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

Jul 5, 2018

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, $\left(5 x + 2 y\right) - \left(3 x + 2 y\right) = - 4 - 0$ or

$5 x + \cancel{2 y} - 3 x - \cancel{2 y} = - 4$ or

$2 x = - 4 \mathmr{and} x = - 2$. Putting $x = - 2$ in equation (2) we get,

$3 \cdot \left(- 2\right) + 2 y = 0 \mathmr{and} 2 y = 6 \mathmr{and} y = 3 \therefore \left(x = - 2 , y = 3\right)$

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

Jul 5, 2018

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. $y = a x + c$ or $x = b y + d$
($a , b , c , d$ are constants)

$2.5 x + y = - 2$

y=(-2-2.5x)

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

$3 x + 2 y = 0$

$3 x + 2 \cdot \left(- 2 - 2.5 x\right) = 0$

multiply out and simplify

$3 x + \left(- 4 - 5 x\right) = 0$

$3 x - 5 x - 4 = 0$

$- 2 x = 4$

$x = - 2$

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

eqn1

$2.5 \cdot \left(- 2\right) + y = - 2$

$- 5 + y = - 2$

$y = - 2 + 5 = 3$

or eqn2

$3 \cdot \left(- 2\right) + 2 y = 0$

$- 6 + 2 y = 0$

$2 y = 6$

$y = 3$ (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.