Question 1206158
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An informal solution (good mental exercise!!)....<br>
They won twice as many games as they drew.
For each "group" of two wins and one draw, they earned 2(3)+1(1) = 6+1 = 7 points.
All together, they earned 70 points; the number of those "groups" must have been 70/7 = 10.
So they must have won 10*2 = 20 games and drawn 10*1 = 10 games.
That's a total of 20+10 = 30 games in which they earned points.
Since the total number games was 38, the number of games they lost was 38-30 = 8.<br>
ANSWER: 8<br>
The EXACT same solution, formally....<br>
let x = # of games they drew
then 2x = # of games they won<br>
At 3 points for each win and 1 point for each draw (and 0 points for each loss), they earned a total of 70 points:<br>
{{{2x(3)+x(1)=70}}}
{{{6x+x=70}}}
{{{7x=70}}}
{{{x=70/7=10}}}<br>
# of games won: 2x = 2(10) = 20
# of games drawn: x = 10
# of games in which they earned points: 20+10 = 30
# of games they lost: 38-30 = 8<br>