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Solving Systems of Equations By the Graphing Method A linear equation in two variables has an infinite number of solutions. The set of all solutions to a linear equation forms a line in a rectangular coordinate system. Two or more linear equations from a system of linear equations. Infinite number of solutions A solution to a system of linear equations is an ordered pair that is a solution to each individual linear equation. Determining Solutions to a System of Linear Equations x+y=4 a). (3, 1) It is important to test an ordered pair in both the -2x + y = -5 equations to determine if Solution: the ordered pair is a solution a. Substitute the ordered pair (3, 1) x+y=4 -2x + y = -5 (3) + (1) = 4 -2(3) + (1) = -5 TRUE TRUE Because the ordered pair (3, 1) is a solution to each equation, it is a solution to the system of equations Determining Solutions to a System of Linear Equations x+y=4 -2x + y = -5 b). (0, 4) Solution: a. Substitute the ordered pair (0, 4) x+y=4 -2x + y = -5 (0) + (4) = 4 -2(0) + (4) = -5 TRUE False Because the ordered pair (0, 4) is not a solution to each equation, it is not a solution to the system of equations A solution to a system of two linear equations may be interpreted graphically as a point of intersection between the two lines. Using slope-intercept form to graph the lines from Example 1, we have x+y=4 -2x + y = -5 y = -x + 4 y = 2x - 5 Notice that the lines intersect at (3, 1) Two Equations Containing Two Variables The first two cases are called consistent since there are solutions. The last case is called inconsistent. The solution will be one of three cases: 1. Exactly one solution, an ordered pair (x, y) 2. A dependent system with infinitely many solutions 3. No solution independent With two equations and two variables, the graphs are lines and the solution (if there is one) is where the lines intersect. Let’s look at different possibilities. dependent consistent Case 1: The lines intersect at a point (x, y), the solution. inconsistent consistent Case 2: The lines coincide and there are infinitely many solutions (all points on the line). Case 3: The lines are parallel so there is no solution. Example 2 y = 2x y=2 The lines intersect at (1, 2) Solution: The equation y = 2x is written in slope-intercept form as y = 2x + 0. The line passes through the origin, with a slope of 2 The line y = 2 is a horizontal line and has a slope of 0. EXAMPLE 3 x – 2y = -2 -3x + 2y = 6 Equation 1. x – 2y = -2 -2y = -x -2 -2y = -x – 2 -2 -2 -2 y = 1x +1 2 Solution: To graph each equation, write the equation in slope-intercept form: y = mx + b Equation 2. -3x + 2y = 6 2y = 3x + 6 2y = 3x +6 2 2 2 y = 3x +3 2 The point (-2, 0) appears to be the point of intersection. This can be confirmed by substituting x = -2 and y = 0 into both equations. x – 2y = -2 -3x + 2y = 6 (-2) – 2(0) = -2 True -3(-2) + 2(0) = 6 True EXAMPLE 4 -x + 3y = -6 6y = 2x + 6 Equation 1. -x + 3y = -6 3y = x -6 3y = x – 6 3 3 3 y=1x-2 3 Solution: To graph each equation, write the equation in slope-intercept form: y = mx + b Equation 2. 6y = 2x + 6 6y = 2x +6 6 6 6 y = 1x +1 3 Because the lines have the same slope but different y-intercepts, they are parallel. Two parallel line do not intersect, which implies that the system has no solution. The system is inconsistent. EXAMPLE 5 x +4y = 8 y = -1 x + 2 4 Equation 1. x + 4y = 8 4y = -x + 8 4y = -x + 8 4 4 4 y = -1x +2 4 Solution: To graph each equation, write the equation in slope-intercept form: y = mx + b Equation 2. y = -1x + 2 4 Notice that the slopeintercept forms of the two lines are identical. Therefore the equations represent the same line. The system is dependent, and the solution to the system of equations is the set of all points Solving System of Equations by the Substitution Method Solve the system using the substitution method x+y=4 -2x + y = -5 Solution: x+y=4 x=4-y -2 x + y = -5 -2( 4 – y ) + y = -5 This equation now contains only one variable. Solve the resulting equation -8 +2y +y = -5 -8 + 3y = -5 3y = -5 + 8 y=1 To find x, substitute y = 1 back into the first equation x=4-y x = 4 – (1) x=3 The ordered pair (3, 1) can be checked in the original equations to verify the answer x+y=4 -2x + y = -5 (3) + (1) = 4 -2(3) + (1) = -5 The solution is (3, 1) True True Solving a System of Linear Equations Using the Substitution Method 1. Isolate one of the variables from one equation 2. Substitute the quantity found in Step 1 into the other equation. 3. Solve the resulting equation. 4. Substitute the value found in Step 3 back into the equation in Step 1 to find the value of the remaining variable. 5. Check the solution in both original equations and write the answer as an ordered pair. If we have two equations and variables and we want to solve them, graphing would not always be very accurate so we will solve algebraically for exact solutions. We'll look at two methods. The first is solving by substitution. The idea is to solve for one of the variables in one of the equations and substitute it in for that variable in the other equation. 3x 5 y 17 Let's solve for y in the second equation. You can pick either variable and either equation, but go for the easiest one (getting the y alone in the second equation will not involve fractions). 2 x y 6 yy22 x1 66 4 Substitute this for y in the first equation. Easiest here since we already solved for y. 3x 5 2 x 6 17 13x 13 x 1 Now we only have the x variable and we solve for it. Substitute this for x in one of the equations to find y. 3x 5 y 17 2 x y 6 1, 4 So our solution is 1, 4 This means that the two lines intersect at this point. Let's look at the graph. We can check this by subbing in these values for x and y. They should make each equation true. 3 1 5 4 17 2 1 4 6 3 20 17 2 4 6 Yes! Both equations are satisfied. Solving an Inconsistent System Using Substitution The idea is to solve for one of the variables in one of the equations and substitute it in for that variable in the other equation. 2x 3y 6 2 y x4 3 The second equation is in the slope-intercept form. Substitute this for y in the first equation. 2 2x 3 x 4 6 3 2x 2x 12 6 12 6 Contradiction There is no solution 2x 3y 6 2 y x4 3 There is no solution This means that the two lines do not intersect at any point. Let's look at the graph. We can check this by writing each equation in slope-intercept form and graphing the lines. 2x 3y 6 3 y 2 x 6 2 y x6 3 2 y x4 3 Yes! Both equations have the same slope but different y-intercepts Solving a Dependent System Using Substitution ½x - ¼y = 1 6x – 3y = 12 ½x - ¼y = 1 Clear the fractions 4(½x) -4( ¼y) =4(1) 2x – y = 4 Now the system becomes: 2x – y = 4 The y-variable in the first equation is the easiest 6x – 3y = 12 to isolate because its coefficient is -1 The idea is to solve for one of the variables in one of the equations and substitute it in for that variable in the other equation. 2x y 4 6 x 3 y 12 y 2x 4 Let's solve for y in the first equation. You can pick either variable and either equation, but go for the easiest one (getting the y alone in the first equation will not involve fractions). Substitute this for y in the second equation. Now we only have the x variable and we solve for it. 6 x 3 2 x 4 12 6 x 6 x 12 12 12 12 Because the equation produces an identity, all values of x make this equation true. 2x y 4 6 x 3 y 12 There is an infinite number of solutions This means that the two lines intersect at every point. Let's look at the graph. We can check this by writing each equation in slope-intercept form and graphing the lines. 2x y 4 6 x 3 y 12 y 2 x 4 3 y 6 x 12 y 2x 4 y 2x 4 Yes! Both equations have the same slope and y-intercepts. Now let's look at the second method, called the method of elimination. The idea is to multiply one or both equations by a constant (or constants) so that when you add the two equations together, one of the variables is eliminated. 2 x 3 y 1 3 4x y 3 3 12 x 3 y 9 2 x 3 y 1 14 x 8 4 x 7 Let's multiply the bottom equation by 3. This way we can eliminate y's. (we could instead have multiplied the top equation by -2 and eliminated the x's) Add first equation to this. The y's are eliminated. 4 4 y 3 7 5 y 7 Substitute this for x in one of the equations to find y. Solving Systems of Equations by the Addition Method Notice that the coefficients of the y-variables are opposites: Coefficient is 1. x + y = -2 x – y = -6 Coefficient is -1 Because the coefficients of the y-variables are opposites, we can add the two equations to eliminate the y-variable. x + y = -2 x – y = -6 2x = -8 After adding the equations, we have one equation and one variable. Solve the resulting equation. 2x = -8 2 2 x = -4 To find the value of y, substitute x = -4 into either of the original equations x + y = -2 First equation (-4) + y = -2 The solution y = -2 + 4 is (-4, 2) y=2 Solving a System of Equations by the Addition Method • Write both equations in standard form: Ax + By = C • Clear fractions or decimals • Multiply one or both equations by nonzero constants to create opposite coefficients for one of the variables. • Add the equations from Step 3 to eliminate one variable. • Solve for the remaining variable. • Substitute the known value from Step 5 into one of the original equations to solve for the other variable. • Check the solution in both equations So we can arrive at the same answer with either method, but which method should you use? It depends on the problem. If substitution would involve messy fractions, it is generally easier to use the addition method. However, if one variable is already or easily solved for, substitution is generally quicker. With either method, we may end up with a surprise. Let's see what this means. 3 x 6 y 15 3 x 2 y 5 3 3 x 6 y 15 3 x 6 y 15 00 Let's multiply the second equation by 3 and add to the first equation to eliminate the x's. Everything ended up eliminated. This tells us the equations are dependent. This is Case 2 where the lines coincide and all points on the line are solutions. 3 x 6 y 15 x 2 y 5 x 2y 5 Let's solve the second equation for x. (Solving for x in either equation will give the same result) Now to get a solution, you chose any real number for y and x depends on that choice. y If y is 0, x is -5 so the point (-5, 0) is a solution to both equations. If y is 2, x is -1 so the point (-1, 2) is a solution to both equations. Any point on this line is a solution x So we list the solution as: x 2 y 5 where x is any real number Example 6 Solve the system using the addition method 3(x – 10) = 7y + 11 -2(x – y) = 2x - 18 Step 1. Write the equations in standard form. Clear the parentheses 3x –-10) 3(x 30 =7y = 7y++1111 -2(x+–2y -2x y) == 2x 2x --18 18 Write as Ax + By = C 3x – 7y = 41 -4x + 2y = -18 Step 2. There are no fractions or decimals. No need to clear the fractions or decimals in this equation. Notice that neither the coefficients of x nor the coefficients of y are opposites. However, it is possible to change the coefficients of x to 12 and -12 (notice that 12 is the LCM of 3 and 4). This is accomplished by multiplying the first equation by 4 and the second equation by 3 Step 3: Create opposite coefficients of x Step 4: Add the equations. Step 5: Solve the resulting equation. 12x – 28yMultiply = 164 by 4 -12x +28y 6y==41-54 12x 3x–-7y 164 110 -12x -4x++-22y 6y 2y===-54 -18 Multiply by 3 -22y = 110 -22 -22 y = -5 3 x – 7 -5 y = 41 First equation 3x + 35 = 41 3x + 35 -35 =41 -35 3x = 6 3 3 x=2 3(x– –10) 10) 7y ++11 3(2 = =7(-5) 11 -2(x – y) == 2(2) 2x - 18 2[2 – (-5)] - 18 Step 6: Substitute y = -5 -5 into one of the equations. The solution is (2, -5) Step 7: Check the solution in the original equations. -24 = -24 -14 = -14 True True Summary of Methods for Solving Linear Equations in Two Variables If no method of solving a system of linear equations is specified, you may use the method of your choice. However, it is recommended you use the following guidelines: 1. If one of the equations is written with a variable isolated, the substitution method is a good choice. 2x + 5y = 2 x=y-6 2. If both equations are written in standard form, Ax + By = C, where none of the variables has coefficients of 1 or -1, then the addition method is a good choice. 4x + 5y = 12 5x + 3y = 15 3. If both equations are written in standard form, Ax + By = C, and at least one variable has a coefficient of 1 or -1, then either the substitution method or the addition method is a good choice.