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You are solving for the variable that makes the equation true (equal). When you solve one variable equations 2x - 4 = 6 You are solving for the variable that makes the equation true (equal). When you solve one variable equations 2x - 4 = 6 +4 +4 Add 4 to both sides You are solving for the variable that makes the equation true (equal). When you solve one variable equations 2x - 4 = 6 2x = 10 +4 +4 Fill in the correct value You are solving for the variable that makes the equation true (equal). When you solve one variable equations 2x - 4 = 6 2x = 10 +4 2 +4 2 You are solving for the variable that makes the equation true (equal). When you solve one variable equations 2x - 4 = 6 2x = 10 +4 2 x = 5 +4 2 You are solving for the variable that makes the equation true (equal). When you solve one variable equations 2x - 4 = 6 2x = 10 +4 2 x = 5 +4 2 You are solving for the number that can go in for x When you solve one variable equations 2x - 4 = 6 2x - 4 = 6 2x = 10 2 x = 5 x = 5 +4 2 +4 2( 5 ) -4 = 6 Plug in 5 for x in the original equation to check There are many sets of possible solutions For a two variable equations x + y = 5 such as: x = 1 There are many sets of possible solutions y = 4 For a two variable equations For a two variable equations because 1 + 4 = 5 x + y = 5 such as: x =2 x = 1 Fill in the correct value There are many sets of possible solutions y = 3 y = 4 For a two variable equations For a two variable equations because 2 + 3 = 5 because 1 + 4 = 5 x + y = 5 such as: x =2 x = 100 x = 1 There are many sets of possible solutions y = 3 y = 4 y = -95 For a two variable equations For a two variable equations because 2 + 3 = 5 because 1 + 4 = 5 because 100 - 95 = 5 x + y = 5 Fill in the correct value such as: x =2 x = 100 x = 1 In fact, the number of solutions are infinite There are many sets of possible solutions y = 3 y = 4 y = -95 For a two variable equations For a two variable equations because 2 + 3 = 5 because 1 + 4 = 5 because 100 - 95 = 5 x + y = 5 such as: But if I have two equations each with two variables x + y = 5 x - y = -3 I can solve for the x and y values that work for BOTH The x and y values that make both equations true But if I have two equations each with two variables x + y = 5 x - y = -3 I can solve for the x and y values that work for BOTH Solutions Possible But if I have two equations each with two variables x + y = 5 x - y = -3 I can solve for the x and y values that work for BOTH x = 0 y = 5 Solutions Possible But if I have two equations each with two variables x + y = 5 x - y = -3 x = 1 y = 4 I can solve for the x and y values that work for BOTH x = 0 y = 5 Solutions Fill in the correct value Possible But if I have two equations each with two variables x + y = 5 x - y = -3 x = 2 y = 3 x = 1 y = 4 I can solve for the x and y values that work for BOTH x = 0 y = 5 Solutions Possible Fill in the correct value But if I have two equations each with two variables x + y = 5 x - y = -3 x = 2 y = 3 x = 1 y = 4 I can solve for the x and y values that work for BOTH x = 0 y = 5 Solutions Possible But if I have two equations each with two variables x + y = 5 x - y = -3 Solutions Possible x = 2 y = 3 x = 1 y = 4 I can solve for the x and y values that work for BOTH x = 0 y = 5 Solutions Possible But if I have two equations each with two variables x + y = 5 x - y = -3 x = 0 y = 3 Solutions Possible x = 2 y = 3 x = 1 y = 4 I can solve for the x and y values that work for BOTH x = 0 y = 5 Solutions Possible But if I have two equations each with two variables x + y = 5 x - y = -3 x = 0 y = 3 x = 1 y = 4 Solutions Possible x = 2 y = 3 x = 1 y = 4 I can solve for the x and y values that work for BOTH x = 0 y = 5 Solutions Possible But if I have two equations Drag in the values each with two variables x + y = 5 x - y = -3 x = y = x = 0 y = 3 x = 1 y = 4 Solutions Possible 2 ? 5 ? x = 2 y = 3 x = 1 y = 4 I can solve for the x and y values that work for BOTH x = 0 y = 5 Solutions Possible But if I have two equations The set that matches each with two variables is the answer x + y = 5 x - y = -3 x = 2 y = 5 x = 0 y = 3 x = 1 y = 4 Solutions Possible x = 2 y = 3 x = 1 y = 4 I can solve for the x and y values that work for BOTH x = 0 y = 5 Solutions Possible But if I have two equations The set that MATCHES each with two variables is the answer x + y = 5 x - y = -3 x = 2 y = 5 x = 0 y = 3 x = 1 y = 4 Solutions Possible When x=1 and y=4 which equation(s) are true (equal)? 2nd equation 1st equation x = 1 and y = 4 is the solution to this system 1st equation only 2nd equation only Neither equation Both equations x + y = 5 x - y = -3 Substitute x=4, y=1 (4,1) into both equations to see if (4,1) is a solution of the system of equations. 2nd equation 1st equation When x=4 and y=1 which equation(s) are true? 2x - y = 4 x + y = 5 1st equation only 2nd equation only Neither equation Both equations Substitute x=4, y=1 (4,1) into both equations to see if (4,1) is a solution of the system of equations. 2nd equation 1st equation When x=4 and y=1 which equation(s) are true? 2x - y = 4 x + y = 5 Point (4,1) only Sat-is-fies the 1st equation. (4,1) is NOT a solution to the entire system Substitute x=2, y=0 (2,0) into both equations to see if (2,0) is a solution of the system of equations. 2nd equation 1st equation When x=2 and y=0 which equation(s) are true? 2x - y = 4 x + y = 5 1st equation only 2nd equation only Neither equation Both equations Substitute x=2, y=0 (2,0) into both equations to see if (2,0) is a solution of the system of equations. 2nd equation 1st equation When x=2 and y=0 which equation(s) are true? 2x - y = 4 x + y = 5 Point (2,0) only Sat-is-fies the 2nd equation. (2,0) is NOT a solution to the entire system Substitute x=3, y=2 (3,2) into both equations to see if (3,2) is a solution of the system of equations. 2nd equation 1st equation When x=3 and y=2 which equation(s) are true? 2x - y = 4 x + y = 5 1st equation only 2nd equation only Neither equation Both equations Substitute x=3, y=2 (3,2) into both equations to see if (3,2) is a solution of the system of equations. 2nd equation 1st equation When x=3 and y=2 which equation(s) are true? 2x - y = 4 x + y = 5 Both equations are true. (3,2) is a solution of this system of equations The End |