Verify Solutions to an Inequality in Two Variables

We have learned how to solve inequalities in one variable. Now, we will look at inequalities in two variables. Inequalities in two variables have many applications. If you ran a business, for example, you would want your revenue to be greater than your costs - so that your business would make a profit.

Linear Inequality

A linear inequality is an inequality that can be written in one of the following forms:

\(A x+B y>C \quad \qquad A x+B y \geq C \quad \qquad A x+B y < C \quad \qquad A x+B y \leq C \)

where \( A\) and \(B\) are not both zero.

Do you remember that an inequality with one variable had many solutions? The solution to the inequality \(x>3\) is any number greater than 3. We showed this on the number line by shading in the number line to the right of 3, and putting an open parenthesis at 3. See Figure 4.30.

The figure shows a number line extending from negative 5 to 5. A parenthesis is shown at positive 3 and an arrow extends for
Figure 4.30


Similarly, inequalities in two variables have many solutions. Any ordered pair \((x,y)\) that makes the inequality true when we substitute in the values is a solution of the inequality.

Solution of a Linear Inequality

An ordered pair \((x,y)\) is a solution of a linear inequality if the inequality is true when we substitute the values of \(x\) and \(y\).

Example 4.69

Determine whether each ordered pair is a solution to the inequality \( y > x+4\):

  1. \((0,0)\)
  2. \((1,6)\)
  3. \((2,6)\)
  4. \((−5,−15)\)
  5. \((−8,12)\)
Solution

  1. \((0,0)\) \(y>x+4\)
    Substitute \(0\) for \(x\) and \(0\) for \(y\). \(0\stackrel{?}{>} 0+4\)
    Simplify. \(0 \not{>} 4\)
    So, (0,0) is not a solution to \(y>x+4\).


  2. \((1,6)\) \(y>x+4\)
    Substitute \(1\) for \(x\) and 6 for \(y\).
    		 \(6\stackrel{?}{>} 1+4\)
    Simplify.

    \(6>5\)

    So, \((1,6)\) is a solution to \(y>x+4\).


  3. \((2,6)\) \(y>x+4\)
    Substitute \(2\) for \(x\) and \(6\) for \(y\). \(6\stackrel{?}{>} 2+4\)
    Simplify.

    \(6 \not{>}6\)

    So, \((2,6)\) is not a solution to \(y>x+4\).


  4. \((−5,−15)\) \( y>x+4 \)
    Substitute \(-5\) for \(x\) and \(-15\) for \(y\). \(-15\stackrel{?}{>}-5+4\)
    Simplify. \(-15 \not{>} -1\)
    So, \((−5,−15)\) is not a solution to \(y>x+4\).


  5. \((−8,12)\) \( y>x+4 \)
    Substitute \(-8\) for \(x\) and \(12\) for \(y\). \(12 \stackrel{?}{>}-8+4\)
    Simplify. \(12>-4\)
    So, \((−8,12)\) is a solution to \(y>x+4\).
Try It 4.137

Determine whether each ordered pair is a solution to the inequality \(y>x−3\):

  1. \((0,0)\)
  2. \((4,9)\)
  3. \((−2,1)\)
  4. \((−5,−3)\)
  5. \((5,1)\)
Try It 4.138

Determine whether each ordered pair is a solution to the inequality \(y<x+1\):

  1. \((0,0)\)
  2. \((8,6)\)
  3. \((−2,−1)\)
  4. \((3,4)\)
  5. \((−1,−4)\)
Back to '5.9: Graphing Linear Inequality of Two Variables on the Coordinate Plane'