has a side of 2 m, which equals 200 cm, giving a total volume of 200³ = 8,000,000 cm³. Each small cube has a side of 20 cm, yielding a volume of 20³ = 8,000 cm³. Dividing the volume of the box by the volume of one cube gives 8,000,000 / 8,000 = 1,000 small cubes packed tightly inside.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 14 Math of Space: Surface Area and Volume Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-14/
See less