Important Linear Equation Questions

Question 1
Deleted for CBSE Board 2025 Exams

Question 2 Deleted for CBSE Board 2025 Exams

Question 3 Deleted for CBSE Board 2025 Exams

Question 4 Important Deleted for CBSE Board 2025 Exams

Question 5 Deleted for CBSE Board 2025 Exams

Question 6 Important Deleted for CBSE Board 2025 Exams

Question 7 (i) Deleted for CBSE Board 2025 Exams

Question 7 (ii) Important Deleted for CBSE Board 2025 Exams You are here

Question 7 (iii) Deleted for CBSE Board 2025 Exams

Question 7 (iv) Deleted for CBSE Board 2025 Exams

Question 7 (v) Important Deleted for CBSE Board 2025 Exams

Question 8 Deleted for CBSE Board 2025 Exams

Chapter 3 Class 10 Pair of Linear Equations in Two Variables

Serial order wise

Last updated at April 16, 2024 by Teachoo

Question 7 Solve the following pair of linear equations: (ii) ax + by = c bx + ay = 1 + c Solving equations ax + by = c Multiplying both sides by a a (ax + by) = ac a2x + aby = ac bx + ay = 1 + c Multiplying both sides by b b (bx + ay) = b (1 + c) b2x + aby = b + bc Hence, the equations are a2x + aby = ac …(1) b2x + aby = b + bc …(2) From equation (1) a2x + aby = ac aby = ac − a2x y = (𝑎𝑐 − 𝑎^2 𝑥)/𝑎𝑏 Putting y = (𝑎𝑐 − 𝑎^2 𝑥)/𝑎𝑏 in equation (2) b2𝑥 + aby = b + bc b2𝑥 + ab((𝑎𝑐 − 𝑎^2 𝑥)/𝑎𝑏) = b + bc b2𝑥 + ac − a2𝑥 = b + bc b2𝑥 − a2𝑥 = b + bc − ac (b2 − a2) 𝑥 = b + c (b − a) 𝑥 = (𝑏 + 𝑐(𝑏 − 𝑎))/(𝑏2 − 𝑎2) 𝑥 = (𝑏 − 𝒄(𝒂 − 𝒃))/(𝑏2 − 𝑎2) 𝑥 = (𝑏 − 𝑐(𝑎 − 𝑏))/(−(𝒂𝟐 − 𝒃𝟐) ) 𝑥 = (−(−𝒃 + 𝒄(𝒂 − 𝒃)))/(−(𝑎2 − 𝑏2) ) 𝑥 = (−𝑏 + 𝑐(𝑎 − 𝑏))/(𝑎2 − 𝑏2) 𝒙 = (𝒄(𝒂 − 𝒃) − 𝒃)/(𝒂𝟐 − 𝒃𝟐) Put 𝑥 = ( 𝑐(𝑎 − 𝑏) − 𝑏)/(𝑎2 − 𝑏2) in equation (1) a2𝑥 + aby = ac a2((𝒄(𝒂 − 𝒃) − 𝒃)/(𝒂𝟐 − 𝒃𝟐)) + aby = ac a2((𝒄(𝒂 − 𝒃) − 𝒃)/(𝒂𝟐 − 𝒃𝟐)) + aby = ac a2((𝑐(𝑎 − 𝑏) − 𝑏)/(𝑎2 − 𝑏2)) + aby = ac a2((𝑎𝑐 − 𝑐𝑏 − 𝑏)/(𝑎2 − 𝑏2)) + aby = ac (𝒂^𝟐 (𝒂𝒄 − 𝒄𝒃 − 𝒃))/(𝑎2 − 𝑏2) + aby = ac (𝑎^3 𝑐 − 𝑎^2 𝑐𝑏 − 𝑎^2 𝑏)/(𝑎2 − 𝑏2) + aby = ac Multiplying a2 – b2 both sides (a2 – b2) (𝑎^3 𝑐 − 𝑎^2 𝑐𝑏 − 𝑎^2 𝑏)/(𝑎2 − 𝑏2) + (a2 – b2) aby = (a2 – b2) ac 𝒂𝟑𝒄 −𝒂𝟐𝒃𝒄 −𝒂𝟐𝒃 + (a2 – b2) aby = (a2 – b2) ac 𝑎3𝑐 −𝑎2𝑏𝑐 −𝑎2𝑏 + (a2 – b2) aby =𝒂^𝟑 𝒄−〖𝒂𝒃〗^𝟐 𝒄 (a2 – b2) aby =𝑎^3 𝑐−〖𝑎𝑏〗^2 𝑐 − 𝑎3𝑐+𝑎2𝑏𝑐+𝑎2𝑏 (a2 – b2) aby =−〖𝑎𝑏〗^2 𝑐 + 𝑎2𝑏𝑐+𝑎2𝑏 Dividing whole equation by ab (𝑎^2−〖 𝑏〗^2 )𝑎𝑏𝑦/𝑎𝑏 = (−𝑎𝑏2𝑐)/𝑎𝑏 + (𝑎^2 𝑏𝑐)/𝑎𝑏 + (𝑎^2 𝑏)/𝑎𝑏 (𝑎^2−〖 𝑏〗^2 )𝑦 = −bc + ac + a " " (𝑎^2−〖 𝑏〗^2 )𝑦" = ac − bc + a" " " (𝑎^2−〖 𝑏〗^2 )𝑦" = c(a − b) + a" y = (𝒄(𝒂 − 𝒃) + 𝒂)/(𝒂𝟐 − 𝒃𝟐) Thus, 𝒙 = (𝒄(𝒂 − 𝒃) − 𝒃)/(𝒂𝟐 − 𝒃𝟐) y = (𝒄(𝒂 − 𝒃) + 𝒂)/(𝒂𝟐 − 𝒃𝟐)