Last updated at August 2, 2026 by Teachoo
Transcript
Question 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (v) (7š„ ā 2š¦)/š„š¦ = 5 (8š„ + 7š¦)/š„š¦ = 15 Given (7š„ ā 2š¦)/š„š¦ = 5 (7š„ )/š„š¦ ā (2š¦ )/š„š¦ = 5 (7 )/š¦ ā(2 )/š„ = 5 (āš )/š +(š )/š = 5 (8š„ + 7š¦)/š„š¦ = 15 (8š„ )/š„š¦ + (7š¦ )/š„š¦ = 15 (8 )/š¦ +(7 )/š„ = 15 (š )/š +(š )/š = 15 Our equations are (ā2 )/š„ +(7 )/š¦ = 5 ā¦(1) (7 )/š„ +(8 )/š¦ = 15 ...(2) So, our equations become ā2u + 7v = 5 7u + 8v = 15 Hence, we solve ā2u + 7v = 5 ā¦(3) 7u + 8v = 15 ā¦(4) From (3) ā2u + 7v = 5 7v = 5 + 2u v = (5 + 2š¢)/7 Putting value of v in (4) 7u + 8v = 15 7u + 8((5 + 2š¢)/7) = 15 Multiplying 7 both sides 7 Ć 7u + 7 Ć 8 ((5 + 2š¢)/7) = 7 Ć 15 49u + 8(5 + 2u) = 105 49u + 40 + 16u = 105 49u + 16u = 105 ā 40 65u = 65 u = 65/65 u = 1 Putting value of u in (3) ā2u + 7v = 5 ā2(1) + 7v = 5 ā2 + 7v = 5 7v = 5 + 2 7v = 7 v = 7/7 v = 1 Hence, u = 1, v = 1 But we have to find x & y We know that u = š/š 1 = 1/š„ x = 1 v = š/š 1 = 1/š¦ y = 1 Hence, x = 1 , y = 1 is the solution of the given equation Question 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (vi) 6x + 3y = 6xy 2x + 4y = 5xy Given 6x + 3y = 6xy Diving whole equation by xy (6š„ + 3š¦)/š„š¦ = 6š„š¦/š„š¦ 6š„/š„š¦ +3š¦/š„š¦ = 6 š/š +š/š = 6 2x + 4y = 5xy Diving whole equation by xy (2š„ + 4š¦)/š„š¦ = 5š„š¦/š„š¦ 2š„/š„š¦ +4š¦/š„š¦ = 5 š/š +š/š = 5 Hence, our equations are 6/š¦ +3/š„ = 6 ā¦(1) 2/š¦ +4/š„ = 5 ā¦(2) So, our equations become 6v + 3u = 6 2v + 4u = 5 Now, we solve 6v + 3u = 6 ā¦(3) 2v + 4u = 5 ā¦(4) From (3) 6v + 3u = 6 6v = 6 ā 3u v = (6 ā 3š¢)/6 Putting value of v in (4) 2v + 4u = 5 2((6 ā 3š¢)/6) + 4u = 5 ((6 ā 3š¢)/3) + 4u = 5 Multiplying both sides by 3 3 Ć ((6 ā3š¢)/3) + 3 Ć 4u = 3 Ć 5 (6 ā 3u) + 12u = 15 ā3u + 12u = 15 ā 6 9u = 9 u = 9/9 u = 1 Putting u = 1 in (3) 6v + 3u = 6 6v + 3(1) = 6 6v + 3 = 6 6v = 6 ā 3 6v = 3 v = 3/6 v = š/š Hence, u = 1 , v = 1/2 But we have to find x & y Now, u = š/š 1 = 1/š„ x = 1 v = š/š 1/2 = 1/š¦ y = 2 Hence, x = 1 , y = 2 is the solution of the given equation