`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`
`11x` `=` `-50`
And, isolate x.
Then, substitute this value of x to either of the original equation. So, plug-in x=-50/11 to EQ2.
Hence, the solution to the given system of equations is `(-50/11,-152/11)` .