Question

Solve this pleasee!!!!!!!!!!!

Answer 1

Answer:

1) Solution (2,-3)

2) Solution (-1, $-\frac{3}{2}$)

3) Solution ($\mathbf{-\frac{1}{2},-\frac{5}{2}}$)

Step-by-step explanation:

We need to solve equations using elimination

1) $2x+2y=-2\\3x-2y=12$

Let:

$2x+2y=-2--eq(1)\\3x-2y=12--eq(2)$

Adding both equations:

$2x+2y=-2\\3x-2y=12\\--------\\5x=10\\x=\frac{10}{5}\\x=2$

We get value of x = 2

Now, put value of x into equation 1 to find value of y

$2x+2y=-2\\Put\:x=2\\2(2)+2y=-2\\4+2y=-2\\2y=-2-4\\2y=-6\\y=-6/2\\y=-3$

So, we get y = -3

Solution (2,-3)

2) $4x-2y=-1\\-4x+4y=-2$

Let:

$4x-2y=-1--eq(1)\\-4x+4y=-2--eq(2)$

Adding both equations

$4x-2y=-1\\-4x+4y=-2\\------\\2y=-3\\y=\frac{-3}{2}$

We get: y= -3/2

Now find value of x by putting value of y in eq(1)

$4x-2y=-1\\Put\:y=\frac{-3}{2}\\4x-2( \frac{-3}{2})=-1\\4x+3=-1\\4x=-1-3\\4x=-4\\x=\frac{-4}{4}\\x=-1$

We get x = -1

Solution (-1, $-\frac{3}{2}$)

3) $x-y=2\\x+y=-3$

let:

$x-y=2--eq(1)\\x+y=-3--eq(2)$

Add both equations:

$x-y=2\\x+y=-3\\-----\\2x=-1\\x=-\frac{1}{2}$

We get x = -1/2

Now, put value of x in equation 2

$x+y=-3\\-\frac{1}{2}+y=-3\\y=-3+ \frac{1}{2}\\y=\frac{-6+1}{2}\\y=\frac{-5}{2}$

We get y = -5/2

Solution ($\mathbf{-\frac{1}{2},-\frac{5}{2}}$)