`13x - 5y = 10` (Let this be EQ1.)

`3x - 2y =14` (Let this be EQ2.)

To solve for values of x and y, elimination method can be applied. So, let's try to eliminate the variable y.

To eliminate y, multiply EQ1 by 2 and EQ3 by -5. So, that the coefficients of y become the same but with different signs.

2*EQ1: `26x - 10y=20`

-5*EQ2: `-15x + 10y = -70`

Then, add the products of 2EQ1 and -5EQ2.

`26x - 10y =20`

`(+)` `-15x+10y=-70`

________________________

`11x` `=` `-50`

And, isolate x.

`(11x)/11=(-50)/11`

`x=-50/11`

Then, substitute this value of x to either of the original equation. So, plug-in x=-50/11 to EQ2.

`3x-2y=14`

`3(-50/11)-2y=14`

`-150/11-2y=14`

`-2y=14+150/11`

`-2y=154/11+150/11`

`-2y=304/11`

`y=(304/11)/-2`

`y=-304/22`

`y=-152/11`

**Hence, the solution to the given system of equations is `(-50/11,-152/11)` .**