A international phone call cost a fixed amount for the first minute and a certain rate per each additional minute. If a 7 minute phone call cost $10 and a $4 call cost &6.40, find the fixed charge and the rate per each additional minute.
Variables,
X=first, (fixed) rate
y=second, rate(/minute)
t=time of call
z=total cost
Equation,
z=x+((t-1)*y)
Values,
7minutes is $10, 4minutes is $6.40
Plug them in, isolate one variable, set them equal to each other, and solve for the remaining variable,
10=x+((7-1)*y), 6.4=x+((4-1)*y)
10=x+(6*y), 6.4=x+(3*y)
10=x+(6*y), 6.4=x+(3*y)
-6y= -6y, -3y= -3y
10-6y=x, 6.4-3y=x
10-6y= 6.4-3y
-6.4=-6.4
3.6-6y=-3y
+6y=+6y
3.6=3y
(3.6)/3=(3y)/3
1.2=y
pick one original equation, plug in your new value, and solve for the remaining variable
10-6y=x
10-6(1.2)=x
10-7.2=x
2.8=x
Fixed first minute is $2.80, with $1.20 per each additional minute
I gave you the equations so that you could have them to help you on later problems. And I hope the directions helped.
To solve this problem, let's assign variables to the fixed charge and the rate per each additional minute.
Let's say the fixed charge is represented by 'F' (in dollars) and the rate per each additional minute is represented by 'R' (in dollars per minute).
Now, let's set up two equations using the given information:
Equation 1: For a 7-minute phone call: F + 6R = 10
Equation 2: For a 4-minute phone call: F + 3R = 6.40
We now have a system of two equations with two unknowns. To solve this system, we can use a method called substitution.
Let's solve for F in Equation 2:
F = 6.40 - 3R
Now substitute this value of F into Equation 1:
(6.40 - 3R) + 6R = 10
Simplifying this equation gives:
6.40 + 3R = 10
Subtracting 6.40 from both sides gives:
3R = 3.60
Dividing by 3:
R = 1.20
Now, substitute this value of R back into Equation 1 to solve for F:
F + 6(1.20) = 10
F + 7.20 = 10
Subtracting 7.20 from both sides gives:
F = 2.80
Therefore, the fixed charge (F) is $2.80 and the rate per each additional minute (R) is $1.20.