\(-x + \dfrac 2 5 = -\dfrac 3 5 y\\ x = \dfrac 2 5 + \dfrac 3 5 y = \dfrac 1 5 (2+3y)\\ \text{now look at the second equation}\\ 3y = -\dfrac{18}{11}x + \dfrac{51}{11}\\ \text{substitute in the first result}\\ 3y = -\dfrac{18}{11}\left(\dfrac 1 5(2+3y)\right)+\dfrac{51}{11}\)
\(3y = -\dfrac{18}{55}(2+3y) + \dfrac{51}{11}\\ 3y+\dfrac{18\cdot 3}{55}y=-\dfrac{18\cdot 2}{55}+\dfrac{51}{11}\\ \dfrac{219}{55}y = \dfrac{219}{55}\\ y=1\\ x = \dfrac 1 5 (2+3y) = 1\\ (x,y) = (1,1)\)
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