The solution is x=3 and y=2
[tex]\begin{gathered} 4x+2y=16\ldots\ldots eqn(i) \\ 3x+3y=15\ldots\ldots eqn(ii) \\ By\text{ elimination method of solving simultaneous equations, we have,} \\ 3(4x+2y=16) \\ 4(3x+3y=15) \\ 12x+6y=48\ldots\ldots.eqn(iii) \\ 12x+12y=60\ldots\ldots eqn(iv) \\ \text{Subtracting eqn(i}ii)\text{ from eqn(iv), we have,} \\ (12x-12x)+(12y-6y)=(60-48) \\ 6y=12 \\ y=\frac{12}{6} \\ y=2 \\ \text{Substituting 2 for y in eqn(i) , we get,} \\ 4x+2(2)=16 \\ 4x=16-4 \\ 4x=12 \\ x=\frac{12}{4} \\ x=3 \\ \text{Hence, x=3 and y=2} \end{gathered}[/tex]