A car tracked 640 miles at a certain speed. If the speed of had been 8 mph faster, the trip could have been made in 4 hours less time. Find the speed.

Answer:[tex]\large \boxed{\text{32 mi/h}}[/tex]Step-by-step explanation:Distance = rate × time, or  d=rt, so[tex]t = \dfrac{d}{r}[/tex]          Let x = original speed.Then x + 8 = faster speed        Let t₁  = time at original speed       And t₂ = time at faster speed     Then t₂ = t ₁ - 4[tex]\begin{array}{rcl}t_{2} & = & t_{1} - 4\\\dfrac{640}{x + 8 } & = & \dfrac{640}{x} - 4\\\\640& = & \dfrac{640(x + 8)}{x} - 4(x + 8)\\\\640x& = & 640(x + 8) - 4x(x + 8)\\640x& = & 640x + 5120 - 4x^{2} - 32x\\4x^{2} + 32x - 5120& = & 0\\x^{2} + 8x - 1280& = & 0\\x + 40 = 0 & \text{or} & x - 32 = 0\\x = -40 & \text{or} & \mathbf{x = 32}\\\end{array}\\\text{We reject the negative speed, so $\mathbf{x = 32}$}\\\text{The original speed was $\large \boxed{\textbf{32 mi/h}}$}[/tex]Check:[tex]\begin{array}{rcl}\dfrac{640}{x + 8 } & = & \dfrac{640}{x} - 4\\\\\dfrac{640}{32 + 8} & = & \dfrac{640}{32 } - 4\\\\\dfrac{640}{40} & = & 20 - 4\\\\16 & = & 16\\\end{array}[/tex]OK.