Let t be the time in minutes after 7:10 at which train B catches train A.
The distance (in funny units) traveled by train A from the station to the meeting point is given by
d = t*100
The distance traveled by train B from the station to the meeting point is
d = (t-12)120
The two distances are equal (and we don't care what they are), so we can write
t*100 = (t-12)*120
100t = 120t - 1440 (use the distributive property)
0 = 20t - 1440 (subtract 100t from both sides)
0 = t - 72 (divide both sides by 20. 0/20 = 0)
72 = t (add 72 to both sides)
7:10 + 72 minutes = 7:10 + 1:12 = 8:22
Train B will catch train A at 8:22 pm.
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The term 100t has units (miles/hour)(minutes). This is distance multiplied by (minutes/hour), or 60 times the distance in miles. Both sides of the equation have these funny units, so no harm is done. Usually, it is a good idea to pay attention to the units of the problem. It helps you make sure you're performing the arithmetic correctly.
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If you are not intimidated by numbers in different number bases, you can work with the time directly.
100(t - 7:10) = 120(t - 7:22) (equation where t is actual clock time)
5t - 5(7:10) = 6t - 6(7:22) (divide both sides by 20 to make the numbers manageable)
6(7:22) - 5(7:10) = t (subtract 5t-6*7:22 from both sides)
(5+1)(7:10+0:12) - 5(7:10) = t (rearrange the product on the left. This makes it possible to cancel the 5*7:10 term. It is really an unnecessary step, but it makes the calculation a little more straightforward.)
5(7:10) + 5(0:12) + 1(7:10) + 1(0:12) - 5(7:10) = t (use the distributive law)
5(0:12) + 7:22 = t (collect terms. Recognize that 7:10 + 0:12 is 7:22)
1:00 + 7:22 = t (recognize that 5*0:12 is 0:60, one hour)
8:22 = t (clock time at which the trains meet)
The distance (in funny units) traveled by train A from the station to the meeting point is given by
d = t*100
The distance traveled by train B from the station to the meeting point is
d = (t-12)120
The two distances are equal (and we don't care what they are), so we can write
t*100 = (t-12)*120
100t = 120t - 1440 (use the distributive property)
0 = 20t - 1440 (subtract 100t from both sides)
0 = t - 72 (divide both sides by 20. 0/20 = 0)
72 = t (add 72 to both sides)
7:10 + 72 minutes = 7:10 + 1:12 = 8:22
Train B will catch train A at 8:22 pm.
_____
The term 100t has units (miles/hour)(minutes). This is distance multiplied by (minutes/hour), or 60 times the distance in miles. Both sides of the equation have these funny units, so no harm is done. Usually, it is a good idea to pay attention to the units of the problem. It helps you make sure you're performing the arithmetic correctly.
_____
If you are not intimidated by numbers in different number bases, you can work with the time directly.
100(t - 7:10) = 120(t - 7:22) (equation where t is actual clock time)
5t - 5(7:10) = 6t - 6(7:22) (divide both sides by 20 to make the numbers manageable)
6(7:22) - 5(7:10) = t (subtract 5t-6*7:22 from both sides)
(5+1)(7:10+0:12) - 5(7:10) = t (rearrange the product on the left. This makes it possible to cancel the 5*7:10 term. It is really an unnecessary step, but it makes the calculation a little more straightforward.)
5(7:10) + 5(0:12) + 1(7:10) + 1(0:12) - 5(7:10) = t (use the distributive law)
5(0:12) + 7:22 = t (collect terms. Recognize that 7:10 + 0:12 is 7:22)
1:00 + 7:22 = t (recognize that 5*0:12 is 0:60, one hour)
8:22 = t (clock time at which the trains meet)