Question 131349
{{{d}}}=distance in mi
{{{r}}}=rate in mi/hr
{{{t}}}=time in hours
{{{d = r*t}}}
write an equation for both of them
{{{d[c] = r[c]*t[c]}}} This is Cindy's equation
{{{d[b] = r[b]*t[b]}}} This is Bill's equation
{{{d[c] = d[b]}}} They ran the same distance, {{{d = 6}}}mi.
{{{t[c] = t[b] - .25}}}
{{{r[c] = r[b] + 2}}}
{{{d[c] = r[c]*t[c]}}}
{{{6 = (r[b] + 2)(t[b] - .25)}}}
{{{r[b]*t[b] = 6}}}
{{{t[b] = 6/r[b]}}}
{{{6 = (r[b] + 2)((6/r[b]) - .25)}}}
{{{6 = 6 + (12/r[b]) - .25*r[b] - .5}}}
{{{(12/r[b]) - .25*r[b] - .5 = 0}}}
multiply both sides by {{{r[b]}}}
{{{12 - .25*(r[b])^2 - .5*r[b] = 0}}}
{{{.25*(r[b])^2 + .5*r[b] - 12 = 0}}}
multiply both sides by {{{4}}}
{{{r[b]^2 + 2*r[b] - 48 = 0}}}
Solve by completing the square
{{{r[b]^2 + 2*r[b] + 1 = 48 + 1}}}
{{{(r[b] + 1)^2 = 49}}}
{{{r[b] + 1 = 7}}} only the + root makes sense here
{{{r[b] = 6}}} mi/hr answer
check answer:
{{{6 = r[b]*t[b]}}}
{{{6 = 6*(t[b])}}}
{{{t[b] = 1}}} hr
{{{r[c] = r[b] + 2}}}
{{{r[c] = 8}}}mi/hr
{{{t[c] = t[b] - .25}}}
{{{t[c] = 1 - .25}}}
{{{t[c] = .75}}}hr
{{{6 = r[c]*t[c]}}}
{{{6 = 8*.75}}}
{{{6 = 6}}}
OK