HaCkEd By KaSpEr511

Worksheet 5 (Inequalities) Solution

September 16th, 2009

Solve the linear inequality. Express the solution using interval notation.
1. $3x+11<5$

Solution

$\begin{align*} 3x+11&<5\\ 3x&<-6\\ x&<-2 \end{align*}$
Therefore $x$ is in the interval $(-\infty,-2)$ .

2. $\frac{2}{5}x+1<\frac{1}{5}-2x$

Solution

Multiply both sides by 5, we have $2x+5<1-10x$
Now
$\begin{align*} 2x+5&<1-10x\\ 12x+5&<1\\ 12x&<-4\\ x&<-4/12\\ x&<-1/3 \end{align*}$
So $x$ is in $(-\infty,-1/3)$ .

3. $2(7x-3)\leq 12x+16$

Solution

Divide both sides by 2, we have
$\begin{align*} 7x-3&\leq 6x+8\\ x-3&\leq 8\\ x&\leq 11 \end{align*}$
So $x$ is in $(-\infty,11]$ .

4. $5\leq 3x-4\leq 14$

Solution

Add 4 to each term, we have
$\begin{align*} 9\leq &3x\leq 18\\ 3\leq &x\leq 6 \end{align*}$
So $x$ is in $[3,6]$ .

5. $-\frac{1}{2}<\frac{4-3x}{5}\leq \frac{1}{4}$

Solution

Multiply by 5 we have $-5/2<4-3x <5/4$ . Subtract 4 to get $(5/2)-4<-3x<(5/4)-4$ . The last inequality can be simplify into
$-13/2<-3x<11/4$ .Divide by -3 we have $13/6>x>-11/12″ /> or reading from the right to the left <img src=$ . Therefore the solution is $(-11/12,13/6)$ .

Solve the nonlinear inequality. Express the solution using interval notation.
6. $(x-5)(x+4)\geq 0$

Solution

$(-\infty,-4]\cup [5,\infty)$

7. $5x^2+3x\geq 3x^2+2$

Solution

Rewrite the inequality as $(5x^2-3x^2)+3x-2\geq 0$
$\begin{align*} 2x^2+3x-2&\geq 0\\ 2x^2+4x-x-2&\geq 0\\ 2x(x+2)-(x+2)&\geq 0\\ (2x-1)(x+2)&\geq 0 \end{align*}$
Hence $x$ is in the interval $(-infty,-2]\cup [1/2,\infty)$