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` ...

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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`

3x-4y=9-------------(1)

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

multiply (2) by 4

8x+4y=24----------(3)

Then add (1)+(3)

3x-4y+8x+4y=9+24

11x=33

x=33/11

x=3

Let's use x=3 in (2)

2(3)+y=6

6+y=6

y=6-6

y=0

So the answer is

x=3 and 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**