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3 8 linear inequalities (optional) 3 8 linear inequalities (optional) Presentation Transcript

  • Linear Inequalities (Optional) Frank Ma © 2011
  • The graphs of the linear equations are straight lines. Linear Inequalities
  • The graphs of the linear equations are straight lines. Example A. Graph of the linear equations y = x. Linear Inequalities
  • The graphs of the linear equations are straight lines. Example A. Graph of the linear equations y = x. x –3 0 3 y –3 0 3 Linear Inequalities
  • The graphs of the linear equations are straight lines. Example A. Graph of the linear equations y = x. x –3 0 3 y –3 0 3 Linear Inequalities
  • The graphs of the linear equations are straight lines. Example A. Graph of the linear equations y = x. x –3 0 3 y –3 0 3 Linear Inequalities y = x
  • The graphs of the linear equations are straight lines. Example A. Graph of the linear equations y = x. x –3 0 3 y –3 0 3 Linear Inequalities The points on the line fit the description that y = x. y = x
  • The graphs of the linear equations are straight lines. Example A. Graph of the linear equations y = x. x –3 0 3 y –3 0 3 Linear Inequalities The points on the line fit the description that y = x. So for any point that off the line, y = x. y = x
  • The graphs of the linear equations are straight lines. Example A. Graph of the linear equations y = x. x –3 0 3 y –3 0 3 Linear Inequalities The points on the line fit the description that y = x. So for any point that off the line, y = x. Specifically, the line divides the plane into two half-planes. y = x
  • The graphs of the linear equations are straight lines. Example A. Graph of the linear equations y = x. x –3 0 3 y –3 0 3 Linear Inequalities The points on the line fit the description that y = x. So for any point that off the line, y = x. Specifically, the line divides the plane into two half-planes. One of them fits the relation that y < x and the other fits y > x. y = x
  • The graphs of the linear equations are straight lines. Example A. Graph of the linear equations y = x. x –3 0 3 y –3 0 3 Linear Inequalities The points on the line fit the description that y = x. So for any point that off the line, y = x. Specifically, the line divides the plane into two half-planes. One of them fits the relation that y < x and the other fits y > x. y = x To match which half-plane fits which inequality, we sample a point,
  • The graphs of the linear equations are straight lines. Example A. Graph of the linear equations y = x. x –3 0 3 y –3 0 3 Linear Inequalities The points on the line fit the description that y = x. So for any point that off the line, y = x. Specifically, the line divides the plane into two half-planes. One of them fits the relation that y < x and the other fits y > x. y = x To match which half-plane fits which inequality, we sample a point, any point, from these two regions.
  • The graphs of the linear equations are straight lines. Example A. Graph of the linear equations y = x. x –3 0 3 y –3 0 3 Linear Inequalities The points on the line fit the description that y = x. So for any point that off the line, y = x. Specifically, the line divides the plane into two half-planes. One of them fits the relation that y < x and the other fits y > x. y = x To match which half-plane fits which inequality, we sample a point, any point, from these two regions. Let’s take the point (0, 5),
  • The graphs of the linear equations are straight lines. Example A. Graph of the linear equations y = x. x –3 0 3 y –3 0 3 Linear Inequalities The points on the line fit the description that y = x. So for any point that off the line, y = x. y = x Specifically, the line divides the plane into two half-planes. One of them fits the relation that y < x and the other fits y > x. To match which half-plane fits which inequality, we sample a point, any point, from these two regions. Let’s take the point (0, 5),
  • The graphs of the linear equations are straight lines. Example A. Graph of the linear equations y = x. x –3 0 3 y –3 0 3 Linear Inequalities The points on the line fit the description that y = x. So for any point that off the line, y = x. y = x Specifically, the line divides the plane into two half-planes. One of them fits the relation that y < x and the other fits y > x. To match which half-plane fits which inequality, we sample a point, any point, from these two regions. Let’s take the point (0, 5), since 5 > 0, it fits the relation y > x.
  • The graphs of the linear equations are straight lines. Example A. Graph of the linear equations y = x. x –3 0 3 y –3 0 3 Linear Inequalities The points on the line fit the description that y = x. So for any point that off the line, y = x. y = x Specifically, the line divides the plane into two half-planes. One of them fits the relation that y < x and the other fits y > x. To match which half-plane fits which inequality, we sample a point, any point, from these two regions. Let’s take the point (0, 5), since 5 > 0, it fits the relation y > x. So the side that contains (0, 5) is y > x,
  • The graphs of the linear equations are straight lines. Example A. Graph of the linear equations y = x. x –3 0 3 y –3 0 3 Linear Inequalities The points on the line fit the description that y = x. So for any point that off the line, y = x. y = x y > x Specifically, the line divides the plane into two half-planes. One of them fits the relation that y < x and the other fits y > x. To match which half-plane fits which inequality, we sample a point, any point, from these two regions. Let’s take the point (0, 5), since 5 > 0, it fits the relation y > x. So the side that contains (0, 5) is y > x,
  • The graphs of the linear equations are straight lines. Example A. Graph of the linear equations y = x. x –3 0 3 y –3 0 3 Linear Inequalities The points on the line fit the description that y = x. So for any point that off the line, y = x. y = x y > x y < x Specifically, the line divides the plane into two half-planes. One of them fits the relation that y < x and the other fits y > x. To match which half-plane fits which inequality, we sample a point, any point, from these two regions. Let’s take the point (0, 5), since 5 > 0, it fits the relation y > x. So the side that contains (0, 5) is y > x, and the other half-plane must be y < x.
  • In general, the points that fit the linear inequality Ax + By > C or Ax + By < C corresponds to one of the two half-planes on each side of the line Ax + By = C. Linear Inequalities
  • In general, the points that fit the linear inequality Ax + By > C or Ax + By < C corresponds to one of the two half-planes on each side of the line Ax + By = C. To identify which half-plane fits the given inequality, sample any point in a half-plane by plugging in the values of x and y. Linear Inequalities
  • In general, the points that fit the linear inequality Ax + By > C or Ax + By < C corresponds to one of the two half-planes on each side of the line Ax + By = C. To identify which half-plane fits the given inequality, sample any point in a half-plane by plugging in the values of x and y. If the point satisfies the inequality, then the entire half-plane that contains the sample point fits that inequality. Linear Inequalities
  • In general, the points that fit the linear inequality Ax + By > C or Ax + By < C corresponds to one of the two half-planes on each side of the line Ax + By = C. To identify which half-plane fits the given inequality, sample any point in a half-plane by plugging in the values of x and y. If the point satisfies the inequality, then the entire half-plane that contains the sample point fits that inequality. Otherwise, the other side fits the inequality. Linear Inequalities
  • In general, the points that fit the linear inequality Ax + By > C or Ax + By < C corresponds to one of the two half-planes on each side of the line Ax + By = C. To identify which half-plane fits the given inequality, sample any point in a half-plane by plugging in the values of x and y. If the point satisfies the inequality, then the entire half-plane that contains the sample point fits that inequality. Otherwise, the other side fits the inequality. Linear Inequalities Remark: If (0, 0) is not on the line, sample (0, 0).
  • In general, the points that fit the linear inequality Ax + By > C or Ax + By < C corresponds to one of the two half-planes on each side of the line Ax + By = C. To identify which half-plane fits the given inequality, sample any point in a half-plane by plugging in the values of x and y. If the point satisfies the inequality, then the entire half-plane that contains the sample point fits that inequality. Otherwise, the other side fits the inequality. Linear Inequalities Remark: If (0, 0) is not on the line, sample (0, 0). Example B. Shade 2x + 3y > 12.
  • In general, the points that fit the linear inequality Ax + By > C or Ax + By < C corresponds to one of the two half-planes on each side of the line Ax + By = C. To identify which half-plane fits the given inequality, sample any point in a half-plane by plugging in the values of x and y. If the point satisfies the inequality, then the entire half-plane that contains the sample point fits that inequality. Otherwise, the other side fits the inequality. Linear Inequalities Remark: If (0, 0) is not on the line, sample (0, 0). Example B. Shade 2x + 3y > 12. Graph 2x + 3y = 12.
  • In general, the points that fit the linear inequality Ax + By > C or Ax + By < C corresponds to one of the two half-planes on each side of the line Ax + By = C. To identify which half-plane fits the given inequality, sample any point in a half-plane by plugging in the values of x and y. If the point satisfies the inequality, then the entire half-plane that contains the sample point fits that inequality. Otherwise, the other side fits the inequality. Linear Inequalities Remark: If (0, 0) is not on the line, sample (0, 0). Example B. Shade 2x + 3y > 12. Graph 2x + 3y = 12. Use intercepts to graph.
  • In general, the points that fit the linear inequality Ax + By > C or Ax + By < C corresponds to one of the two half-planes on each side of the line Ax + By = C. To identify which half-plane fits the given inequality, sample any point in a half-plane by plugging in the values of x and y. If the point satisfies the inequality, then the entire half-plane that contains the sample point fits that inequality. Otherwise, the other side fits the inequality. Linear Inequalities Remark: If (0, 0) is not on the line, sample (0, 0). Example B. Shade 2x + 3y > 12. Graph 2x + 3y = 12. Use intercepts to graph. x y 0 0
  • In general, the points that fit the linear inequality Ax + By > C or Ax + By < C corresponds to one of the two half-planes on each side of the line Ax + By = C. To identify which half-plane fits the given inequality, sample any point in a half-plane by plugging in the values of x and y. If the point satisfies the inequality, then the entire half-plane that contains the sample point fits that inequality. Otherwise, the other side fits the inequality. Linear Inequalities Remark: If (0, 0) is not on the line, sample (0, 0). Example B. Shade 2x + 3y > 12. Graph 2x + 3y = 12. Use intercepts to graph. x y 0 4 0
  • In general, the points that fit the linear inequality Ax + By > C or Ax + By < C corresponds to one of the two half-planes on each side of the line Ax + By = C. To identify which half-plane fits the given inequality, sample any point in a half-plane by plugging in the values of x and y. If the point satisfies the inequality, then the entire half-plane that contains the sample point fits that inequality. Otherwise, the other side fits the inequality. Linear Inequalities Remark: If (0, 0) is not on the line, sample (0, 0). Example B. Shade 2x + 3y > 12. Graph 2x + 3y = 12. Use intercepts to graph. x y 0 4 6 0
  • In general, the points that fit the linear inequality Ax + By > C or Ax + By < C corresponds to one of the two half-planes on each side of the line Ax + By = C. To identify which half-plane fits the given inequality, sample any point in a half-plane by plugging in the values of x and y. If the point satisfies the inequality, then the entire half-plane that contains the sample point fits that inequality. Otherwise, the other side fits the inequality. Linear Inequalities Remark: If (0, 0) is not on the line, sample (0, 0). Example B. Shade 2x + 3y > 12. Graph 2x + 3y = 12. Use intercepts to graph. x y 0 4 6 0
  • In general, the points that fit the linear inequality Ax + By > C or Ax + By < C corresponds to one of the two half-planes on each side of the line Ax + By = C. To identify which half-plane fits the given inequality, sample any point in a half-plane by plugging in the values of x and y. If the point satisfies the inequality, then the entire half-plane that contains the sample point fits that inequality. Otherwise, the other side fits the inequality. Linear Inequalities Remark: If (0, 0) is not on the line, sample (0, 0). Example B. Shade 2x + 3y > 12. Graph 2x + 3y = 12. Use intercepts to graph. Draw a dotted line because the line itself is not part of the answer. x y 0 4 6 0 2x + 3y = 12
  • Linear Inequalities To match the region that fits 2x + 3y > 12, sample (0, 0). 2x + 3y = 12
  • Linear Inequalities To match the region that fits 2x + 3y > 12, sample (0, 0). 2x + 3y = 12
  • Linear Inequalities To match the region that fits 2x + 3y > 12, sample (0, 0). Plug in (0, 0) into the inequality, we’ve 0 + 0 > 12 2x + 3y = 12
  • Linear Inequalities To match the region that fits 2x + 3y > 12, sample (0, 0). Plug in (0, 0) into the inequality, we’ve 0 + 0 > 12 which is false. 2x + 3y = 12
  • Linear Inequalities To match the region that fits 2x + 3y > 12, sample (0, 0). Plug in (0, 0) into the inequality, we’ve 0 + 0 > 12 which is false. Hence the half-plane that fits 2x + 3y > 12 must be the other side of the line. 2x + 3y = 12
  • Linear Inequalities To match the region that fits 2x + 3y > 12, sample (0, 0). Plug in (0, 0) into the inequality, we’ve 0 + 0 > 12 which is false. 2x + 3y > 12 2x + 3y = 12 Hence the half-plane that fits 2x + 3y > 12 must be the other side of the line.
  • Linear Inequalities To match the region that fits 2x + 3y > 12, sample (0, 0). Plug in (0, 0) into the inequality, we’ve 0 + 0 > 12 which is false. 2x + 3y > 12 2x + 3y = 12 Hence the half-plane that fits 2x + 3y > 12 must be the other side of the line. If the inequality is > or < , then the boundary is also part of the solution and we draw a solid line to express that.
  • Linear Inequalities To match the region that fits 2x + 3y > 12, sample (0, 0). Plug in (0, 0) into the inequality, we’ve 0 + 0 > 12 which is false. 2x + 3y > 12 2x + 3y = 12 To find the region that fits a system of two linear inequalities, graph the equations first. Hence the half-plane that fits 2x + 3y > 12 must be the other side of the line. If the inequality is > or < , then the boundary is also part of the solution and we draw a solid line to express that.
  • Linear Inequalities To match the region that fits 2x + 3y > 12, sample (0, 0). Plug in (0, 0) into the inequality, we’ve 0 + 0 > 12 which is false. 2x + 3y > 12 2x + 3y = 12 To find the region that fits a system of two linear inequalities, graph the equations first. In general, we get two intersecting lines that divide the plane into 4 regions. Hence the half-plane that fits 2x + 3y > 12 must be the other side of the line. If the inequality is > or < , then the boundary is also part of the solution and we draw a solid line to express that.
  • Linear Inequalities To match the region that fits 2x + 3y > 12, sample (0, 0). Plug in (0, 0) into the inequality, we’ve 0 + 0 > 12 which is false. 2x + 3y > 12 2x + 3y = 12 To find the region that fits a system of two linear inequalities, graph the equations first. In general, we get two intersecting lines that divide the plane into 4 regions. Then we sample to determine the two half-planes that fit the two inequalities. Hence the half-plane that fits 2x + 3y > 12 must be the other side of the line. If the inequality is > or < , then the boundary is also part of the solution and we draw a solid line to express that.
  • Linear Inequalities To match the region that fits 2x + 3y > 12, sample (0, 0). Plug in (0, 0) into the inequality, we’ve 0 + 0 > 12 which is false. 2x + 3y > 12 2x + 3y = 12 To find the region that fits a system of two linear inequalities, graph the equations first. In general, we get two intersecting lines that divide the plane into 4 regions. Then we sample to determine the two half-planes that fit the two inequalities. The overlapped region of the two half-planes is the region that fits the system. Hence the half-plane that fits 2x + 3y > 12 must be the other side of the line. If the inequality is > or < , then the boundary is also part of the solution and we draw a solid line to express that.
  • Linear Inequalities To match the region that fits 2x + 3y > 12, sample (0, 0). Plug in (0, 0) into the inequality, we’ve 0 + 0 > 12 which is false. 2x + 3y > 12 2x + 3y = 12 To find the region that fits a system of two linear inequalities, graph the equations first. In general, we get two intersecting lines that divide the plane into 4 regions. Then we sample to determine the two half-planes that fit the two inequalities. The overlapped region of the two half-planes is the region that fits the system. To give the complete solution, we need to locate the tip of the region by solving the system. Hence the half-plane that fits 2x + 3y > 12 must be the other side of the line. If the inequality is > or < , then the boundary is also part of the solution and we draw a solid line to express that.
  • Example C. Shade the system 2x – y < –2 x + y < 5 Find and label the tip of the region. { Linear Inequalities
  • Example C. Shade the system 2x – y < –2 x + y < 5 Find and label the tip of the region. { We are looking for the region that satisfies both inequalities. Linear Inequalities
  • Example C. Shade the system 2x – y < –2 x + y < 5 Find and label the tip of the region. { We are looking for the region that satisfies both inequalities. For 2x – y < –2, graph the equation 2x – y = –2. Linear Inequalities
  • Example C. Shade the system 2x – y < –2 x + y < 5 Find and label the tip of the region. { We are looking for the region that satisfies both inequalities. For 2x – y < –2, graph the equation 2x – y = –2. x y 0 0 Linear Inequalities Find the intercepts.
  • Example C. Shade the system 2x – y < –2 x + y < 5 Find and label the tip of the region. { We are looking for the region that satisfies both inequalities. For 2x – y < –2, graph the equation 2x – y = –2. x y 0 2 –1 0 Linear Inequalities Find the intercepts.
  • Example C. Shade the system 2x – y < –2 x + y < 5 Find and label the tip of the region. { We are looking for the region that satisfies both inequalities. For 2x – y < –2, graph the equation 2x – y = –2. x y 0 2 –1 0 Linear Inequalities Find the intercepts.
  • Example C. Shade the system 2x – y < –2 x + y < 5 Find and label the tip of the region. { We are looking for the region that satisfies both inequalities. For 2x – y < –2, graph the equation 2x – y = –2. x y 0 2 –1 0 Since the inequality includes “ = ”, use a solid line for the graph. Linear Inequalities Find the intercepts.
  • Example C. Shade the system 2x – y < –2 x + y < 5 Find and label the tip of the region. { We are looking for the region that satisfies both inequalities. For 2x – y < –2, graph the equation 2x – y = –2. x y 0 2 –1 0 Since the inequality includes “ = ”, use a solid line for the graph. Linear Inequalities Find the intercepts.
  • Example C. Shade the system 2x – y < –2 x + y < 5 Find and label the tip of the region. { We are looking for the region that satisfies both inequalities. For 2x – y < –2, graph the equation 2x – y = –2. x y 0 2 –1 0 Since the inequality includes “ = ”, use a solid line for the graph. Test (0, 0), Linear Inequalities Find the intercepts.
  • Example C. Shade the system 2x – y < –2 x + y < 5 Find and label the tip of the region. { We are looking for the region that satisfies both inequalities. For 2x – y < –2, graph the equation 2x – y = –2. x y 0 2 –1 0 Since the inequality includes “ = ”, use a solid line for the graph. Test (0, 0), it does not fit. Linear Inequalities Find the intercepts.
  • Example C. Shade the system 2x – y < –2 x + y < 5 Find and label the tip of the region. { We are looking for the region that satisfies both inequalities. For 2x – y < –2, graph the equation 2x – y = –2. x y 0 2 –1 0 Since the inequality includes “ = ”, use a solid line for the graph. Test (0, 0), it does not fit. Hence, the other side fits the inequality 2x – y < –2. Linear Inequalities Find the intercepts.
  • Example C. Shade the system 2x – y < –2 x + y < 5 Find and label the tip of the region. { We are looking for the region that satisfies both inequalities. For 2x – y < –2, graph the equation 2x – y = –2. x y 0 2 –1 0 Since the inequality includes “ = ”, use a solid line for the graph. Test (0, 0), it does not fit. Hence, the other side fits the inequality 2x – y < –2. Shade it. Linear Inequalities Find the intercepts.
  • Example C. Shade the system 2x – y < –2 x + y < 5 Find and label the tip of the region. { We are looking for the region that satisfies both inequalities. For 2x – y < –2, graph the equation 2x – y = –2. x y 0 2 –1 0 Since the inequality includes “ = ”, use a solid line for the graph. Test (0, 0), it does not fit. Hence, the other side fits the inequality 2x – y < –2. Shade it. Linear Inequalities Find the intercepts.
  • For x + y < 5, graph x = y = 5 Linear Inequalities
  • For x + y < 5, graph x = y = 5 x y 0 0 Linear Inequalities
  • For x + y < 5, graph x = y = 5 x y 0 5 5 0 Linear Inequalities
  • For x + y < 5, graph x = y = 5 x y 0 5 5 0 Linear Inequalities
  • For x + y < 5, graph x = y = 5 x y 0 5 5 0 Since the inequality is strict, use a dotted line for the graph. Linear Inequalities
  • For x + y < 5, graph x = y = 5 x y 0 5 5 0 Since the inequality is strict, use a dotted line for the graph. Linear Inequalities
  • For x + y < 5, graph x = y = 5 x y 0 5 5 0 Since the inequality is strict, use a dotted line for the graph. Test (0, 0), Linear Inequalities
  • For x + y < 5, graph x = y = 5 x y 0 5 5 0 Since the inequality is strict, use a dotted line for the graph. Test (0, 0), it fits x + y < 5. Linear Inequalities
  • For x + y < 5, graph x = y = 5 x y 0 5 5 0 Since the inequality is strict, use a dotted line for the graph. Test (0, 0), it fits x + y < 5. Hence, this side fits x + y < 5. Linear Inequalities
  • For x + y < 5, graph x = y = 5 x y 0 5 5 0 Since the inequality is strict, use a dotted line for the graph. Test (0, 0), it fits x + y < 5. Hence, this side fits x + y < 5. Shade it. Linear Inequalities
  • x y 0 5 5 0 Since the inequality is strict, use a dotted line for the graph. Test (0, 0), it fits x + y < 5. Hence, this side fits x + y < 5. Shade it. Linear Inequalities For x + y < 5, graph x = y = 5
  • x y 0 5 5 0 Since the inequality is strict, use a dotted line for the graph. Test (0, 0), it fits x + y < 5. Hence, this side fits x + y < 5. Shade it. The region that fits the system is the region has both shading. Linear Inequalities For x + y < 5, graph x = y = 5
  • x y 0 5 5 0 Since the inequality is strict, use a dotted line for the graph. Test (0, 0), it fits x + y < 5. Hence, this side fits x + y < 5. Shade it. The region that fits the system is the region has both shading. Linear Inequalities For x + y < 5, graph x = y = 5
  • x y 0 5 5 0 Since the inequality is strict, use a dotted line for the graph. Test (0, 0), it fits x + y < 5. Hence, this side fits x + y < 5. Shade it. The region that fits the system is the region has both shading. Linear Inequalities 2x – y = –2 x + y = 5{ For x + y < 5, graph x = y = 5 To find the tip of the region, we solve the system of equations for the point of intersection.
  • x y 0 5 5 0 Since the inequality is strict, use a dotted line for the graph. Test (0, 0), it fits x + y < 5. Hence, this side fits x + y < 5. Shade it. The region that fits the system is the region has both shading. Linear Inequalities 2x – y = –2 x + y = 5{ Add these equations to remove the y. For x + y < 5, graph x = y = 5 To find the tip of the region, we solve the system of equations for the point of intersection.
  • Linear Inequalities 2x – y = –2 x + y = 5+)
  • Linear Inequalities 2x – y = –2 x + y = 5+) 3x = 3
  • Linear Inequalities 2x – y = –2 x + y = 5+) 3x = 3 x = 1
  • Linear Inequalities Set x = 1 in x + y = 5, 2x – y = –2 x + y = 5+) 3x = 3 x = 1
  • Linear Inequalities Set x = 1 in x + y = 5, we get 1 + y = 5  y = 4. 2x – y = –2 x + y = 5+) 3x = 3 x = 1
  • Linear Inequalities Set x = 1 in x + y = 5, we get 1 + y = 5  y = 4. 2x – y = –2 x + y = 5+) 3x = 3 x = 1 Hence the tip of the region is (1, 4). (1, 4)
  • Exercise A. Shade the following inequalities in the x and y coordinate system. 1. x – y > 3 2. 2x ≤ 6 3. –y – 7 ≥ 0 4. 0 ≤ 8 – 2x 5. y < –x + 4 6. 2x/3 – 3 ≤ 6/5 7. 2x < 6 – 2y 8. 4y/5 – 12 ≥ 3x/4 9. 2x + 3y > 3 10. –6 ≤ 3x – 2y 11. 3x + 2 > 4y + 3x 12. 5x/4 + 2y/3 ≤ 2 Linear Inequalities 16.{–x + 2y ≥ –12 2x + y ≤ 4 Exercise B. Shade the following regions. Label the tip. 13.{x + y ≥ 3 2x + y < 4 14. 15.{x + 2y ≥ 3 2x – y > 6 {x + y ≤ 3 2x – y > 6 17. {3x + 4y ≥ 3 x – 2y < 6 18. { x + 3y ≥ 3 2x – 9y ≥ –4 19.{–3x + 2y ≥ –1 2x + 3y ≤ 5 20. {2x + 3y > –1 3x + 4y ≥ 2 21. {4x – 3y ≤ 3 3x – 2y > –4 { x – y < 3 x – y ≤ –1 3 2 2 3 1 2 1 4 22. { x + y ≤ 1 x – y < –1 1 2 1 5 3 4 1 6 23.