Question

How do you solve the system `5x-10y=15` and `3x-2y=3` by multiplication?

Answer 1

See explanation.

Explanation:

The system is:

`{(5x-10y=15),(3x-2y=3):}`

First we can see that all terms in the first equation are divisible by `5`. So we can divide both sides bu `5` to get the lower numbers:

`{(x-2y=3),(3x-2y=3):}`

Now the coefficients of `y` are the same in both equations `(-2)`, so if we multiply any of the equations by `-1` we get the opposite coefficients:

`{(-x+2y=-3),(3x-2y=3):}`

Now if we add both sides of both equations we get an equation with one unknown only:

`2x=0`

`x=0`

Now we can substitute the calculated alue of `x` into any of the previous equations to calculate `y`:

`3*0-2y=3`

`-2y=3`

`y=-3/2`

Now we can write the answer:

The system has one solution:

`{(x=0),(y=-3/2):}`