Question

How do you solve the following system: `-2x + 5y = 20, 2x – 5y = 5`?

Answer 1

No solution

Explanation:

Some( in fact many) system of equations doesn't have answers... as you'll solve this you'll always get `20=5` which is impossible

it is useful in `11^(th)` standard... so wait till then

Answer 2

No solutions

Explanation:

`-2x + 5y =20`
`2x - 5y = 5`

We need to solve `-2x+5y=20` for `x`

`-2x + 5y = 20`

`-2x = 20 - 5y`

`x=(20-5y)/(-2)`

`x=5/2 y -10`

Now we can substitute `5/2y-10` for `x` in `2x-5y=5`

`2x - 5y=5`

`2(5/2y-10)-5y=5`

Distribute

`(2)(5/2y)+(2)(-10)-5y=5`

`(10/2y)-20-5y=5`

`5y - 20 - 5y = 5`

Combine like terms

`cancel(5y) cancel(-5y) - 20 = 5`

`-20 = 5`

Add `20` to both sides

`-20+20=5+20`

`0=25`

Thus,

There are no solutions!