Question

Solve for x and y:

3(2x + y) = 7xy, 3(x + 3y) = 11xy (x ≠ 0 and y ≠ 0)

Answer 1

1. Simplify the equations

First, expand the brackets on the left side of both equations to simplify them:

3(2x + y) = 7xy

⇒ 6x + 3y = 7xy   ...(Equation 1)

3(x + 3y) = 11xy

⇒ 3x + 9y = 11xy   ...(Equation 2)

2. Divide by xy

Since it is given that x ≠ 0 and y ≠ 0, we can divide both sides of both equations by xy to eliminate the product term on the right:

From Equation 1:

`(6x)/(xy) + (3y)/(xy) = (7xy)/(xy)`

⇒ `6/y + 3/x = 7`   ...(Equation 3)

From Equation 2:

`(3x)/(xy) + (9y)/(xy) = (11xy)/(xy)`

⇒ `3/y + 9/x = 11`   ...(Equation 4)

3. Substitute variables

To convert this into a standard system of linear equations, substitute new variables for the reciprocal fractions.

Let `u = 1/x` and `v = 1/y`:

3u + 6v = 7   ...(Equation 5)

9u + 3v = 11   ...(Equation 6)

4. Eliminate a variable

Multiply Equation 5 by 3 so that the coefficients of u match up in both equations:

3 × (3u + 6v = 7)

⇒ 9u + 18v = 21   ...(Equation 7)

Now, subtract Equation 6 from Equation 7 to eliminate u:

(9u + 18v) – (9u + 3v) = 21 – 11

15v = 10

`v = 10/15`

`v = 2/3`

5. Solve for u

Substitute the value of `v = 2/3` back into Equation 5 to solve for u:

`3u + 6(2/3) = 7`

3u + 4 = 7

3u = 3

⇒ u = 1

6. Convert back to x and y

Now, reverse the initial substitution to find the actual values of x and y:

`x = 1/u`

= `1/1`

= 1

`y = 1/v`

= `1/(2/3)`

= `3/2`