Equations reduced to pair of linear equations     This video is only available for Teachoo black users     This video is only available for Teachoo black users

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

### Transcript

Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (i) 1/2𝑥 + 1/3𝑦 = 2 1/3𝑥 + 1/2𝑦 = 13/6 1/2𝑥 + 1/3𝑦 = 2 1/3𝑥 + 1/2𝑦 = 13/6 Let 1/𝑥 = u 1/𝑦 = v So, our equations become 1/2 u + 1/3 v = 2 (3𝑢 + 2𝑣)/(2 × 3) = 2 3u + 2v = 12 1/3 u + 1/2 v = 13/6 (2𝑢 +3𝑣)/(2 × 3) = 13/6 2u + 3v = 13 Our equations are 3u + 2v = 12 …(3) 2u + 3v = 13 …(4) From (3) 3u + 2v = 12 3u = 12 – 2v u = (12 − 2𝑣)/3 Putting value of u in (4) 2u + 3v = 13 2 ((12 −2𝑣)/3) + 3v = 13 Multiplying both sides by 3 3 × 2((12 − 2𝑣)/3) + 3 × 3v = 3 × 13 2(12 – 2v) + 9v = 39 24 – 4v + 9v = 39 – 4v + 9v = 39 – 24 5v = 15 v = 15/5 v = 3 Putting v = 3 in (3) 3u + 2v = 12 3u + 2(3) = 12 3u + 6 = 12 3u = 12 – 6 3u = 6 u = 6/3 u = 2 Hence, v = 3, u = 2 But we have to find x & y We know that u = 𝟏/𝒙 2 = 1/𝑥 x = 𝟏/𝟐 v = 𝟏/𝒚 3 = 1/𝑦 y = 𝟏/𝟑 So, x = 𝟏/𝟐 , y = 𝟏/𝟑 is the solution of the given equation Question 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (ii) 2/√𝑥 + 3/√𝑦 = 2 4/√𝑥 − 9/√𝑦 = –1 2/√𝑥 + 3/√𝑦 = 2 4/√𝑥 − 9/√𝑦 = −1 So, our equations become 2u + 3v = 2 4u – 9v = –1 Our equations 2u + 3v = 2 …(3) 4u – 9v = –1 …(4) From (3) 2u + 3v = 2 2u = 2 – 3v u = (2 − 3𝑣)/2 Putting value of u in (4) 4u – 9v = – 1 4 ((2 − 3𝑣)/2) – 9v = –1 2(2 – 3v) – 9v = –1 4 – 6v – 9v = –1 – 6v – 9v = –1 – 4 –15v = – 5 v = (−5)/(−15) v = 𝟏/𝟑 Putting v = 1/3 in (3) 2u + 3v = 2 2u + 3 (1/3) = 2 2u + 1 = 2 2u = 2 – 1 u = 𝟏/𝟐 Hence, u = 1/2 & v = 1/3 But, we need to find x & y u = 𝟏/√𝒙 1/2 = 1/√𝑥 √𝑥 = 2 Squaring both sides (√𝑥)2 = (2)2 x = 4 v = 𝟏/√𝒚 1/3 = 1/√𝑦 √𝑦 = 3 Squaring both sides (√𝑦)2 = (3)2 y = 9 Therefore, x = 4, y = 9 is the solution of the given equation 