Question

1. geometry write and solve an equation to find the value of x so that the figures have the same area.

Explanation:

Step1: Calculate area of first figure

The first figure is a rectangle with length $(5 + x)$ ft and width $9$ ft. Its area $A_1=9(5 + x)=45+9x$.

Step2: Calculate area of second figure

The second figure is a rectangle with length $15\frac{3}{7}=\frac{108}{7}$ ft and width $(x + 6)$ ft. Its area $A_2=\frac{108}{7}(x + 6)=\frac{108x}{7}+\frac{648}{7}$.

Step3: Set areas equal and solve for x

Set $A_1 = A_2$. So $45+9x=\frac{108x}{7}+\frac{648}{7}$. Multiply through by 7 to clear the fraction: $7\times45+7\times9x = 108x+648$. Which simplifies to $315 + 63x=108x+648$. Subtract $63x$ from both sides: $315=108x - 63x+648$, or $315 = 45x+648$. Subtract 648 from both sides: $315 - 648=45x$, so $- 333 = 45x$. Then $x=\frac{-333}{45}=-\frac{37}{5}=-7.4$.

Answer:

$x=-7.4$