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?

Answer 1

Answer:

Step-by-step explanation:

Let X be the at home games and Y be the away games

then  

X+Y=49     (1)      

and

(2x/3 + 2y/5)=26

15(2x/3 + 2y/5)=15*26

(15*2x/3) + (15*2y/5)=390

5*2x + 3*2y=390

10x + 6y=390

5x + 3y=195      (2)

3X+3Y=147        3*(1)          (1b)

(2)-(1b)

2X=48

X=24

X+Y=49

24+Y=49

Y=49-24

y=25

24 home games      25 away games.

Answer 2

Answer:

24 Step by step is Latex form or somethin

Step-by-step explanation:

We can introduce a variable, say, $h$, for the number of home games Lisa's team has played. Since the total number of games they have played is $49$, they must have played $49-h$ away games.

Thus, Lisa's team has won $\frac{2}{3}h$ home games and $\frac{2}{5}(49-h)$ away games. We know they have won a total of $26$ games, so

$$\frac{2}{3}h + \frac{2}{5}(49-h) = 26.$$We can eliminate denominators from the problem by multiplying both sides of the equation by $3\cdot 5$:

$$\frac{2\cdot 3\cdot 5}{3}h + \frac{2\cdot 3\cdot 5}{5}(49-h) = 26\cdot 3\cdot 5.$$Simplifying this equation, we have

$$10h + 6\cdot (49-h) = 390.$$Expanding the left side using the distributive property, we have

$$10h + 294 - 6h = 390.$$Subtracting $294$ from both sides and simplifying $10h-6h$ to $4h$, we have

$$4h = 96.$$Finally, dividing both sides by $4$, we have $h=24$. Therefore, Lisa's team has played $\boxed{24}$ home games.

We can check that this answer is correct. If Lisa's team has played $24$ home games, then they have won $\frac 23\cdot 24 = 16$ of those games; also, they have played $49-24=25$ away games, and they have won $\frac 25\cdot 25 = 10$ of those games. So, they have won $16+10=26$ games in all, which is what we were expecting.