Question 772331
{{{speed*times=distance}}}
 
THE ALGEBRA WAY:
Since the units agree,
{{{(25x+1)(4x-6)=102}}} --> {{{100x^2-150x+4x-6=102}}} --> {{{100x^2-146x-6=102}}} --> {{{100x^2-146x-6-102=0}}} --> {{{100x^2-146x-108=0}}}
Now we solve that quadratic equation.
Using the quadratic formula, {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} ,
{{{x = (-(-146) +- sqrt((-146)^2-4*100*(-108)))/(2*100) }}}-->{{{x = (146 +- sqrt(21316+43200))/200 }}}-->{{{x = (146 +- sqrt(64516))/200 }}}-->{{{x = (146 +- 254)/200 }}}
{{{x = (146 - 254)/200=-108/200=-0.54}}} would give a negative {{{4x-6=-8.16}}} number for the hours driving.
That does not make sense, so we can only use the other solution.
{{{x = (146 + 254)/200 }}}-->{{{x = 400/200 }}}-->{{{x = 2}}}
So the speed (in km/h) is {{{25x+1=25*2+1=50+1=51}}}
The speed is {{{highlight(51km*h^(-1))}}}
 
THE FIFTH GRADER WAY:
If {{{25x+1}}} is the speed, {{{x}}} must be a small number, maybe 0, or 1, or 2, or 3.
The numbers 0 and 1, give believable speeds, but the time driving turns out negative.
For {{{x=0}}}, {{{4x-6=-6}}}, and for {{{x=1}}}, {{{4x-6=4-6=-2}}}
Let's try {{{x=2}}}
That makes the speed {{{25x+1=25*2+1=50+1=51}}} km/h.
It makes the time {{{4x-6=4*2-6=8-6=2}}} hours.
That makes the distance (2 hours) X (51 km/h)= 102 km, so {{{highlight{x=2)}}} and the speed is {{{highlight(51km*h^(-1))}}}