Slide17.JPG

Slide18.JPG
Slide19.JPG
Slide20.JPG

 

Share on WhatsApp

Transcript

Ex 3.2, 1 Solve the following pair of linear equations by the substitution method. (vi) 3𝑥/2−5𝑦/3=−2 𝑥/3+𝑦/2=13/6 Removing fractions from both equations 𝟑𝒙/𝟐−𝟓𝒚/𝟑=−2 Multiplying both equations by 6 6 × 3𝑥/2−"6 ×" 5𝑦/3="6 ×"−2 9x – 10y = − 12 Hence, our equations are 9x – 10y = −12 …(1) 2x + 3y = 13 …(2) From (1) 9x – 10y = –12 9x = –12 + 10y x = ((−𝟏𝟐 + 𝟏𝟎𝒚)/𝟗) Substituting value of x in (2) 2x + 3y = 13 2((−12 + 10𝑦)/9)+3𝑦=13 Multiplying both sides by 9 9 × 2((−12 + 10𝑦)/9) + 9 × 3y = 9 × 13 (2(−12 + 10𝑦) + 9 × 3𝑦)/9=13 2(–12 + 10y) + 27y = 13×9 – 24 + 20y + 27y = 117 – 24 + 20y + 27y – 117 = 0 47y – 141 = 0 47y = 141 y = 141/47 y = 3 Putting y = 3 in (2) 2x + 3y = 13 2x + 3(3) = 13 2x + 9 = 13 2x = 13 – 9 2x = 4 x = 4/2 x = 2 Hence, x = 2, y = 3 is the solution of equation .

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo