Move all terms not containing `y` to the right-hand side of the equation.

`3x-4y=9`

`y=6-2x`

Since the equation was solved for `y` , replace all occurrences of `y` in the other equations with the solution `(6-2x)` .

`3x-4(6-2x)=9 `

Multiply `-4` by each term inside the parentheses.

`3x+8x-24=9`

Since `3x` and `8x` are like terms, add `8x` to `3x` to get `11x` .

`11x-24=9`

Move all terms not containing `x` to the right-hand side of the equation.

`11x=33 `

Since the equation was solved for `x` , replace all occurrences of `x ` in the other equations with the solution `(3)` .

`x=3`

Since the equation was solved for ` x` , replace all occurrences of ` x` in the other equations with the solution `(3).`

`y=6-2(3) `

Multiply `-2` by each term inside the parentheses.

`y=6-6`

Subtract 6 from 6 to get 0.

`y=0 `

This is the solution to the system of equations.

`x=3`

`y=0`

The system of equations 3x - 4y = 9, 2x + y = 6 has to be solved by substitution.

3x - 4y = 9

=> 3x = 9 + 4y

=> `x = 3 + (4/3)*y`

Substitute for x in 2x + y = 6

=> `2*(3 + (4/3)*y) + y = 6`

=> `6 + 8/3*y + y = 6`

=> y = 0

x = 3

**The solution of the given set of equations is x = 3 and y = 0**

3x - 4y = 9 ----(i)

2x + y = 6 ----(ii)

These simultaneous equations can be solved with the help of substitution method, consider eq(ii)

2x + y = 6

y = 6 - 2x

Input the value of y in eq(i)

3x - 4y = 9

3x - 4(6 - 2x) = 9

3x - 24 + 8x = 9

3x -24 + 24 + 8x = 9 + 24 Add 24 to both sides

11x = 33

11x/11 = 33/11 Divide 11 by both sides

**x = 3**

Now input this value in eq(ii)

2x + y = 6

2(3) + y = 6

6 + y = 6

6 + y - 6 = 6 - 6

**y = 0 **

Now substitute both values in eq(i)

3x - 4y = 9

3(3) - 4(0) = 9

9 - 0 = 9

9 = 9

LHS = RHS

Proved.

**x = 3**

**y = 0 Answer.**