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 1

We have $\begin{align*}2x+1 = 8 &\iff 2x = 7 \\ &\iff x = \frac{7}{2},\end{align*}$ so $\begin{align*}4x+1 &= 4\left(\frac{7}{2}\right)+1 \\ &= 2(7)+1 \\ &= \boxed{\text{(A)} \ 15}.\end{align*}$

Solution 2

From $2x = 7$ (as above), we can directly compute $\begin{align*}4x &= 2(2x) \\ &= 2(7) \\ &= 14,\end{align*}$ so $4x+1 = 14+1 = \boxed{\text{(A)} \ 15}$ .

1985 AHSME (Problems • Answer Key • Resources)
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1985 AHSME Problems/Problem 1

Contents

Problem

Solution 1

Solution 2

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