Solve the following system of equations:x/3 + y/4 = 11; 5x/6 − y/3

1.	Solve the following system of equations:x/3 + y/4 = 11; 5x/6 − y/3 = −7
Answer» The given pair of equations are: x/3 + y/4 = 11………(i) 5x/6 − y/3 = −7…….(ii) From (i), we get x/3 + y/4 = 11 ⇒ 4x + 3y = (11 x 12) [After taking LCM] ⇒ 4x =132 – 3y ⇒ x = \(\frac{(132 – 3y)}{4}\)……. (iv) From (ii), we get 5x/6 − y/3 = −7 ⇒ 5x – 2y = -42 ………(iii) [After taking LCM] Now, substituting x in equation (iii) we get, 5[\(\frac{(132 − 3y)}{4}\)] – 2y = -42 ⇒ 660 – 15y – 8y = -42 x 4 [After taking LCM] ⇒ 660 + 168 = 23y ⇒ 23y = 828 ⇒ y = 36 Now, putting the value of y in the equation (iv) x = \(\frac{(132 – 3(36))}{4}\) ⇒ x = \(\frac{(132 − 108)}{4}\) = \(\frac{24}{4}\) ∴ x = 6 Thus, the value of x and y obtained are 6 and 36 respectively.

Solve the following system of equations:x/3 + y/4 = 11; 5x/6 − y/3 = −7

Answer»

The given pair of equations are:

x/3 + y/4 = 11………(i)

5x/6 − y/3 = −7…….(ii)

From (i), we get

x/3 + y/4 = 11

⇒ 4x + 3y = (11 x 12) [After taking LCM]

⇒ 4x =132 – 3y

⇒ x = \(\frac{(132 – 3y)}{4}\)……. (iv)

From (ii), we get

5x/6 − y/3 = −7

⇒ 5x – 2y = -42 ………(iii) [After taking LCM]

Now, substituting x in equation (iii) we get,

5[\(\frac{(132 − 3y)}{4}\)] – 2y = -42

⇒ 660 – 15y – 8y = -42 x 4 [After taking LCM]

⇒ 660 + 168 = 23y

⇒ 23y = 828

⇒ y = 36

Now, putting the value of y in the equation (iv)

x = \(\frac{(132 – 3(36))}{4}\)

⇒ x = \(\frac{(132 − 108)}{4}\) = \(\frac{24}{4}\)

∴ x = 6

Thus, the value of x and y obtained are 6 and 36 respectively.