4x 3y 6 2y 14

Ex 3.6, 1 (three) and (4) - Chapter three Form ten Pair of Linear Equations in Two Variables (Term i)

Concluding updated at May 19, 2021 by Teachoo

Ex 3.6, 1 (iii) & (iv) : 4/x + 3y = 14 , 3/x - 4y = 23 - Ex 3.6

Ex 3.6, 1 (iii) and (iv) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 2

Ex 3.6, 1 (iii) and (iv) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 3

Ex 3.6, 1 (iii) and (iv) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 4

Ex 3.6, 1 (iii) and (iv) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 5

Ex 3.6, 1 (iii) and (iv) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 6

Ex 3.6, 1 (iii) and (iv) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 7

Ex 3.6, 1 (iii) and (iv) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 8

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Transcript

Ex 3.6, 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 = fourteen 3/𝑥 – 4y = 23 Then, our equations become 4u + 3y = fourteen 3u – 4y = 23 Now, our equations are 4u + 3y = 14 …(3) 3u – 4y = 23 …(4) From (3) 4u + 3y = xiv 4u = fourteen – 3y u = (xiv −3𝑦)/4 Putting value of u in (4) 3u – 4y = 23 3 ((14 − three𝑦)/iv) – 4y = 23 Multiplying both sides past 4 four × 3 ((14 − 3𝑦)/4) – 4 × 4y = 4 × 23 3(14 – 3y) − 16y = 92 42 – 9y − 16y = 92 – 9y – 16y = 92 – 42 –25y = fifty y = l/(−25) y = – 2 Putting y = – 2 in equation (three) 4u + 3y = fourteen 4u + 3(–2) = 14 4u – half-dozen = fourteen 4u = 14 + six u = 20/4 u = v But u = ane/𝑥 five = 1/𝑥 x = 𝟏/𝟓 Hence, ten = 𝟏/𝟓, y = –ii is the solution of the given equation Ex three.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (iv) 5/(𝑥 − 1) + 1/(𝑦 − 2) = two 6/(𝑥 − 1) − three/(𝑦 − two) = i 5/(𝑥 − 1) + one/(𝑦 − two) = 2 6/(𝑥 − 1) − 3/(𝑦 − two) = 1 So, our equations become 5u + v = 2 6u – 3v = i Our equations are 5u + v = 2 …(3) 6u – 3v = 1 …(4) From (3) 5u + v = ii five = 2 – 5u Putting value of five in (4) 6u – 3v = 1 6u – iii(2 – 5u) = 1 6u – vi + 15u = one 6u + 15u = 1 + 6 21u = 7 u = 7/21 u = 𝟏/𝟑 Putting u = ane/3 in (3) 5u + 5 = ii 5 (1/3) + five = 2 v/3 + v = 2 v = 2 – v/3 five = (2(3) − 5)/3 v = 𝟏/𝟑 Hence, u = 1/3 & 5 = one/3 Nosotros need to find x & y We know that u = 𝟏/(𝒙 − 𝟏) 1/3 = 1/(𝑥 − 1) ten – 1 = 3 x = iii + 1 ten = four five = 𝟏/(𝒚 − 𝟐) i/3 = 1/(𝑦 −two) y – ii = 3 y = 3 + 2 y = v And then, 10 = 4, y = 5 is the solution of our equations