The sides of an equilateral triangle are (x + 3y) cm, (3x + 2y − 2) cm and $\Big(4x + \dfrac{y}{2} + 1\Big)$ cm. Find the length of each side.

Given,

In an equilateral triangle, all three sides are equal.

∴ x + 3y = 3x + 2y - 2 = 4x + $\dfrac{y}{2}$ + 1

Solving L.H.S of the above equation, we get :

⇒ x + 3y = 3x + 2y - 2

⇒ x - 3x + 3y - 2y = -2

⇒ -2x + y = -2

⇒ y = 2x - 2     …..(1)

Solving R.H.S of the above equation, we get :

$\Rightarrow 3x + 2y - 2 = 4x + \dfrac{y}{2} + 1 \\[1em] \Rightarrow 3x + 2y = 4x + \dfrac{y}{2} + 1 + 2 \\[1em] \Rightarrow 3x - 4x + 2y - \dfrac{y}{2} = 3 \\[1em] \Rightarrow -x + \dfrac{4y - y}{2} = 3 \\[1em] \Rightarrow -x + \dfrac{3y}{2} = 3 \\[1em] \Rightarrow \dfrac{3y - 2x}{2} = 3 \\[1em] \Rightarrow 3y - 2x = 3 \times 2 \\[1em] \Rightarrow 3y - 2x = 6 \text{ ………(2)}$

Substituting value of y from equation (1) in (2), we get :

⇒ 3(2x - 2) - 2x = 6

⇒ 6x - 6 - 2x = 6

⇒ 4x - 6 = 6

⇒ 4x = 6 + 6

⇒ 4x = 12

⇒ x = $\dfrac{12}{4}$

⇒ x = 3.

Substituting value of x in equation (1), we get :

⇒ y = 2(3) - 2

⇒ y = 6 - 2

⇒ y = 4.

Substituting value of x and y in x + 3y, we get :

⇒ x + 3y = 3 + 3(4) = 3 + 12 = 15 cm.

Since, all sides are equal.

Hence, the length of each side is 15 cm.