The length of the first rectangle is 32 cm and the length of the second one is 15 cm. the width of the second rectangle is 6 cm longer than the width of the first rectangle. find their areas if it is given that the area of the first rectangle is 46 cm2 greater than the area of the second rectangle.

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

The answer would be 484

24 students walk to school