ISHIK UNIVERSITY FACULTY OF EDUCATION Mathematics Education Department

Presentation on theme: "ISHIK UNIVERSITY FACULTY OF EDUCATION Mathematics Education Department"— Presentation transcript:

1 ISHIK UNIVERSITY FACULTY OF EDUCATION Mathematics Education Department
MATH-203 Linear Algebra-I System of Linear Equations and Solutions Orhan TUĞ (PhDc) 2/19/2018 MATH-203

2 Systems of Linear Equations and Their Solutions
We have seen that all equations in the form Ax + By = C are straight lines when graphed. Two such equations, such as those listed below, 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). 2/19/2018 MATH-203

3 Example: Determining Whether an Ordered Pair Is a Solution of a System
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 = 2 x – 2y = 6 4 + 2(-1) = 2 4 – 2(-1) = 6 4 + (-2) = 2 4 – (-2) = 6 2 = 2 true = 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)}. ? 2/19/2018 MATH-203

4 The Number of Solutions to a System of Two Linear Equations
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 solution The two lines intersect at one point. No solution The two lines are parallel. Infinitely many solutions The two lines are identical. y x Exactly one solution y x No Solution (parallel lines) y x Infinitely many solutions (lines coincide) 2/19/2018 MATH-203

5 Determining Types of solutions
One way to determine the type of solution you expect to get is by looking at the coefficients of each variable in the two equations. Consider the general systems: Compare the corresponding coefficients: Same line: Parallel lines: Unique Solution: 2/19/2018 MATH-203

6 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. 2/19/2018 MATH-203

7 Example: Solving a System by Substitution
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 2/19/2018 MATH-203

8 Example: Solving a System by Substitution
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. 2/19/2018 MATH-203 more

9 Example: Solving a System by Substitution
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)}. 2/19/2018 MATH-203

10 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. 2/19/2018 MATH-203

11 Example: Solving a System by the Addition Method
Solve by the addition method: 2x = 7y - 17 5y = x. 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 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. 2/19/2018 MATH-203

12 Steps 3 and 4 Add the equations and solve for the remaining variable.
Solution Multiply by 3. Multiply by -2. 2x – 7y = -17 3x + 5y 17 3•2x – 3•7y = 3(-17) -2•3x + (-2)5y -2(17) 6x – 21y = -51 -6x + 10y -34 Steps 3 and 4 Add the equations and solve for the remaining variable. 6x – 21y = -51 -6x 10y -34 Add: -31y = -85 -31y = -85 -31 Divide both sides by -31. Simplify. y = 85/31 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. 2/19/2018 MATH-203

13 Solution Multiply by 5. Multiply by 7. 2x – 7y = -17 3x + 5y 17 5•2x – 5•7y = 5(-17) 7•3x + 7•5y 7(17) 10x – 35y = -85 21x + 119 Add: 31x = 34 x = 34/31 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)}. 2/19/2018 MATH-203

14 Examples Determine the type of solution, then solve. 2/19/2018
MATH-203