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Algebra 22 Online
OpenStudy (taz007):

Where do I begin with this question? This season, Lisa's lacrosse team has won 2/3 of their home games (games played at Lisa's school), but just 2/5 of their away games (games played at other schools). In total, Lisa's team has won 26 games out of 49 games they have played. How many home games has Lisa's team played? Explain how you solved the problem.

jimthompson5910 (jim_thompson5910):

Let x = number of home games y = number of away games We know that they have played a total of 49 games. So that means the sum of x and y must be 49, ie x+y = 49 Solve for either variable. I'm going to solve for y to get y = 49 - x

jimthompson5910 (jim_thompson5910):

"Lisa's lacrosse team has won 2/3 of their home games" So (2/3)*x is the number of home games the team has won. The team has also won "2/5 of their away games" They won (2/5)*y away games with me so far?

OpenStudy (taz007):

Yes, I am with you.

jimthompson5910 (jim_thompson5910):

so in total, we have (2/3)*x + (2/5)*y games won

jimthompson5910 (jim_thompson5910):

that total is equal to the 26 total games won

jimthompson5910 (jim_thompson5910):

that means (2/3)*x + (2/5)*y = 26

jimthompson5910 (jim_thompson5910):

Now those fractions complicate things, so we can multiply both sides by the LCD 15 to clear them out. (2/3)*x + (2/5)*y = 26 15*[ (2/3)*x + (2/5)*y ] = 15*26 15*[ (2/3)*x ] + 15*[ (2/5)*y ] = 15*26 10x + 6y = 390

jimthompson5910 (jim_thompson5910):

So we have gone from (2/3)*x + (2/5)*y = 26 to 10x + 6y = 390

jimthompson5910 (jim_thompson5910):

See where to go from here?

OpenStudy (taz007):

Got it!

jimthompson5910 (jim_thompson5910):

ok great. Tell me what you get

OpenStudy (taz007):

10h = 390 -6a?

jimthompson5910 (jim_thompson5910):

so you let h = # of home games a = # of away games just so we're on the same page

OpenStudy (taz007):

yes

jimthompson5910 (jim_thompson5910):

we have this 10h + 6a = 390

jimthompson5910 (jim_thompson5910):

previously I solved for y, which is also 'a' (two names for the same thing) I got y = 49 - x which is the same as a = 49 - h

jimthompson5910 (jim_thompson5910):

You then take that and plug it into 10h + 6a = 390 10h + 6a = 390 10h + 6(49 - h) = 390 ... everywhere you see an 'a', replace it with 49-h now solve for h

jimthompson5910 (jim_thompson5910):

That substitution only works because a = 49 - h

OpenStudy (taz007):

10h + 294 -6h = 390 4h = 96 h=96/4 h=24

OpenStudy (taz007):

So, the total number of home games is 24

jimthompson5910 (jim_thompson5910):

very good

OpenStudy (taz007):

Thank you, thank you, THANK YOU!

jimthompson5910 (jim_thompson5910):

that leaves 49-h = 49-24 = 25 away games notice how (2/3)*h = (2/3)*24 = 16 games were won at home (2/5)*a = (2/5)*25 = 10 games were won on the road So 16+10 = 26 games were won total (that works out)

jimthompson5910 (jim_thompson5910):

you're welcome

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