Respuesta :
Since the distance = speed*time we have:
(1/2)*x = (3/4)(x - 2)
........
........
x = 6 mi
You ran 6 miles on the treadmill on Monday.
recall your d = rt, distance = rate * time.
so, the distance "d" miles for each day is the same, however the rate on Monday is "x", whilst the rate on Tuesday is "x - 2".
[tex]\bf \begin{array}{lccclll} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ Monday&d&x&\frac{1}{2}\\\\ Tuesday&d&x-2&\frac{3}{4} \end{array} \\\\\\ \begin{cases} d=\frac{x}{2}\implies 2d=\boxed{x}\\\\ d=\frac{3}{4}(x-2)\\ ----------\\ d=\frac{3}{4}\left( \boxed{2d}-2 \right) \end{cases} \\\\\\ d=\cfrac{3(2d-2)}{4}\implies 4d=6d-6\implies 6=2d\implies \cfrac{6}{2}=d \\\\\\ 3=d[/tex]
so, the distance "d" miles for each day is the same, however the rate on Monday is "x", whilst the rate on Tuesday is "x - 2".
[tex]\bf \begin{array}{lccclll} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ Monday&d&x&\frac{1}{2}\\\\ Tuesday&d&x-2&\frac{3}{4} \end{array} \\\\\\ \begin{cases} d=\frac{x}{2}\implies 2d=\boxed{x}\\\\ d=\frac{3}{4}(x-2)\\ ----------\\ d=\frac{3}{4}\left( \boxed{2d}-2 \right) \end{cases} \\\\\\ d=\cfrac{3(2d-2)}{4}\implies 4d=6d-6\implies 6=2d\implies \cfrac{6}{2}=d \\\\\\ 3=d[/tex]
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