The cost of a taxi ride is calculated by adding a fixed amount to a charge for each 1/4 of the trip. If one mile costs $4 and 3/2 miles costs $5.50, what is the cost of a three-mile trip?
I made a table:
This question came from the chapter titled linear equations in two variables.
After doing the math, I got $11 for 3 miles but the math book tells me that the correct answer is $10 for 3 miles.
How do I set up a system of linear equations for this question? How do I find $10?
a little help
This does not require a system of equations, that's a little overkill.
So, when you are setting up linear equations, you should know that a general linear equation looks like this: y=mx+b, where m is the slope and b is the y-intercept. So, you need to find the slope here, which you find using the equation
So, what points will you use for this little formula above?
The linear equation in 2 variables they are talking about is the above, as opposed to the x=4 or y=10 linear equations.
Once you find out the slope here, you can use the point-slope formula to find your equation
Ok, so you have your equation. Now just plus in 3 for the variable you used for miles and simplify.
Let me know if that helped
I do not see the connection between my question and the slope of lines.
Originally Posted by Mirado
Can you show me your method?
The slope is basically defined as the change that is going on. You said "The cost of a taxi ride is calculated by adding a fixed amount to a charge for each 1/4 of the trip" which means that this is the slope. It is going up a fixed amount for every quarter mile. So, basically, just use the chart that you made and use the 'y' as the cost and the 'x' as the distance traveled. This will give you the slope. Try it out.
Not sure how to do it the previously mentioned way. But, I've come up with $10, quite easily my way.
I thought of it this way.
1 mile is $4.00
1 and 1/2 mile is $5.50.
The problem says it is a FIXED amount and a charge for every 1/4 mile.
Divide the $4.00 by 4, that's $0.25 cents every 1/4 mile.
But you'll notice the increase from 1 mile to 1 and a half is a $1.50 increase, which can't be possible at $0.25 every 1/4mile. So the initial FIXED amount has to be $4.00 for the FIRST MILE, then $0.75 for every 1/4 mile after that.
So 1 mile = $4.00
2 miles contains 8 one-fourth miles. So $0.75 * 8 = $6
$4.00 + $6.00 = $10.00 = 3 mile trip.
that's how I achieved the answer. Not sure about the above way.
Since this comes from the linear equations in two variables, you want two equations with two unknowns. The 1/4 isn't needed though....
Originally Posted by magentarita
Let x=the fixed charge, and y=price per mile. Then the two equations become 4=x+y and 5.50=x+1.5y. That can be solve by substitution, or by subtraction. These two equations are your system, both of the form
cost=fixed amount + miles*(charge per mile).
Solving, you get y=3, and x =1. Thus, for 3 miles, cost =x+3y=1+3(3)=10.
if you really want the 1/4 mile in there, you measure miles in 1/4's, and you get
4=x+4y and 5.50=x+6y.... cost=fixed amount + (number of quarter miles) *(charge per mile). Your y then should be (3/4).
Ok, first off, it's completely unnecessary to use 2 equations.
Use the formulas I posted and get it that way.
if you wanted to do it without them, then you can see that the slope is for every half mile is $1.50, or every mile is $3.
Hence, to then figure out the base cost, is going to be $4 - $3 = $1 (which is the y-intercept).
Then, you use the idea of the slope is $3/mile (our slope, m)
So, when we put this stuff back into the equation (which is what you are supposed to learn from these exercises) is that y=mx+b or by substitution is y = 3x + 1; where y is the cost of the trip and x is the mileage.
So, when you use the formula y=3x+1, and plug in for 3 miles, you will get $10, or to use the units, it'll be $y = ($3/mile) \times X miles + $1
which means the miles cancels out and you are left with dollars, or $y = $3 \times X + $1
Thank you for breaking this question down for me.
Originally Posted by IAmLegendToo
Thank you for breaking down this question.
Originally Posted by woof