Teacher – Mrs. Volynskaya 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 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. y x Exactly one solution y x No Solution (parallel lines) y x Infinitely many solutions (lines coincide)

Example:Solving a System by Substitution Solve by the substitution method: 5x – 4y = 9 x – 2y = -3 Solution Rewrite second equation as x =2y -3 and Sub. into first equation This gives us an equation in one variable, 5(2y - 3) - 4y = 9. 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. mor e

Example:Solving a System by Substitution Solve by the substitution method: 5x – 4y = 9 x – 2y = -3. Solution 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). Check. Take a moment to show that (5, 4) satisfies both given equations. The solution set is {(5, 4)}.

Example:Solving a System by the Addition Method Solve by the addition method: 2x = 7y y = 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 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. mor e

Solution Steps 3 and 4Add the equations and solve for the remaining variable. 2x2x–7y7y=-17 3x3x+5y5y=17 32x–37y=3(-17) -23x+(-2)5y=-2(17) 6x6x–21y=-51 -6x+10y=-34 6x6x–21y=-51 -6x–10y= y= y= y= 85/31 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. mor e

Solution 2x2x–7y7y=-17 3x3x+5y5y=17 52x–57y=5(-17) 73x+75y=7(17) 10x–35y=-85 21x+35y=119 31x=34 x= 34/31 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)}.

Examples Determine the type of solution, then solve.