In the 1st question , we are given with a rectangle whose breadth is 4x and length is 2x+5y and we need to find it's perimeter in simplified form . Now , as we knows that for any rectangle with Length 'l' and breadth 'b' it's perimeter is given by 2(l+b) . Now , applying the same concept in our question , we have
[tex]{:\implies \quad \sf Perimeter=2(2x+5y+4x)}[/tex]
[tex]{:\implies \quad \sf Perimeter=2(6x+5y)}[/tex]
[tex]{:\implies \quad \sf Perimeter=2\times 6x+2\times 5y}[/tex]
[tex]{:\implies \quad \bf \therefore \quad \underline{\underline{Perimeter=12x+10y}}}[/tex]
Now , in 2nd question , we are given That Jeanelle bought 3 chocolate bars more than as of Eashan , and Rameen bought twice as many as Jeanelle . So , now here assume that Eashan bought x chocolate bars . So , Jeanelle have (x+3) bars and Rameen have 2(x+3) bars and all chocolate bars add up to 17 .Now , According to question
[tex]{:\implies \quad \sf x+(x+3)+2(x+3)=17}[/tex]
[tex]{:\implies \quad \sf x+x+3+2x+6=17}[/tex]
[tex]{:\implies \quad \sf (2x+x+x)+(3+6)=17}[/tex]
[tex]{:\implies \quad \sf 4x+9=17}[/tex]
Subtracting 9 from both sides
[tex]{:\implies \quad \sf 4x+\cancel{9}-\cancel{9}=17-9}[/tex]
[tex]{:\implies \quad \sf 4x=8}[/tex]
[tex]{:\implies \quad \sf x=\dfrac{8}{4}}[/tex]
[tex]{:\implies \quad \bf \therefore \quad \underline{\underline{x=2}}}[/tex]
Now ,