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  1. Chapter 3 Class 10 Pair of Linear Equations in Two Variables
  2. Serial order wise
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Ex 3.7, 7 Solve the following pair of linear equations: px + qy = p – q qx + py = p + q Given Hence, our equations are pqx + q2y = pq − q2 …(1) pqx + p2y = p2 + pq …(2) px + qy = p − q Multiplying both sides by q q(px + qy) = q(p – q) pqx + q2y = pq − q2 qx − py = p + q Multiplying both sides by p p(qx + py) = p(p + q) pqx + p2y = p2 + pq From equation (1), pqx + q2y = pq − q2 q2y = pq − q2 − pqx y = (𝑝𝑞 − 𝑞^2 − 𝑝𝑞𝑥)/𝑞^2 y = (𝑞(𝑝 − 𝑞 − 𝑝𝑥))/𝑞^2 y = (𝑝 − 𝑞 − 𝑝𝑥)/𝑞 Putting y = (𝑝 − 𝑞 − 𝑝𝑥)/𝑞 in equation (2) pqx − p2y = p2 + pq pqx − p2[(𝑝 − 𝑞 − 𝑝𝑥)/𝑞] = p2 + pq pqx − [(𝒑^𝟐 (𝒑 − 𝒒 − 𝒑𝒙))/𝑞] = p2 + pq pqx − [(𝑝^3 − 𝑝^2 𝑞 − 𝑝^3 𝑥)/𝑞]= p2 + pq Multiplying q both sides q(pqx) − q × [(𝑝^3 − 𝑝^2 𝑞 − 𝑝^3 𝑥)/𝑞]= q × p2 + q × pq pq2x − (𝑝^3−𝑝^2 𝑞−𝑝^3 𝑥) = 𝑝2𝑞 + pq2 pq2x − 𝑝^3+"p2q + " 𝑝^3 𝑥=𝒑𝟐𝒒 + pq2 pq2x − 𝑝^3 " + " 𝑝^3 𝑥=𝒑𝟐𝒒 – 𝒑𝟐𝒒 + pq2 pq2x − 𝑝^3 " + " 𝑝^3 𝑥=pq2 "pq2x + " 𝑝^3 𝑥="pq2 "+ 𝑝3 "(pq2 + " 𝑝^3 ")" 𝑥="pq2 "+ 𝑝3 𝑥 = (𝑝𝑞^2+〖 𝑝〗^3)/(𝑝𝑞^2+〖 𝑝〗^3 ) 𝑥 = 1 Put 𝑥 = 1 in equation (1) pqx + q2y = pq − q2 pq(1) + q2y = pq − q2 pq + q2y = − q2 + pq q2y = −q2 y = (−𝑞^2)/𝑞^2 y = −1 Hence, x = 1 and y = −1 Ex 3.7, 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 q 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) Ex 3.7, 7 Solve the following pair of linear equations: (iii) 𝑥/𝑎 – 𝑦/𝑏 = 0 ax + by = a2 + b2 𝑥/𝑎 − 𝑦/𝑏 = 0 …(1) ax + by = 𝑎2+𝑏2 …(2) Solving Equation (1) 𝑥/𝑎 − 𝑦/𝑏 = 0 (𝑏𝑥 − 𝑎𝑦)/𝑎𝑏 = 0 bx − ay = 0 bx − ay = 0 Multiplying both sides by b b(bx – ay) = b × 0 b2x − aby = 0 ax + by = a2 + b2 Multiplying both sides by b a(ax + by) = a(a2 + b2) a2x + aby = a3 + ab2 Now, solving equations (2) & (3) Hence, our equations are b2x − aby = 0 …(4) a2x + aby = a3 + ab2 …(5) From equation (4) b2x − aby = 0 b2x = aby aby = b2x y = (𝑏^2 𝑥)/𝑎𝑏 y = 𝑏𝑥/𝑎 Putting y = 𝑏𝑥/𝑎 in equation (5) a2x + aby = a3 + ab2 a2𝑥 + ab(𝑏𝑥/𝑎) = a3 + ab2 a2𝑥 + b2𝑥 = a3 + ab2 (a2 + b2) 𝑥 = a (a2 + b2) 𝑥 = (𝑎(𝑎^2 + 𝑏^2))/((𝑎^2 + 𝑏^2)) 𝑥 = a Putting 𝑥 = a in equation (4) b2𝑥 − aby = 0 b2(a) − aby = 0 ab2 − aby = 0 ab2 = aby aby = ab2 y = (𝑎𝑏^2)/𝑎𝑏 y = b Therefore, 𝒙 = a and y = b Ex 3.7, 7 Solve the following pair of linear equations: (iv) (𝑎−𝑏)𝑥+(𝑎+𝑏)𝑦=𝑎2 −2𝑎𝑏−𝑏2 (𝑎 + b) (x + y) = 𝑎2 + 𝑏2 (a − b) 𝑥 + (a + b) y = a2 − 2ab − b2 …(1) (a + b) (𝑥 + y) y = a2 − 2ab − b2 …(2) Solving equation (2) (a + b) (𝑥 + y) = a2 + b2 (a + b)𝒙 + (a + b)y = a2 + b2 (a + b) y = a2 + b2 − (a + b) 𝑥 y = (𝑎^2 + 𝑏^2 − (𝑎 + 𝑏)𝑥)/((𝑎 + 𝑏)) Putting y = (𝑎^2 + 𝑏^2 − (𝑎 + 𝑏)𝑥)/((𝑎 + 𝑏)) in equation (1) (a − b) 𝑥 + (a + b) y = a2 − 2ab − b2 (a − b) x + (a + b) [(𝑎^2 + 𝑏^2 − (𝑎 + 𝑏)𝑥)/((𝒂 + 𝒃))] = 𝑎2−2𝑎𝑏−𝑏2 (a − b)x + 𝒂𝟐 + 𝒃𝟐 − (a + b) x = 𝑎2−2𝑎𝑏−𝑏2 ax − bx + b2 − ax − bx = 2𝑎𝑏−𝑏2 − bx − bx + b2 = 2𝑎𝑏−𝑏2 -2bx + b2 = −2ab − b2 -2bx = −2ab − b2 − b2 -2bx = −2ab − 2b2 -2bx = −(2ab + 2b2) 2bx = 2ab + 2b2 2bx = 2b (a + b) x = (2𝑏(𝑎 + 𝑏))/2𝑏 x = a + b Put x = a + b in equation (3) y = (𝑎2+ 𝑏2− (𝑎 + 𝑏)𝑥)/(𝑎 + 𝑏) y = (𝑎2+ 𝑏2 − (𝑎 + 𝑏)(𝑎 + 𝑏))/(𝑎 + 𝑏) y = (𝑎2+ 𝑏2 − (𝑎 + 𝑏)^2)/(𝑎 + 𝑏) y = (𝑎2 + 𝑏2 − 𝑎2 − 𝑏2 − 2𝑎𝑏)/(𝑎 + 𝑏) y = (−2𝑎𝑏)/(𝑎 + 𝑏) Therefore, x = a + b and y = (−𝟐𝒂𝒃)/(𝒂 + 𝒃). Ex 3.7, 7 Solve the following pair of linear equations: (v) 152x – 378y = – 74 –378x + 152y = – 604 152x − 378y = −74 …(1) −378x + 152y = −604 …(1) From equation (1) 152x – 378y = −74 378y = 152x + 74 y = (152𝑥 + 74)/378 y = (2(76𝑥 + 37))/(2(189)) y = (76𝑥 + 37)/189 Put y = (76𝑥 + 37)/189 in equation (2) −378x + 152y = −604 −378x + 152[(76𝑥 + 37)/189] = −604 −378x + [(152(76𝑥) +152 × 37)/189] = −604 −378x + [(11552𝑥 + 5624)/189] = −604 Multiplying 189 both sides 189(−378x) + 189[(11552𝑥 + 5624)/189] = 189(−604) −71442x + 11552x + 5624 = −114156 11552x − 71442 x = −114156 − 5624 −59890x = −119780 x = (−119780)/(−59890) x = 2 Putting x = 2 in equation (3) y = (76 × 2 + 37)/189 y = (152 + 37)/189 y = 189/189 y = 1 Therefore, x = 2 and y = 1. −59890x = −119780 x = (−119780)/(−59890) x = 2 Putting x = 2 in equation (3) y = (76 × 2 + 37)/189 y = (152 + 37)/189 y = 189/189 y = 1 Therefore, x = 2 and y = 1.

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