Difference between revisions of "2006 AMC 10B Problems/Problem 12"
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<math> \textbf{(A) } 0\qquad \textbf{(B) } \frac{3}{4}\qquad \textbf{(C) } 1\qquad \textbf{(D) } 2\qquad \textbf{(E) } \frac{9}{4} </math> | <math> \textbf{(A) } 0\qquad \textbf{(B) } \frac{3}{4}\qquad \textbf{(C) } 1\qquad \textbf{(D) } 2\qquad \textbf{(E) } \frac{9}{4} </math> | ||
− | == Solution == | + | == Solution 1 == |
Since <math>(1,2)</math> is a solution to both equations, plugging in <math>x=1</math> and <math>y=2</math> will give the values of <math>a</math> and <math>b</math>. | Since <math>(1,2)</math> is a solution to both equations, plugging in <math>x=1</math> and <math>y=2</math> will give the values of <math>a</math> and <math>b</math>. | ||
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<math> b = \frac{7}{4} </math> | <math> b = \frac{7}{4} </math> | ||
− | So | + | So <math> a+b = \frac{1}{2} + \frac{7}{4} = \boxed{\textbf{(E) }\frac{9}{4}}</math>. |
+ | == Solution 2 == | ||
+ | Substituting <math>x=1</math> and <math>y=2</math> into the equations gives <math>1=\frac{2}{4}+a\quad\text{and}\quad 2=\frac{1}{4}+b. | ||
+ | </math> It follows that <math>a+b=\left(1-\frac{2}{4}\right)+\left(2-\frac{1}{4}\right)=3 - \frac{3}{4}=\boxed{\frac{9}{4}}.</math> | ||
+ | |||
+ | == Solution 3== | ||
+ | Because <math>a=x- \frac{y}{4}\quad\text{and}\quad b=y-\frac{x}{4} | ||
+ | \quad\text{we have}\quad a +b = \frac{3}{4}(x+y).</math> Since <math>x=1</math> when <math>y=2</math>, this implies that <math>a+b = \frac{3}{4}(1 + 2) = \boxed{\frac{9}{4}}</math>. | ||
+ | == Video Solution == | ||
+ | |||
+ | https://www.youtube.com/watch?v=q1mxhiyciSk ~David | ||
== See Also == | == See Also == |
Latest revision as of 19:07, 27 November 2023
Problem
The lines and intersect at the point . What is ?
Solution 1
Since is a solution to both equations, plugging in and will give the values of and .
So .
Solution 2
Substituting and into the equations gives It follows that
Solution 3
Because Since when , this implies that .
Video Solution
https://www.youtube.com/watch?v=q1mxhiyciSk ~David
See Also
2006 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 10 Problems and Solutions |
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