Question

linear equation problems
name:
period (circle): 2a 3a 4a 1b 2b 4b
problem 1: cell phone plan
sarah is choosing between two cell phone plans. plan a charges $30 per month plus $0.10 per text message. plan b charges $45 per month with unlimited texting.
part a: write linear equations for the total monthly cost of each plan, where x represents the number of text messages sent.
part b: what does the slope represent in plan as equation?
part c: how many text messages would sarah need to send for both plans to cost the same amount?
problem 2: car rental service
a car rental company charges $25 per day plus $0.15 per mile driven. the total cost can be represented by the equation c = 25d + 0.15m, where c is the total cost, d is the number of days, and m is the number of miles.
part a: if someone rents a car for 3 days, write a simplified equation showing cost based only on miles driven.
part b: what does the coefficient 0.15 represent in real - world terms?
part c: if the total bill was $89.50 for a 3 - day rental, how many miles were driven?

Explanation:

Problem 1

Part A

• Step1: Analyze Plan A

The total cost \(y_A\) for Plan A has a fixed cost of \(30\) dollars and a variable cost of \(0.10\) dollars per text message. Using the slope - intercept form \(y = mx + b\) (where \(m\) is the slope and \(b\) is the y - intercept), the equation is \(y_A=0.1x + 30\).

• Step2: Analyze Plan B

Plan B has a fixed cost of \(45\) dollars with no variable cost (since texting is unlimited). So the equation is \(y_B = 45\).

Part B

In the linear equation \(y=mx + b\) (for Plan A: \(y = 0.1x+30\)), the slope \(m = 0.1\). In the context of the cell - phone plan, the slope represents the cost per text message.

Part C

• Step1: Set the two equations equal

Set \(y_A=y_B\), so \(0.1x + 30=45\).

• Step2: Solve for \(x\)

Subtract \(30\) from both sides: \(0.1x=45 - 30\). Then \(0.1x=15\). Divide both sides by \(0.1\): \(x=\frac{15}{0.1}=150\).

Problem 2

Part A

• Step1: Substitute \(d = 3\) into the cost equation

Given \(C = 25d+0.15m\), when \(d = 3\), we have \(C=25\times3+0.15m\).

• Step2: Simplify

\(C = 75+0.15m\).

Part B

In the equation \(C = 25d+0.15m\), the coefficient \(0.15\) represents the cost per mile driven.

Part C

• Step1: Substitute \(C = 89.5\) and \(d = 3\) into the cost equation

We know \(C = 25d+0.15m\), so \(89.5=25\times3+0.15m\).

• Step2: Simplify the right - hand side

\(25\times3 = 75\), so the equation becomes \(89.5=75 + 0.15m\).

• Step3: Solve for \(m\)

Subtract \(75\) from both sides: \(89.5−75=0.15m\), which gives \(14.5 = 0.15m\). Then \(m=\frac{14.5}{0.15}=\frac{1450}{15}=\frac{290}{3}\approx96.67\).

Answer:

Problem 1

Part A

Plan A: \(y = 0.1x + 30\), Plan B: \(y = 45\)

Part B

The cost per text message.

Part C

\(150\) text messages.

Problem 2

Part A

\(C=75 + 0.15m\)

Part B

The cost per mile driven.

Part C

\(\frac{290}{3}\approx96.67\) miles.