# Thread: Using an algebraic method to determine... (simultaneous equations???)

1. ## Using an algebraic method to determine... (simultaneous equations???)

Hi I think I've figured out question 1, but I don't know how to start question 2.

This is the first question, and how I answered it:

Two transport companies advertise their rates for the cost of a trip:

Blue Taxis: Flagfall = $2.50 and Cost per km =$1.50 per km
Red Taxis: Flagfall = $1.25 and Cost per km=$2.25 per km

Using an algebraic method, develop a linear function for the cost (y) of a trip of distance x km for each company. Show the strategies used to find each of your equations.

Using y=mx+c

Blue Taxis
m=cost per km (y-intercept)
c=flagfall (constant)

Given m=1.5, c=2.5

y=1.5x+2.5

Red Taxis
m=cost per km (y-intercept)
c=flagfall (constant)

Given m= 2.25, c= 1.25

y=2.25x+1.25

Firstly, is that right?

This is the question I need help on:

Using an algebraic method involving your equations from question 1 determine at what distance the cost to the customer would be the same for both companies. WHat is that cost?

2. Originally Posted by bemypenguinxx
Hi I think I've figured out question 1, but I don't know how to start question 2

This is the first question, and how I answered it:
Two transport companies advertise their rates for the cost of a trip:

Blue Taxis: Flagfall = $2.50 and Cost per km =$1.50 per km
Red Taxis: Flagfall = $1.25 and Cost per km=$2.25 per km

Using an algebraic method, develop a linear function for the cost (y) of a trip of distance x km for each company. Show the strategies used to find each of your equations.

Using y=mx+c

Blue Taxis
m=cost per km (y-intercept)
c=flagfall (constant)

Given m=1.5, c=2.5
]y=1.5x+2.5

Red Taxis
m=cost per km (y-intercept)
c=flagfall (constant)

Given m= 2.25, c= 1.25

y=2.25x+1.25

Firstly, is that right?
Yes, it is correct.

Originally Posted by bemypenguinxx
This is the question I need help on:

Using an algebraic method involving your equations from question 1 determine at what distance the cost to the customer would be the same for both companies. WHat is that cost?

Ther are saying at which point cost will be the same. Cost is your $y$ variable so equate it.

$\textcolor{blue}{\text{Blue}} \rightarrow y=1.5x+2.5$
$\textcolor{red}{\text{Red}} \rightarrow y=2.25x+1.25$

$y$ variable are equal so...

$1.5x+2.5 = 2.25x+1.25$

Now solve for $x$ to get the distance in km at which the cost will be the same. After you find $x$ (the distance), substitute back into the original equation (any, Blue or Red equation that you found earlier) to get the cost.

3. Thankyou soo much! I think I've got it - can you check?

Blue taxis: y= 1.5x+2.5
Red taxis: y= 2.25x+1.25

1.5x+2.5 = 2.25x+1.25

Using graphics calculator to solve for x

*screenshot of graph*

y=5, x=1.67

*the x= should be an approximate equals sign)

The distance at which the cost will be the same for the two taxi companies (x) is approximately $1.67 Substitute x back into an original equation to find the cost of trip Blue taxis: 1.5x+2.5=y (1.5x1.67)+2.5=y =5.01 (again, approximate equals sign same for equation below) Verify using equation for red taxis Red taxis: 2.25x+1.25=y (2.25x1.67)+1.25=y =5.01 Is that right, or if not am I on the right track? I'm using the TI-84 graphics calculator and got the y and x values by graphing them and going to 2nd-calc-5:intersect which gave me the values from the graph of the two lines. the x was 1.6666667 and y was 5 because i rounded it to 1.67 and used that number for my equations, instead of the more precise answer of 1.6666667 is my answer wrong? sorry if that dosen't make sense. if that does make the answer wrong, is there a way i can insert that value into my equation when solving it in my calculator? sorry if that question was a bit vague. thanks for your help 4. Originally Posted by bemypenguinxx Blue taxis: 1.5x+2.5=y (1.5x1.67)+2.5=y =5.01 (again, approximate equals sign same for equation below) Verify using equation for red taxis Red taxis: 2.25x+1.25=y (2.25x1.67)+1.25=y =5.01 Is that right, or if not am I on the right track? I'm using the TI-84 graphics calculator and got the y and x values by graphing them and going to 2nd-calc-5:intersect which gave me the values from the graph of the two lines. the x was 1.6666667 and y was 5 because i rounded it to 1.67 and used that number for my equations, instead of the more precise answer of 1.6666667 is my answer wrong? sorry if that dosen't make sense. if that does make the answer wrong, is there a way i can insert that value into my equation when solving it in my calculator? sorry if that question was a bit vague. thanks for your help Yes, it is correct. Keep the accurate answer of $y=5$ when$ $1.67$. Keeping $y=5$ shows you have used the full value of $x = 1.666...$ and it shows that you have rounded correctly after calculation to appropriate significant figures.

5. Yay thank you so much - I'll be sure to come on here tomorrow for further help

Thaaaank youu!