Example — Carrying out the rectangle problem

From the system {2x+2y=24xy=32\begin{cases} 2x+2y=24 \\ xy=32 \end{cases}, I isolate yy from the first equation: y=12x.y = 12 - x. I substitute into the second: x(12x)=32    x2+12x32=0    x212x+32=0.x\,(12-x) = 32 \;\Longrightarrow\; -x^2+12x-32=0 \;\Longrightarrow\; x^2-12x+32=0. I use the quadratic formula: x=12±1441282=12±42.x = \frac{12\pm\sqrt{144-128}}{2} = \frac{12\pm 4}{2}. Hence x=8x=8 or x=4x=4. In the two cases, y=12xy=12-x gives y=4y=4 or y=8y=8. The rectangle has sides 4 cm\boxed{4\ \text{cm}} and 8 cm\boxed{8\ \text{cm}} (the two solutions correspond to the same rectangle, seen from its base or from its height).

Topics: Working method
Concepts: Write-up
Skills: Calculating · Solving equations · Solving systems