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# Difference between revisions of "1985 AHSME Problems/Problem 1"

## Problem

If $2x+1=8$, then $4x+1=$

$\mathrm{(A)\ } 15 \qquad \mathrm{(B) \ }16 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ } 18 \qquad \mathrm{(E) \ }19$

## Solution

### Solution 1

From $2x+1=8$, we subtract $1$ from both sides to get $2x=7$. We can divide both sides now by $2$ to get $x=\frac{7}{2}$. Now we can substitute this into $4x+1$ to get $4x+1=4\left(\frac{7}{2}\right)+1=(2)(7)+1=14+1=15, \boxed{\text{A}}$.

### Solution 2

We proceed from $2x=7$ as above. Notice that $4x=2(2x)=2(7)=14$, so $4x+1=14+1=15, \boxed{\text{A}}$.

## See Also

 1985 AHSME (Problems • Answer Key • Resources) Preceded byFirst Problem Followed byProblem 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 • 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.

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