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Section 2.5 Linear Inequalities in Two Variables

Subsection 1. Solve a linear inequality

Before we solve inequalities in two variables, let's review solving linear inequalities in one variable.

Subsubsection Example

Example 2.29.

Solve \(~3k-13 \lt 5+6k\)

Solution

We begin just as we do to solve an equation. The only difference is that we must reverse the direction of the inequality if we multiply or divide by a negative number.

\begin{align*} 3k-13 \amp \lt 5+6k \amp \amp \blert{\text{Subtract}~6k~\text{from both sides.}} \\ -3k \amp \lt 18 \amp \amp \blert{\text{Divide both sides by}~-3.}\\ k \amp \gt -6 \amp \amp \blert{\text{Don't forget to reverse the inequality.}} \end{align*}

In interval notation, the solution set is \((-6,\infty)\text{.}\)

Subsubsection Exercise

Solve \(~~4(3a-7) \gt -18+2a.~~\) Write the solution with interval notation.

Answer
\((1, \infty) \)

Solve \(~~4 \le \dfrac{-3x}{4}-2.~~\) Write the solution with interval notation.

Answer
\((-\infty,8] \)

Solve \(~~15 \ge -6+3m \ge -6.~~\) Write the solution with interval notation.

Answer
\([0,7] \)

Solve \(~~\dfrac{-9}{2} \lt \dfrac{5-2n}{-4} \le -1.~~\) Write the solution with interval notation.

Answer
\(\left(\dfrac{-13}{2}, \dfrac{1}{2}\right] \)

Subsection 2. Graph a line

The boundary of the solution set for a linear inequality in two variables is made up of portions of straight lines.

Subsubsection Examples

Example 2.34.

Use the most convenient method to graph the equation.

  1. \(\displaystyle 5x-10y=750\)
  2. \(\displaystyle y=400-25x\)
Solution
  1. This equation is in the form \(Ax+By=C\text{,}\) so the intercept method of graphing is convenient. The intercepts are \((150,0)\) and \((0,-75)\text{.}\) The graph is shown below.

    line
  2. This equation is in the form \(y=mx+b\text{,}\) so the slope-intercept method of graphing is convenient. The \(y\)-intercept is \((0,400)\text{,}\) and the slope is \(-25\text{.}\) The graph is shown below.

    line

Subsubsection Exercise

Graph the equation \(~~24x+9y=432\)

Graph the equation \(~~y=12x+60\)

Graph the equation \(~~y=600-1.25x\)

Graph the equation \(~~45x-30y=15\)

Subsection 3. Solve a 2x2 system

To find the vertices of the boundary of the solution set, we solve a linear 2x2 system.

Subsubsection Examples

Example 2.39.

Use substitution to solve the system:

\begin{align*} 3y-2x \amp = 3\\ x-2y \amp = -2 \end{align*}
Solution

We start by solving the second equation for \(x\) to get \(x=2y-2\text{.}\) Then we substitute this expression for \(x\) into the first equation, which gives us

\begin{equation*} 3y-2(2y-2)=3 \end{equation*}

We solve this equation for \(y\) to find \(y=1\text{.}\) Finally, we substitute \(y=1\) into our first step to find

\begin{equation*} x=2(1)-2=0 \end{equation*}

The solution is \(x=0,~y=1\text{,}\) or \((0,1)\text{.}\)

Example 2.40.

Use elimination to solve the system:

\begin{align*} 2x+3y \amp = -1\\ 3x+5y \amp = -2 \end{align*}
Solution

We multiply the first equation by 3 and the second equation by \(-2\) in order to eliminate \(x\text{.}\)

\begin{align*} 6x+9y \amp = -3\\ -6x-10y \amp = 4 \end{align*}

Adding these two equations gives us \(-y=1\text{,}\) or \(y=-1\text{.}\) Finally, we substitute \(y=-1\) into either equation (we choose the first equation), and solve for \(x\text{.}\)

\begin{align*} 2x+3(-1) \amp = -1\\ 2x-3 \amp = -1 \end{align*}

We find\(x=1\text{,}\) so the solution is \(x=1,~y=-1\text{,}\) or \((1,-1)\text{.}\)

Subsubsection Exercise

Solve the system:

\begin{align*} y \amp = 2x+1\\ 2x+3y \amp = -21 \end{align*}
Answer
\((-3,-5)\)

Solve the system:

\begin{align*} x+4y \amp = 1\\ 2x+3y \amp = -3 \end{align*}
Answer
\((-3,1)\)

Solve the system:

\begin{align*} 2x+7y \amp = -19\\ 5x-3y \amp = 14 \end{align*}
Answer
\((1,-3)\)

Solve the system:

\begin{align*} 4x+3y \amp = -19\\ 5x+15 \amp = -2y \end{align*}
Answer
\((-1,-5)\)