Say that:
[tex]\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 10,12 \right) \\ \\ \left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 15,18 \right) \\ \\ \therefore \quad \frac { \delta y }{ \delta x } =\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { 18-12 }{ 15-10 } =\frac { 6 }{ 5 } \\ \\ [/tex]
If this is the case,
[tex]y=\frac { 6 }{ 5 } x[/tex]
Another way to solve the problem,
[tex]y\propto x\\ \\ \therefore \quad y=kx[/tex]
When x=10, y=12, so...
[tex]12=k\cdot 10\\ \\ k=\frac { 12 }{ 10 } =\frac { 6 }{ 5 } \\ \\ \therefore \quad y=\frac { 6 }{ 5 } x[/tex]
