Question

Picture attached

A rectangle has side lengths of (x+y)cm and xcm where x and y are both positive

Answer 1

Check the picture below.

`(x+y)(x)=54\\\\ 2(x+y)+2x=42 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{using the 1st equation}}{(x+y)(x)=54}\implies x^2+yx=54\implies yx=54-x^2\implies y=\cfrac{54-x^2}{x} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 2nd equation}}{2(x+y)+2x=42}\implies 2x+2y+2x=42\implies 4x+2y=42 \\\\\\ 2y=42-4x\implies y=\cfrac{42-4x}{2}\implies y=21-2x \\\\[-0.35em] ~\dotfill`

`\stackrel{\textit{since we know that y = y, we can say that}~\hfill~}{\cfrac{54-x^2}{x}~~ = ~~21-2x\implies 54-x^2=21x-2x^2} \implies 54+x^2=21x \\\\\\ x^2-21x+54=0\implies (x-3)(x-18)=0\implies \boxed{x= \begin{cases} 3\\ 18 \end{cases}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{since we know that}}{y = 21-2x}\implies \begin{cases} y=21-2(3)\implies \boxed{y=15}\\\\ y=21-2(18)\implies \boxed{y=-15} \end{cases}`

both values are valid by the way.