# 1960 AHSME Problems/Problem 26

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## Problem

Find the set of $x$-values satisfying the inequality $|\frac{5-x}{3}|<2$. [The symbol $|a|$ means $+a$ if $a$ is positive, $-a$ if $a$ is negative, $0$ if $a$ is zero. The notation $1 means that a can have any value between $1$ and $2$, excluding $1$ and $2$. ] $\textbf{(A)}\ 1 < x < 11\qquad \textbf{(B)}\ -1 < x < 11\qquad \textbf{(C)}\ x< 11\qquad \textbf{(D)}\ x>11\qquad \textbf{(E)}\ |x| < 6$

## Solutions

### Solution 1

Break up the absolute value into two cases.

For the first case, let $x < 5$, so $\frac{5-x}{3}$ is positive. That means (for $x<5$) $$\frac{5-x}{3} < 2$$ $$5-x<6$$ $$-x<1$$ $$x>-1$$ For the second case, let $x \ge 5$, so $\frac{5-x}{3}$ is negative. That means (for $x \ge 5$) $$\frac{x-5}{3} < 2$$ $$x-5<6$$ $$x<11$$

Combine both cases to get $-1 < x < 11$, which is answer choice $\boxed{\textbf{(B)}}$.

### Solution 2 $[asy]import graph; size(10.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-4.2,xmax=14.2,ymin=-1.2,ymax=4.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); draw((5,0)--(-4,3),EndArrow); draw((5,0)--(14,3),EndArrow); draw((-4,2)--(14,2),Arrows); draw(circle((-1,2),0.2)); draw(circle((11,2),0.2)); [/asy]$

Another way to solve this is to graph $y=|\frac{5-x}{3}|$ and $y=2$. The solution is the areas on the graph where the y-values of $y=|\frac{5-x}{3}|$ are lower than $2$. From the graph, $-1, so the answer is $\boxed{\textbf{(B)}}$.