`4x + 3y = 1 ` (Let this be EQ1.)

`2x - 5y = -19` (Let this be EQ2.)

Note that in addition method, two equations are added to eliminate one of the variables.

For the given equations above, let's eliminate x. But before adding the two equations, multiply EQ2 by -2.

`-2(2x-5y=-19)` `===gt` `-4x+10y=38`

Now the coefficient of x in EQ2 is the same with EQ1 but opposite in sign, proceed to add the two equations to eliminate x.

`4x` `+` `3y` `=` `1`

`(+)` `-4x` `+` `10y` `=` `38`

`---------------`

`0` `+` `13y` `=` `39`

`13y` `=` `39`

Then, divide both sides by 13 to isolate the y.

`(13y)/13=39/13`

`y=3`

Next, substitute this value of y to either of the given equations. And solve for x.

Let's substitute it to EQ2.

`2x-5y=-19`

`2x-5(3)=-19`

`2x-15=-19`

To determine the value of x, add both sides by 15.

`2x-15+15=-19+15`

`2x=-4`

And divide both sides by 2.

`(2x)/2-(-4)/2`

`x=-2`

**Hence, the solution tot the given equation is (-2,3).**