Systems of Linear Equations in Two Variables. We have seen that all equations in the form Ax + By = C are straight lines when graphed. Two such equations,

Presentation on theme: "Systems of Linear Equations in Two Variables. We have seen that all equations in the form Ax + By = C are straight lines when graphed. Two such equations,"— Presentation transcript:

1 Systems of Linear Equations in Two Variables

2 We have seen that all equations in the form Ax + By = C are straight lines when graphed. Two such equations, such as those listed above, are called a system of linear equations. A solution to a system of linear equations is an ordered pair that satisfies all equations in the system. For example, (3, 4) satisfies the system x + y = 7 (3 + 4 is, indeed, 7.) x – y = -1 (3 – 4 is indeed, -1.) Thus, (3, 4) satisfies both equations and is a solution of the system. The solution can be described by saying that x = 3 and y = 4. The solution can also be described using set notation. The solution set to the system is {(3, 4)} - that is, the set consisting of the ordered pair (3, 4). Systems of Linear Equations and Their Solutions

3 Determine whether (4, -1) is a solution of the system x + 2y = 2 x – 2y = 6. Solution Because 4 is the x-coordinate and -1 is the y-coordinate of (4, -1), we replace x by 4 and y by -1. x + 2y = 2x – 2y = 6 4 + 2(-1) = 24 – 2(-1) = 6 4 + (-2) = 24 – (-2) = 6 2 = 2 true 4 + 2 = 6 6 = 6 true The pair (4, -1) satisfies both equations: It makes each equation true. Thus, the pair is a solution of the system. The solution set to the system is {(4, -1)}. ? ? ? ? ? Text Example

4 Solving Linear Systems by Substitution Solve either of the equations for one variable in terms of the other. (If one of the equations is already in this form, you can skip this step.) Substitute the expression found in step 1 into the other equation. This will result in an equation in one variable. Solve the equation obtained in step 2. Back-substitute the value found in step 3 into the equation from step 1. Simplify and find the value of the remaining variable. Check the proposed solution in both of the system's given equations.

5 Solve by the substitution method: 5x – 4y = 9 x – 2y = -3. Solution Step 1 Solve either of the equations for one variable in terms of the other. We begin by isolating one of the variables in either of the equations. By solving for x in the second equation, which has a coefficient of 1, we can avoid fractions. x - 2y = -3 This is the second equation in the given system. x = 2y - 3 Solve for x by adding 2y to both sides. Step 2 Substitute the expression from step 1 into the other equation. We substitute 2y - 3 for x in the first equation. x = 2y – 3 5 x – 4y = 9 Text Example

6 Solve by the substitution method: 5x – 4y = 9 x – 2y = -3. Solution This gives us an equation in one variable, namely 5(2y - 3) - 4y = 9. The variable x has been eliminated. Step 3 Solve the resulting equation containing one variable. 5(2y – 3) – 4y = 9 This is the equation containing one variable. 10y – 15 – 4y = 9 Apply the distributive property. 6y – 15 = 9 Combine like terms. 6y = 24 Add 15 to both sides. y = 4 Divide both sides by 6. Text Example cont.

7 Solve by the substitution method: 5x – 4y = 9 x – 2y = -3. Solution Step 4 Back-substitute the obtained value into the equation from step 1. Now that we have the y-coordinate of the solution, we back-substitute 4 for y in the equation x = 2y – 3. x = 2y – 3 Use the equation obtained in step 1. x = 2 (4) – 3 Substitute 4 for y. x = 8 – 3 Multiply. x = 5 Subtract. With x = 5 and y = 4, the proposed solution is (5, 4). Step 5 Check. Take a moment to show that (5, 4) satisfies both given equations. The solution set is {(5, 4)}. Text Example cont.

8 Solving Linear Systems by Addition If necessary, rewrite both equations in the form Ax + By = C. If necessary, multiply either equation or both equations by appropriate nonzero numbers so that the sum of the x-coefficients or the sum of the y- coefficients is 0. Add the equations in step 2. The sum is an equation in one variable. Solve the equation from step 3. Back-substitute the value obtained in step 4 into either of the given equations and solve for the other variable. Check the solution in both of the original equations.

9 Solve by the addition method: 2x = 7y - 17 5y = 17 - 3x. Solution Step 1 Rewrite both equations in the form Ax + By = C. We first arrange the system so that variable terms appear on the left and constants appear on the right. We obtain 2x - 7y = -17 3x + 5y = 17 Step 2 If necessary, multiply either equation or both equations by appropriate numbers so that the sum of the x-coefficients or the sum of the y-coefficients is 0. We can eliminate x or y. Let's eliminate x by multiplying the first equation by 3 and the second equation by -2. Text Example

10 Solution Steps 3 and 4Add the equations and solve for the remaining variable. 17=5y5y+3x3x -17=7y7y–2x2x -2(17)=(-2)5y+-23x 3(-17)=37y37y–32x32x -34=10y–-6x -51=21y–6x6x -34=10y–-6x -51=21y–6x6x -85=-31y -31 -85=-31y 85/31 =y Multiply by 3. Multiply by -2. Add: Divide both sides by -31. Simplify. Step 5 Back-substitute and find the value for the other variable. Back- substitution of 85/31 for y into either of the given equations results in cumbersome arithmetic. Instead, let's use the addition method on the given system in the form Ax + By = C to find the value for x. Thus, we eliminate y by multiplying the first equation by 5 and the second equation by 7. Text Example cont.

11 Solution 17=5y5y+3x3x -17=7y7y–2x2x 7(17)=75y75y+73x73x 5(-17)=57y57y–52x52x 119=35y+21x -85=35y–10x 34=31x 34/31 =x Multiply by 5. Multiply by 7. Add: Step 6 Check. For this system, a calculator is helpful in showing the solution (34/31, 85/31) satisfies both equations. Consequently, the solution set is {(34/31, 85/31)}. Text Example cont.

12 The number of solutions to a system of two linear equations in two variables is given by one of the following. Number of Solutions What This Means Graphically Exactly one ordered-pair solutionThe two lines intersect at one point. No solution The two lines are parallel. Infinitely many solutionsThe two lines are identical. The number of solutions to a system of two linear equations in two variables is given by one of the following. Number of Solutions What This Means Graphically Exactly one ordered-pair solutionThe two lines intersect at one point. No solution The two lines are parallel. Infinitely many solutionsThe two lines are identical. yyy x Exactly one solution x No Solution (parallel lines) x Infinitely many solutions (lines coincide) The Number of Solutions to a System of Two Linear Equations

13 Example Solve the system2x + 3y = 4 -4x - 6y = -1 Solution: 2 (2x + 3y = 4) multiply the first equation by 2 -4x - 6y = -1 4x + 6y = 8 -4x - 6y = -1 0 = 7 Add the two equations No solution

14 Systems of Linear Equations in Two Variables