By elimination, it means that we should apply algebraic operations so we find the value of one variable. So first, lets multiply the first equation by 5. We get
[tex]5\cdot(2x-3y)=5\cdot-5\text{ = 10x-15y = -25}[/tex]Now, lets multiply by 2 the second equation
[tex]2\cdot(5x+2y)\text{ = 16}\cdot2\text{ = 10x+4y=32}[/tex]With this two equations, lets subtract the second equation from the first equation
[tex]10x+4y-(10x-15y)\text{ = 32-(-25)}[/tex]We get
[tex]19y\text{ = 57}[/tex]If we divide y by 19 we get
[tex]y=\frac{57}{19}=3[/tex]Now, using this value in the second equation we get
[tex]5x+2\cdot3\text{ = 16 }=5x+6[/tex]If we subtract 6 on both sides, we get
[tex]16-6\text{ = 5x = 10}[/tex]Finally, we divide by 5 on both sides and we get
[tex]x=\frac{10}{5}=2[/tex]