Difference between revisions of "1986 AHSME Problems/Problem 2"
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==Solution== | ==Solution== | ||
| − | + | The original slope and <math>y</math>-intercept are <math>\frac{2}{3}</math> and <math>4</math>, so the new ones are <math>\frac{1}{3}</math> and <math>8</math> respectively. Thus, using the slope-intercept form (<math>y = mx+c</math>), the new equation is <math>y=\frac{1}{3}x + 8</math>, which is <math>\boxed{A}</math>. | |
== See also == | == See also == | ||
| − | {{AHSME box|year=1986| | + | {{AHSME box|year=1986|num-b=1|num-a=3}} |
[[Category: Introductory Algebra Problems]] | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 17:01, 1 April 2018
Problem
If the line 
in the 
-plane has half the slope and twice the 
-intercept of the line 
, then an equation for is:
Solution
The original slope and -intercept are 
and 
, so the new ones are 
and 
respectively. Thus, using the slope-intercept form (
), the new equation is 
, which is 
.
See also
| 1986 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
