Here is a more suitable method of solving simultaneous equations than simple substiution.

`3x + 2y = 7`

`2x + 3y = 1`

Multiply 1st equation by 2 and multiply second equation by 3 and substract them.

`2(3x + 2y) -3(2x + 3y) = 2 xx 7 - 3 xx 1`

`6x+4y-6x-9y = 14-3`

`-5y = 11`

`y = -11/5`

You can use this in second equation,

`2x+3(-11/5) = 1`

`2x - 33/5 = 1`

`2x = 1+33/5`

`2x = (5+33)/5`

`2x = 38/5`

`x = 19/5`

This gives `x = 19/5` and `y = -11/5`

The set of linear equations 3x + 2y = 7 and 2x + 3y = 1 has to be solved.

3x + 2y = 7

=> `x = (7 - 2y)/3`

substitute in 2x + 3y = 1

=> `(2/3)*(7 - 2y) + 3y = 1`

=> `2(7 - 2y) = 3(1-3y)`

=> 14 - 4y = 3 - 9y

=> 5y = -11

=> y = -11/5

x = `(7 + 22/5)/3 = 57/15 = 19/5`

**The solution of the set of equations is x = `19/5` and y = `-11/5` **

(1) 3x + 2y = 7

(2) 2x + 3y = 1

(1) x 2:

6x + 4y = 14

(2) x 3:

6x + 9y = 3

Subtract the two equations:

6x + 9y = 3

- 6x + 4y = 14

-----------------------

5y = 11

Solve for y, plug back in to solve for x.

3x + 2y = 7

2x + 3y = 1

First multiply everything in the top equation by 2, and everything in the second equation by 3

By multiplying, your should get

**6x + 4y = 14**

**6x + 9y = 3 **now subtract 6x with 6x ( which means subtract 4y with 9y and 14 with 3 )

By subtracting, you should get

**-5y = 11 **divide both sides by -5

By dividing, you should get

**y = -11/5 **which is your answer for " y "

Now plug -11/5 into one of the equation

**2x + 3 ( -11/5 ) = 1 **Multiply 3 with -11/5

By multiplying your equation should look like

**2x - 33/5 = 1 add** -33/5 on both sides

By adding, you should get

**2x =** **38/5 **now divide both sides by 2

By dividing, you should get

**x = 19/5 **which is your answer of " x "

So your answer is **x = 19/5 ; y = -11/5**