Question

How do you solve the system of equations `2.5x+y=-2` and `3x+2y=0`?

Answer 1

Solution: `x= -2, y=3`

Explanation:

`2.5 x+y= -2 or 5 x + 2 y = -4 ;(1)`

`3 x +2 y =0 ; (2)` Subtracting equation (2) from equation (1)

we get, `(5 x + 2 y) - (3 x +2 y) = -4-0` or

`5 x + cancel(2 y) - 3 x - cancel(2 y) = -4` or

`2 x= -4 or x= -2`. Putting `x=-2` in equation (2) we get,

`3* (-2) +2 y =0 or 2 y=6 or y = 3 :. (x= -2, y=3)`

Solution: `x= -2, y=3` [Ans]

Answer 2

x = -2, y = 3

Explanation:

Rearrange one of the equations to give an expression for one of the variables in terms of the other i.e. `y=ax+c` or `x=by+d`
(`a, b, c, d` are constants)

It doesn't matter which eqn we start with, so rearranging eqn1

`2.5x+y=-2`

`y=(-2-2.5x`)

Put this value for `y` into the other eqn (in this case into eqn2)

`3x+2y=0`

`3x+2*(-2-2.5x)=0`

multiply out and simplify

`3x +(-4-5x)=0`

`3x-5x-4=0`

`-2x=4`

`x=-2`

Now put this value for `x` into eqn1 or eqn2

eqn1

`2.5*(-2)+y=-2`

`-5+y=-2`

`y=-2+5 =3`

or eqn2

`3*(-2) + 2y = 0`

`-6 + 2y = 0`

`2y = 6`

`y = 3` (note: same answer whichever we choose)

We can start with either eqn and find either `x` or `y` in terms of the other, as long as we then use the other eqn to substitute our expression into to solve for the variable.