6-1 Solving Systems by Graphing Holt Algebra 1 Warm Up Lesson Presentation Lesson Quiz
Warm Up Evaluate each expression for  x  = 1 and  y  =–3. 1.   x  – 4 y   2.  –2 x  +  y Write each expression in slope-intercept form. 3.   y  –  x =  1 4.  2 x  + 3 y  =   6 5.  0 = 5 y  + 5 x   13 – 5 y = x +  1 y =   x  + 2 y  =  – x
Identify solutions of linear equations in two variables.   Solve systems of linear equations in two variables by graphing. Objectives
systems of linear equations solution of a system of linear equations Vocabulary
A  system of linear equations  is a set of two or more linear equations containing two or more variables. A  solution of a system of linear equations  with two variables is an ordered pair that satisfies each equation in the system. So, if an ordered pair is a solution, it will make both equations true.
Tell whether the ordered pair is a solution of the given system. Example 1A: Identifying Systems of Solutions (5, 2); The ordered pair (5, 2) makes both equations true. (5, 2) is the solution of the system. Substitute 5 for x and 2 for y in each equation in the system.  3 x  –  y =  13 2 – 2  0 0  0  0 3 (5)  –  2   13 15 – 2  13 13  13  3 x  –  y   13
If an ordered pair does not satisfy the first equation in the system, there is no reason to check the other equations. Helpful Hint
Example 1B: Identifying Systems of Solutions Tell whether the ordered pair is a solution of the given system. (–2, 2); x +  3 y  = 4 – x + y  = 2 Substitute  –2 for x and 2 for y in each equation in the system. The ordered pair (–2, 2) makes one equation true but not the other.  (–2, 2) is not a solution of the system.  – 2  + 3 (2)  4 x  +  3 y  = 4 – 2 + 6  4 4  4 – x  +  y  = 2 – (–2)  +  2   2 4  2
Check It Out!  Example 1a  Tell whether the ordered pair is a solution of the given system. The ordered pair (1, 3) makes both equations true.  Substitute  1 for x and 3 for y in each equation in the system. (1, 3) is the solution of the system. (1, 3);  2 x + y  = 5 – 2 x + y  = 1 2 x  +  y  = 5 2 (1)  +  3   5 2 + 3  5 5  5  – 2 x  +  y  = 1 – 2 (1)  +  3   1 – 2 + 3  1 1  1 
Check It Out!  Example 1b  Tell whether the ordered pair is a solution of the given system. (2, –1);  x  –   2 y  = 4 3 x + y  = 6 The ordered pair (2, –1) makes one equation true, but not the other.  Substitute  2 for x and –1 for y in each equation in the system. (2, –1) is not a solution of the system. 3 x  +  y  = 6 3 (2)   +   (–1)   6 6  –  1  6 5  6 x   – 2 y  = 4 2  – 2 (–1)   4 2 + 2  4  4  4
All solutions of a linear equation are on its graph. To find a solution of a system of linear equations, you need a point that each line has in common. In other words, you need their point of intersection. The point (2, 3) is where the two lines intersect and is a solution of both equations, so (2, 3) is the solution of the systems. y =  2 x  – 1 y =  – x  + 5
Sometimes it is difficult to tell exactly where the lines cross when you solve by graphing. It is good to confirm your answer by substituting it into both equations. Helpful Hint
Solve the system by graphing. Check your answer. Example 2A: Solving a System Equations by Graphing y = x y =  –2 x  – 3 Graph the system. The solution appears to be at (–1, –1). (–1, –1) is the solution of the system. y = x y =  –2 x  – 3 • (–1, –1) Check Substitute (–1, –1) into the system. y  =  x (–1)  (–1) – 1  –1  y =  –2 x   – 3 ( – 1)   –2 ( – 1)   –3 – 1   2   – 3 – 1  – 1 
Solve the system by graphing. Check your answer. Example 2B: Solving a System Equations by Graphing y = x  –   6 Rewrite the second equation in slope-intercept form. Graph using a calculator and then use the intercept command. y +  x =  –1 y = x  –   6 y +  x =  –1 −   x   −   x y =
Solve the system by graphing. Check your answer. Example 2B Continued Check  Substitute  into the system. y = x  –   6 The solution is  .  +  – 1 – 1 – 1 – 1  – 1  y  =  x   –   6 –  6 
Solve the system by graphing. Check your answer. Check It Out!  Example 2a  y  = –2 x  – 1 y  =  x  + 5 Graph the system.  The solution appears to be (–2, 3).  Check  Substitute (–2, 3) into the system. (–2, 3) is the solution of the system.  y  =  x   + 5 3   – 2   + 5 3  3  y  = –2 x   – 1 3   –2 ( – 2)   – 1 3  4   – 1 3  3  y  =  x  + 5 y  = –2 x  – 1
Solve the system by graphing. Check your answer. Check It Out!  Example 2b  2 x + y  = 4 Rewrite the second equation in slope-intercept form. Graph using a calculator and then use the intercept command. 2 x + y  = 4 – 2 x  – 2 x y =  –2 x +  4 2 x + y  = 4
Solve the system by graphing. Check your answer. Check It Out!  Example 2b Continued  2 x + y  = 4 The solution is (3, –2).  Check  Substitute (3, –2)  into the system. 2 x  +  y  =  4 2 (3)  +  (–2)  4 6 – 2  4 4  4  2 x + y  = 4 – 2   (3)  – 3 – 2  1 – 3 – 2  –2  
Lesson Quiz: Part I Tell whether the ordered pair is a solution of the given system.  1.  (–3, 1); 2.  (2, –4); yes no
Lesson Quiz: Part II Solve the system by graphing.   3.   (2, 5) y +  2 x =  9   y =  4 x  – 3

Systems Of Equations

  • 1.
    6-1 Solving Systemsby Graphing Holt Algebra 1 Warm Up Lesson Presentation Lesson Quiz
  • 2.
    Warm Up Evaluateeach expression for x = 1 and y =–3. 1. x – 4 y 2. –2 x + y Write each expression in slope-intercept form. 3. y – x = 1 4. 2 x + 3 y = 6 5. 0 = 5 y + 5 x 13 – 5 y = x + 1 y = x + 2 y = – x
  • 3.
    Identify solutions oflinear equations in two variables. Solve systems of linear equations in two variables by graphing. Objectives
  • 4.
    systems of linearequations solution of a system of linear equations Vocabulary
  • 5.
    A systemof linear equations is a set of two or more linear equations containing two or more variables. A solution of a system of linear equations with two variables is an ordered pair that satisfies each equation in the system. So, if an ordered pair is a solution, it will make both equations true.
  • 6.
    Tell whether theordered pair is a solution of the given system. Example 1A: Identifying Systems of Solutions (5, 2); The ordered pair (5, 2) makes both equations true. (5, 2) is the solution of the system. Substitute 5 for x and 2 for y in each equation in the system. 3 x – y = 13 2 – 2 0 0 0  0 3 (5) – 2 13 15 – 2 13 13 13  3 x – y 13
  • 7.
    If an orderedpair does not satisfy the first equation in the system, there is no reason to check the other equations. Helpful Hint
  • 8.
    Example 1B: IdentifyingSystems of Solutions Tell whether the ordered pair is a solution of the given system. (–2, 2); x + 3 y = 4 – x + y = 2 Substitute –2 for x and 2 for y in each equation in the system. The ordered pair (–2, 2) makes one equation true but not the other. (–2, 2) is not a solution of the system.  – 2 + 3 (2) 4 x + 3 y = 4 – 2 + 6 4 4 4 – x + y = 2 – (–2) + 2 2 4 2
  • 9.
    Check It Out! Example 1a Tell whether the ordered pair is a solution of the given system. The ordered pair (1, 3) makes both equations true. Substitute 1 for x and 3 for y in each equation in the system. (1, 3) is the solution of the system. (1, 3); 2 x + y = 5 – 2 x + y = 1 2 x + y = 5 2 (1) + 3 5 2 + 3 5 5 5  – 2 x + y = 1 – 2 (1) + 3 1 – 2 + 3 1 1 1 
  • 10.
    Check It Out! Example 1b Tell whether the ordered pair is a solution of the given system. (2, –1); x – 2 y = 4 3 x + y = 6 The ordered pair (2, –1) makes one equation true, but not the other. Substitute 2 for x and –1 for y in each equation in the system. (2, –1) is not a solution of the system. 3 x + y = 6 3 (2) + (–1) 6 6 – 1 6 5 6 x – 2 y = 4 2 – 2 (–1) 4 2 + 2 4  4 4
  • 11.
    All solutions ofa linear equation are on its graph. To find a solution of a system of linear equations, you need a point that each line has in common. In other words, you need their point of intersection. The point (2, 3) is where the two lines intersect and is a solution of both equations, so (2, 3) is the solution of the systems. y = 2 x – 1 y = – x + 5
  • 12.
    Sometimes it isdifficult to tell exactly where the lines cross when you solve by graphing. It is good to confirm your answer by substituting it into both equations. Helpful Hint
  • 13.
    Solve the systemby graphing. Check your answer. Example 2A: Solving a System Equations by Graphing y = x y = –2 x – 3 Graph the system. The solution appears to be at (–1, –1). (–1, –1) is the solution of the system. y = x y = –2 x – 3 • (–1, –1) Check Substitute (–1, –1) into the system. y = x (–1) (–1) – 1 –1  y = –2 x – 3 ( – 1) –2 ( – 1) –3 – 1 2 – 3 – 1 – 1 
  • 14.
    Solve the systemby graphing. Check your answer. Example 2B: Solving a System Equations by Graphing y = x – 6 Rewrite the second equation in slope-intercept form. Graph using a calculator and then use the intercept command. y + x = –1 y = x – 6 y + x = –1 − x − x y =
  • 15.
    Solve the systemby graphing. Check your answer. Example 2B Continued Check Substitute into the system. y = x – 6 The solution is . + – 1 – 1 – 1 – 1 – 1  y = x – 6 – 6 
  • 16.
    Solve the systemby graphing. Check your answer. Check It Out! Example 2a y = –2 x – 1 y = x + 5 Graph the system. The solution appears to be (–2, 3). Check Substitute (–2, 3) into the system. (–2, 3) is the solution of the system. y = x + 5 3 – 2 + 5 3 3  y = –2 x – 1 3 –2 ( – 2) – 1 3 4 – 1 3 3  y = x + 5 y = –2 x – 1
  • 17.
    Solve the systemby graphing. Check your answer. Check It Out! Example 2b 2 x + y = 4 Rewrite the second equation in slope-intercept form. Graph using a calculator and then use the intercept command. 2 x + y = 4 – 2 x – 2 x y = –2 x + 4 2 x + y = 4
  • 18.
    Solve the systemby graphing. Check your answer. Check It Out! Example 2b Continued 2 x + y = 4 The solution is (3, –2). Check Substitute (3, –2) into the system. 2 x + y = 4 2 (3) + (–2) 4 6 – 2 4 4 4  2 x + y = 4 – 2 (3) – 3 – 2 1 – 3 – 2 –2 
  • 19.
    Lesson Quiz: PartI Tell whether the ordered pair is a solution of the given system. 1. (–3, 1); 2. (2, –4); yes no
  • 20.
    Lesson Quiz: PartII Solve the system by graphing. 3. (2, 5) y + 2 x = 9 y = 4 x – 3