Question

problem 1: a bowling alley charges a fixed hourly rate for bowling. one customer bowled for 1 hour and spent $4.75 on snacks. another customer bowled for 2.5 hours and received a $5 discount for using their own bowling ball. if both customers spent the same amount, how much did it cost per hour to bowl? problem 2: cable now charges a $40 installation fee plus $70 per month. stay connected cable has no installation fee plus $80 per month. determine the number of months, m, when the cost of setting - up cable will be the same for both companies. problem 3: company a offers a starting salary of $33,550 with a raise of $2,250 each year. company b offers a starting salary of $40,425 with a raise of $1,000 per year. after how many years will the salaries for companies a and b be the same? problem 4: emma and ava are playing video games at the arcade. emma started with $17, and the machine costs $0.25 per game. ava started with $15, and her machine costs $0.50 per game. after how many games will both emma and ava have the same amount of money remaining? solve & write equations variables on both sides

Explanation:

Problem 1:

Step1: Set up cost - equations for each customer

Let $x$ be the hourly rate. For the first customer who bowled for 2.5 hours and got a $5$ discount and spent $47.5$ on snacks, the cost equation is $2.5x - 5+47.5$. For the second customer who bowled for 1 hour and spent $47.5$ on snacks, the cost equation is $x + 47.5$. Since they spent the same amount, we set the two equations equal to each other.
$2.5x-5 + 47.5=x + 47.5$

Step2: Simplify the equation

First, simplify the left - hand side: $2.5x+42.5=x + 47.5$.
Then, subtract $x$ from both sides: $2.5x - x+42.5=x - x + 47.5$, which gives $1.5x+42.5 = 47.5$.

Step3: Solve for $x$

Subtract 42.5 from both sides: $1.5x+42.5-42.5=47.5 - 42.5$, so $1.5x = 5$.
Divide both sides by 1.5: $x=\frac{5}{1.5}=\frac{10}{3}\approx3.33$.

Step1: Set up cost - equations for the cable services

Let $m$ be the number of months. For Cable Now, the cost equation is $C_1 = 40+70m$. For Stay Connected Cable, the cost equation is $C_2 = 80+80m$. We want to find when $C_1 = C_2$.
$40 + 70m=80+80m$

Step2: Solve for $m$

Subtract $70m$ from both sides: $40+70m-70m=80+80m-70m$, which gives $40 = 80 + 10m$.
Then subtract 80 from both sides: $40-80=80 - 80+10m$, so $-40 = 10m$.
Divide both sides by 10: $m=-4$. Since the number of months cannot be negative, there is no solution in the context of this problem (it means the costs will never be the same in a real - world, non - negative number of months situation).

Step1: Set up salary equations for the two companies

Let $y$ be the number of years. For Company A, the salary equation is $S_A=33550 + 1000y$. For Company B, the salary equation is $S_B=40425+2250y$. We want to find when $S_A = S_B$.
$33550+1000y=40425 + 2250y$

Step2: Solve for $y$

Subtract $1000y$ from both sides: $33550+1000y-1000y=40425+2250y-1000y$, which gives $33550=40425 + 1250y$.
Then subtract 40425 from both sides: $33550-40425=40425-40425 + 1250y$, so $-6875 = 1250y$.
Divide both sides by 1250: $y = - 5.5$. Since the number of years cannot be negative in this context, there is no solution (the salaries will never be the same in a real - world, non - negative number of years situation).

Answer:

The hourly rate to bowl is $\$ \frac{10}{3}\approx\$3.33$

Problem 2: