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Difference between revisions of "2018 AMC 12B Problems/Problem 3"

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==Solution 1==
  
  
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Using the slope-intercept form, we get the equations <math>y-30 = 6(x-40)</math> and <math>y-30 = 2(x-40)</math>. Simplifying, we get  <math>6x-y=210</math> and <math>2x-y=50</math>. Letting <math>y=0</math> in both equations gives the <math>x</math>-intercepts: <math>x=35</math> and <math>x=25</math>, respectively. Thus the distance between them is <math>35-25 = 10 \rightarrow (\text{B})</math>
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Using the slope-intercept form, we get the equations <math>y-30 = 6(x-40)</math> and <math>y-30 = 2(x-40)</math>. Simplifying, we get  <math>6x-y=210</math> and <math>2x-y=50</math>. Letting <math>y=0</math> in both equations gives the <math>x</math>-intercepts: <math>x=35</math> and <math>x=25</math>, respectively. Thus the distance between them is <math>35-25 = 10 \Rightarrow (\text{B})
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<math>\indent</math> (Giraffefun)
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\indent</math> (Giraffefun)
  
 
==See Also==
 
==See Also==

Revision as of 14:34, 16 February 2018

A line with slope 2 intersects a line with slope 6 at the point $(40,30)$. What is the distance between the $x$-intercepts of these two lines?

$(\text{A}) 5 \qquad (\text{B}) 10 \qquad (\text{C}) 20 \qquad (\text{D}) 25 \qquad (\text{E}) 50$


Solution 1

Using the slope-intercept form, we get the equations $y-30 = 6(x-40)$ and $y-30 = 2(x-40)$. Simplifying, we get $6x-y=210$ and $2x-y=50$. Letting $y=0$ in both equations gives the $x$-intercepts: $x=35$ and $x=25$, respectively. Thus the distance between them is $35-25 = 10 \Rightarrow (\text{B}) \indent$ (Giraffefun)

See Also

2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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