Difference between revisions of "2030 AMC 8 Problems/Problem 1"
(Created page with "== Problem == The equation is shown as: <math>\frac{24x^2+25x-47}{ax-2}=-8x-3-\frac{53}{ax-2}</math> is true for all values except when <math>x=\frac{2}{a}</math>, where a is...") |
(2030 AMC 8 Problems/Problem 1 Whole Page) |
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<math>\text {(A)}\ -16 \qquad \text {(B)}\ -3 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 3 \qquad \text {(E)} 16</math> | <math>\text {(A)}\ -16 \qquad \text {(B)}\ -3 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 3 \qquad \text {(E)} 16</math> | ||
| − | == Solution | + | == Solution== |
| − | The faster way is to multiply each side of the given equation by <math> | + | The faster way is to multiply each side of the given equation by 𝑎𝑥−2 (so you can get rid of the fraction). When you multiply each side by 𝑎𝑥−2, you should have: |
| − | == | + | <cmath>24x^2+25x-47=(-8x-3)(ax-2)-53</cmath> |
| + | Using the FOIL method, you should then multiply (−8𝑥−3) and (𝑎𝑥−2). After that you should have the following: | ||
| + | <cmath>24x^2+25x-47=-8ax^2-3ax+16x+6-53</cmath> | ||
| + | Then, reduce on the right side of the equation: | ||
| + | <cmath>24x^2+25x-47=-8ax^2-3ax+16x-47</cmath> | ||
| + | Since the coefficients of the x² term has to be equal on both sides of the equation, −8a=24, and that concludes us with 𝑎=−3. | ||
| + | <math>\Rightarrow\boxed{\mathrm{(B)}\ -3}</math>. | ||
| + | |||
| + | == See also == | ||
| + | {{AMC8 box|year=2030|ab=A|num-b=1|num-a=2}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 21:23, 21 June 2024
Problem
The equation is shown as: 
is true for all values except when 
, where a is constant.
What is the value of a?
Solution
The faster way is to multiply each side of the given equation by 𝑎𝑥−2 (so you can get rid of the fraction). When you multiply each side by 𝑎𝑥−2, you should have:
![\[24x^2+25x-47=(-8x-3)(ax-2)-53\]](http://latex.artofproblemsolving.com/a/c/5/ac50cc0b6809140b1cf018b8118258ad039a453f.png)
Using the FOIL method, you should then multiply (−8𝑥−3) and (𝑎𝑥−2). After that you should have the following:
![\[24x^2+25x-47=-8ax^2-3ax+16x+6-53\]](http://latex.artofproblemsolving.com/2/4/3/243f30c8ab30b59203fbd4f570daa3870fa2fa80.png)
Then, reduce on the right side of the equation:
![\[24x^2+25x-47=-8ax^2-3ax+16x-47\]](http://latex.artofproblemsolving.com/7/2/e/72e8b252b223d6f23e50bfac60bc4b5c03a6c952.png)
Since the coefficients of the x² term has to be equal on both sides of the equation, −8a=24, and that concludes us with 𝑎=−3.
.
See also
| 2030 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 2 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
