5Number 5ExplanationThe perimeter of a rectangle is x cmx \text{ cm}x cm. If the length and width are (8y+5) cm(8y + 5) \text{ cm}(8y+5) cm and (6y+1) cm(6y + 1) \text{ cm}(6y+1) cm respectively, then the value of yyy is ....(x28+47) cm\left(\frac{x}{28} + \frac{4}{7}\right) \text{ cm}(28x+74) cm(x28+27) cm\left(\frac{x}{28} + \frac{2}{7}\right) \text{ cm}(28x+72) cm(x28−27) cm\left(\frac{x}{28} - \frac{2}{7}\right) \text{ cm}(28x−72) cm(x28+37) cm\left(\frac{x}{28} + \frac{3}{7}\right) \text{ cm}(28x+73) cm(x28−37) cm\left(\frac{x}{28} - \frac{3}{7}\right) \text{ cm}(28x−73) cmExplanationGiven: Perimeter (KKK) = x cmx \text{ cm}x cm Length (ppp) = (8y+5) cm(8y + 5) \text{ cm}(8y+5) cm Width (lll) = (6y+1) cm(6y + 1) \text{ cm}(6y+1) cm The formula for the perimeter of a rectangle is K=2(p+l)K = 2(p + l)K=2(p+l). We substitute the known values into the formula. x=2((8y+5)+(6y+1))x = 2((8y + 5) + (6y + 1))x=2((8y+5)+(6y+1))x=2(14y+6)x = 2(14y + 6)x=2(14y+6)x=28y+12x = 28y + 12x=28y+12 Next, we rearrange the equation to solve for yyy. 28y=x−1228y = x - 1228y=x−12y=x−1228y = \frac{x - 12}{28}y=28x−12y=x28−1228y = \frac{x}{28} - \frac{12}{28}y=28x−2812y=x28−37y = \frac{x}{28} - \frac{3}{7}y=28x−73 Thus, the value of yyy is (x28−37) cm\left(\frac{x}{28} - \frac{3}{7}\right) \text{ cm}(28x−73) cm.
