<p>1) The total cost of a taxicab ride is the sum of</p>
<pre><code>(1) a basic fixed charge for using the taxicab, and
(2) an additional charge for each 1/4 of a mile that is traveled.
If the total cost to ride 3/4 mile is $4.00 and t he total cost to ride 1.5 miles
$5.50, what is the total cost, in dollars, of a 3-mile ride?
</code></pre>
<p>The answer is 8.5, but how do you you solve this?</p>
<p>Say X is the basic fixed charge and Y is the additional charge per quarter mile.
In the first scenario, the cost for the 3/4 mile cab ride (which is equal to three 1/4 mile segments) is <a href="3">color=red</a>Y + X = 4[/color]. For the second, the cost is <a href="6">color=blue</a>Y + X = 5.5[/color]. Now you can use linear combinations to solve the problem. Let's subtract the second equation from the other:</p>
<p><a href="6">color=blue</a>Y + X = 5.5[/color].
- (3)Y + X = 4
3Y + 0X = 1.5 => Now you know that the cost of traveling 3 1/4 mile segments is $1.50, so Y = 0.50. You can now plug this back into one of the original equations to solve for X (which is $2.50).</p>
<p>So the cost a three mile ride (12 1/4 mile segments) is 12Y + X = 12(0.50) + (2.50) = 6 + 2.50 = $8.50</p>
<p>Hope that makes sense</p>
<p>wait nvm r r r</p>
<p>ok, so this is gonna be long-winded, but it'll explain it thoroughly i hope
i realize halfway through that someone already answered your question, and well, but whatever</p>
<p>the problem seems tricky because you done know what the fixed rate or the rate per 1/4 mile are</p>
<p>so unknowns
primary, is how much it costs for the taxicab ride for 3 miles
secondary, the cost of the fixed rate, and the cost per quartermile</p>
<p>conditions
every cab ride has one and only one fixed rate, and every one has a charge per quartermile</p>
<p>the data</p>
<p>3/4 ride costs 4 dollars
1.5 (six quartermile) ride costs 5.50</p>
<p>so, translate the data into algebraic expressions, where you have two unknowns, because thats what you have, call them x(fixed) and y(per quartermile)</p>
<p>3y + x = 4
6y + x = 5.50</p>
<p>you need to solve for both variables, solve for x first, see the connections between the two equations, one possible way is...</p>
<p>multiply first equation by two and subtract the second from that product
2(3y+x = 4) = 6y + 2x = 8
6y + x = 5.50
-____________
you get x = 2.50
from there, you solve for the other variable, which is y
you get .50 using either equation for substitution</p>
<p>then, you translate a 3 mile ride into quarter miles, then into an algebraic expression</p>
<p>(3 miles) (4 quartermiles/mile) = 12 quartermiles
therefore x+ 12y = cost(let's say it's z)
substitute
1(2.50) + 12(.50) = z
z = 8.50</p>
<p>there's your answer, see if it makes sense, does it cost more than the shorter rides?
is it 3 dollars (6 more quartermiles) more expensive than the 1.5 half ride?</p>
<p>etc...</p>
<p>you wouldnt have time for this much stuff wirtten out and really pondered on the real sat, but sometimes it's helpful to go through every step</p>
<p>i did this for if you really didnt understand the concept or workings of the problem at all, and because i was reading this book on problem solving, so dont be irritated if it sounds like im writing this to a mental midget</p>