Write the available informations∴ The length of first rectangle is 32 cm
l₁ = 32
∴ The length of second rectangle is 15 cm
l₂ = 15 cm
∴
The width of the second rectangle is 6 cm longer than the width of the first rectangle
w₂ = 6 + w₁ (first equation)
∴ The area of first rectangle is 46 cm² greater than the area of the second rectangle
a₁ = 46 + a₂ (second equation)
Solve the equation system
From the second equation, elaborate the area using area formula
area of a rectangle could be determined using formula
a₁ = 46 + a₂
(l₁ × w₁) = 46 + (l₂ × w₂)
plug in the length
32 × w₁ = 46 + (15 × w₂)
32w₁ = 46 + 15w₂
now substitute w₂ in the term of w₁ (see 'first equation')
32w₁ = 46 + 15w₂
32w₁ = 46 + 15(6 + w₁)
32w₁ = 46 + 90 + 15w₁
move the one with coefficient to the left side, add like terms
32w₁ - 15w₁ = 136
17w₁ = 136
w₁ = 136/17
w₁ = 8
Now find w₂ using first equation
w₂ = 6 + w₁
w₂ = 6 + 8
w₂ = 14
Find the area
a₁ = l₁ × w₁
a₁ = 32 × 8
a₁ = 256
a₂ = l₂ × w₂
a₂ = 15 × 14
a₂ = 210
The area of the first rectangle is 256 cm² and the area of the second rectangle is 210 cm²