Solve the following system of equations:7(y + 3) – 2(x + 2) = 14; 4

1.	Solve the following system of equations:7(y + 3) – 2(x + 2) = 14; 4(y – 2) + 3(x – 3) = 2
Answer» The given pair of equations are: 7(y+3) – 2(x+2) = 14……..(i) 4(y-2) + 3(x-3) = 2…….. (ii) From (i), we get 7y + 21 – 2x – 4 = 14 7y = 14 + 4 – 21 + 2x ⇒ y = \(\frac{(2x − 3)}{7}\) From (ii), we get 4y – 8 + 3x – 9 = 2 4y + 3x – 17 – 2 = 0 ⇒ 4y + 3x – 19 = 0 ……(iii) Now, substituting y in equation (iii) 4[\(\frac{(2x − 3)}{7}\)] + 3x – 19 = 0 8x – 12 + 21x – (19 x 17) = 0 [after taking LCM] 29x = 145 ⇒ x = 5 Now, putting the value of x and in the equation (ii) 4(y - 2) + 3(5 - 3) = 2 ⇒ 4y - 8 + 6 = 2 ⇒ 4y = 4 ∴ y = 1 Thus, the value of x and y obtained are 5 and 1 respectively.

Solve the following system of equations:7(y + 3) – 2(x + 2) = 14; 4(y – 2) + 3(x – 3) = 2

Answer»

The given pair of equations are:

7(y+3) – 2(x+2) = 14……..(i)

4(y-2) + 3(x-3) = 2…….. (ii)

From (i), we get

7y + 21 – 2x – 4 = 14

7y = 14 + 4 – 21 + 2x

⇒ y = \(\frac{(2x − 3)}{7}\)

From (ii), we get

4y – 8 + 3x – 9 = 2

4y + 3x – 17 – 2 = 0

⇒ 4y + 3x – 19 = 0 ……(iii)

Now, substituting y in equation (iii)

4[\(\frac{(2x − 3)}{7}\)] + 3x – 19 = 0

8x – 12 + 21x – (19 x 17) = 0 [after taking LCM]

29x = 145

⇒ x = 5

Now, putting the value of x and in the equation (ii)

4(y - 2) + 3(5 - 3) = 2

⇒ 4y - 8 + 6 = 2

⇒ 4y = 4

∴ y = 1

Thus, the value of x and y obtained are 5 and 1 respectively.