Quiz 9 (Solution)

Date: MATH 1090-4 - Fall 2003

1. Solve the inequality $6(x-1)<2x$ and graph the solution.

   We solve for $x$:

   $$\begin{split}6(x-1)&<2x\\ 6x-6&<2x\\ 4x-6&<0\\ 4x&<6\\ x&<\frac{6}{4}=\frac{3}{2} \end{split}$$

   
   
   So that the solution is $x<\frac{3}{2}$ or, equivalently, $(-\infty,\frac{3}{2})$. The graph is shown in Figure 1.

   Figure 1: Solution set of Exercise 1

   

2. Graph the solution to

   $$\begin{cases}2x+y < 3\\ x-2y\geq -1. \end{cases}$$

   
   

   We start by finding the border of the first half-plane: $2x+y<3$. The first step is to replace $<$ with $=$, and so we obtain the line $2x+y=3$. The points $(0,3)$ and $(1,1)$ are on this line, and that suffices to sketch the line. To decide which side of the line corresponds to the half-plane $2x+y<3$ we test with the point $(0,0)$: plugging in we get $2\cdot0+0 = 0 < 3$, so that $(0,0)$ is in the half-plane. Notice that the border line is not part of the region (because we have strict inequality $<$). The results are shown in Figure 2.

   Figure 2: Half-plane corresponding to the first inequality in Exercise 2

   

   We follow the same steps for the half-plane $x-2y\geq -1$. The line is $x-2y=-1$, and it contains the points $(-1,0)$ and $(1,1)$. Testing $(0,0)$ we see that it satisfies the inequality so that it is in the given half-plane. The border line is part of the half plane. The results are shown in Figure 3.

   Figure 3: Half-plane corresponding to the second inequality in Exercise 2

   

   Finally, the points that satisfy both conditions (that is, those that are in both half-spaces simultaneously) are shown in Figure 4. This is the solution of the problem.

   Figure 4: Solution set for Exercise 2