Question 1 (iii) and (iv) - Equations reduced to pair of linear equations - Chapter 3 Class 10 Pair of Linear Equations in Two Variables

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Question 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (iii) 4/𝑥 + 3y = 14 3/𝑥 – 4y = 23 4/𝑥 + 3y = 14 3/𝑥 – 4y = 23 So, our equations become 4u + 3y = 14 3u – 4y = 23 Now, our equations are 4u + 3y = 14 …(3) 3u – 4y = 23 …(4) From (3) 4u + 3y = 14 4u = 14 – 3y u = (14 −3𝑦)/4 Putting value of u in (4) 3u – 4y = 23 3 ((14 − 3𝑦)/4) – 4y = 23 Multiplying both sides by 4 4 × 3 ((14 − 3𝑦)/4) – 4 × 4y = 4 × 23 3(14 – 3y) − 16y = 92 42 – 9y − 16y = 92 – 9y – 16y = 92 – 42 –25y = 50 y = 50/(−25) y = – 2 Putting y = – 2 in equation (3) 4u + 3y = 14 4u + 3(–2) = 14 4u – 6 = 14 4u = 14 + 6 u = 20/4 u = 5 But u = 1/𝑥 5 = 1/𝑥 x = 𝟏/𝟓 Hence, x = 𝟏/𝟓, y = –2 is the solution of the given equation Question 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (iv) 5/(𝑥 − 1) + 1/(𝑦 − 2) = 2 6/(𝑥 − 1) − 3/(𝑦 − 2) = 1 5/(𝑥 − 1) + 1/(𝑦 − 2) = 2 6/(𝑥 − 1) − 3/(𝑦 − 2) = 1 So, our equations become 5u + v = 2 6u – 3v = 1 Our equations are 5u + v = 2 …(3) 6u – 3v = 1 …(4) From (3) 5u + v = 2 v = 2 – 5u Putting value of v in (4) 6u – 3v = 1 6u – 3(2 – 5u) = 1 6u – 6 + 15u = 1 6u + 15u = 1 + 6 21u = 7 u = 7/21 u = 𝟏/𝟑 Putting u = 1/3 in (3) 5u + v = 2 5 (1/3) + v = 2 5/3 + v = 2 v = 2 – 5/3 v = (2(3) − 5)/3 v = 𝟏/𝟑 Hence, u = 1/3 & v = 1/3 We need to find x & y We know that u = 𝟏/(𝒙 − 𝟏) 1/3 = 1/(𝑥 − 1) x – 1 = 3 x = 3 + 1 x = 4 v = 𝟏/(𝒚 − 𝟐) 1/3 = 1/(𝑦 −2) y – 2 = 3 y = 3 + 2 y = 5 So, x = 4, y = 5 is the solution of our equations