Solve graphically by the intersection method. Give the solution in interval notation.

6 x minus 3 less than 2 x minus 7
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Part 1
Graph y 1 equals 6 x minus 3 and y 2 equals 2 x minus 7 using the graphing tool.

Part 2
What is the​ solution?
  
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​(Type your answer in interval​ notation.)

1 answer

To solve the inequality \( 6x - 3 < 2x - 7 \) graphically, we will first rewrite it in a form that allows us to express it as a comparison of two functions. Specifically, we can rewrite the inequality as:

\[ y_1 = 6x - 3 \] \[ y_2 = 2x - 7 \]

Part 1: Graphing

You will need to graph the two equations:

  1. For \( y_1 = 6x - 3 \):

    • The y-intercept is \(-3\).
    • The slope is \(6\), which means for every unit you move to the right (increase x by 1), you move up 6 units.
  2. For \( y_2 = 2x - 7 \):

    • The y-intercept is \(-7\).
    • The slope is \(2\), meaning for every unit you increase x by 1, you move up 2 units.

Finding the intersection

To find the intersection point of the two lines \( y_1 \) and \( y_2 \), set them equal to each other:

\[ 6x - 3 = 2x - 7 \]

Now, solve for \( x \): \[ 6x - 2x = -7 + 3 \] \[ 4x = -4 \] \[ x = -1 \]

Now substitute \( x = -1 \) into either equation to find \( y \): Using \( y_1 \): \[ y_1 = 6(-1) - 3 = -6 - 3 = -9 \]

Thus, the intersection point is \( (-1, -9) \).

Part 2: Intervals

Next, we need to determine where \( 6x - 3 < 2x - 7 \) or equivalently where \( y_1 < y_2 \) on the graph.

  1. Choose test points to the left, directly at, and to the right of \( x = -1 \).

    • Test Point \( x = -2 \):

      • \( y_1 = 6(-2) - 3 = -12 - 3 = -15 \)
      • \( y_2 = 2(-2) - 7 = -4 - 7 = -11 \)
      • Here, \( -15 < -11 \) (True).
    • Test Point \( x = 0 \):

      • \( y_1 = 6(0) - 3 = -3 \)
      • \( y_2 = 2(0) - 7 = -7 \)
      • Here, \( -3 < -7 \) (False).
  2. Since the inequality holds true for \( x < -1 \) and false for \( x > -1 \), we can conclude that the solution to the inequality \( 6x - 3 < 2x - 7 \) is:

Solution in interval notation:

\[ (-\infty, -1) \]

Thus, the answer is:

\[ \boxed{(-\infty, -1)} \]