Answer
$(50,20), (46,28)$
Work Step by Step
Let's note:
$x$=the length of the first square's side
$y$=the length of the second square's side
We can write the system:
$\begin{cases}
x^2+y^2=2900\\
y+3x+(x-y)+2y=240
\end{cases}$
$\begin{cases}
x^2+y^2=2900\\
4x+2y=240
\end{cases}$
$\begin{cases}
x^2+y^2=2900\\
2x+y=120
\end{cases}$
We will use the substitution method. Solve Equation 2 for $y$ and substitute the expression of $y$ in Equation 1 to eliminate $y$ and determine $x$:
$\begin{cases}
x^2+(120-2x)^2=2900\\
y=120-2x
\end{cases}$
$x^2+(120-2x)^2=2900$
$x^2+14400-480x+4x^2=2900$
$5x^2-480x+14400-2900=0$
$5x^2-480x+11500=0$
$5(x^2-96x+2300)=0$
$x^2-96x+2300=0$
$x^2-50x-46x+2300=0$
$x(x-50)-46(x-50)=0$
$(x-50)(x-46)=0$
$x-50=0\Rightarrow x_1=50$
$x-46=0\Rightarrow x_2=46$
Substitute each of the values of $x$ in the expression of $y$ to determine $y$:
$y=120-2x$
$x_1=50\Rightarrow y_1=120-2(50)=20$
$x_2=46\Rightarrow y_2=120-2(46)=28$
The system's solutions are:
$(50,20), (46,28)$
