Section 3.4 Systems of linear equations

Section 3.4 Systems of linear equations. Chapter 3: Lines Section 3.4 Systems of linear equations. Chapter 3: Lines Solve the system of equations by graphing • • • if the lines intersect in a single point write the single point as your answer. if the lines are parallel say the system is inconsistent or simply there is no solution If the lines are the same line say the equations are dependent and write your answer in set builder notation {(𝑥, 𝑦)|𝑤𝑟𝑖𝑡𝑒 𝑒𝑖𝑡ℎ𝑒𝑟 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 ℎ𝑒𝑟𝑒} 1 𝑦 = 2𝑥 + 1 3𝑥 − 2𝑦 = 6 1 Step 1: Graph 𝑦 = 2 𝑥 + 1 Plot the point (0,1) go up 1 right 2 and plot a second point. Then connect the points. Step 2: Graph 3x – 2y = 6 Find the x-intercept 3x – 2(0) = 6 3x = 6 x = 2 x-intercept (2,0) Find the y-intercept 3(x) – 2y = 6 -2y = 6 y = -3 y-intercept (0,-3) Plot the points and draw the line on the same axis. Estimate where the lines cross Answer: (4, 3) I will show the appropriate check a little later in this section when I solve this problem using another method. Section 3.4 Systems of linear equations. Chapter 3: Lines Solve the system of equations by graphing • if the lines intersect in a single point writhe the single point as your answer. • if the line are parallel say the system is inconsistent or simply there is no solution • If the lines are the same line say the equations are dependent and write your answer in set builder notation {(𝑥, 𝑦)|𝑤𝑟𝑖𝑡𝑒 𝑒𝑖𝑡ℎ𝑒𝑟 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 ℎ𝑒𝑟𝑒} 1 𝑦 = 2𝑥 + 1 4𝑦 = 2𝑥 + 4 1 Step 1: Graph 𝑦 = 𝑥 + 1 2 Plot the point (0,1) go up 1 right 2 and plot a second point. Then connect the points. Step 2: Graph 4y = 2x + 4 Find the x-intercept 4(0) = 2x + 4 0 = 2x + 4 -4 = 2x -2 = x x-intercept (-2,0) Find the y-intercept 4y = 2(0) +4 4y = 4 y = 1 y-intercept (0,1) Plot the points and draw the line on the same axis. These graph to be the same line. The system of equations is dependent. The solution is {(𝑥, 𝑦)|4𝑦 = 2𝑥 + 4} (you can write either equation in the set braces) Solving a system of equations by the substitution method Step 1: Pick an equation and solve it for a variable. Step 2: Substitute into the other equation Step 3: Solve the equation Step 4: • If step 3 gives an answer for a single variable, substitute that answer into either equation to solve for the remaining variable and write the solution. • If all the variables drop out and a false statement remains state that there is no solution or that the system is inconsistent • If all the variables drop out and a true statement remains, state that the system is dependent and write your answer using set builder notation. {(𝑥, 𝑦)|𝑤𝑟𝑖𝑡𝑒 𝑒𝑖𝑡ℎ𝑒𝑟 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 ℎ𝑒𝑟𝑒} Step 5: Check • • • If there is a single solution substitute the solution into both equations to check. If the system is inconsistent graph to confirm the lines are parallel If the system is dependent graph to confirm the lines are the same For Example: Solve the system of equations using the substitution method. 1 𝑦 = 2 𝑥 + 1 (This is the same problem as the first example. The answer should (4,3)) 3𝑥 − 2𝑦 = 6 (I should get the same answer regardless of the method used to solve.) Step 1: Pick an equation and solve for a variable. This step is done for me: I pick the top equation and will leave it solved for y. 1 𝑦 = 𝑥+1 2 Step 2: Substitute into the other equation. 1 2 3𝑥 − 2 ( 𝑥 + 1) = 6 1 2 1 2 Note −2 ( 𝑥 + 1) = −2 ( 𝑥) − (2)(1) = −1𝑥 − 2 Step 3: Solve this equation. 3𝑥 − 1𝑥 − 2 = 6 2𝑥 − 2 = 6 2𝑥 = 8 𝑥=4 Step 4: Solve for the other variable, using either equation. I will substitute 𝑥 = 4 into the equation 3𝑥 − 2𝑦 = 6 , but I would get the same answer if I used the other equation. 3(4) − 2𝑦 = 6 12 − 2𝑦 = 6 −2𝑦 = −6 𝑦=3 Answer: (4,3) Step 5: Check by substituting the answer into both equations. Both substitutions must simplify to a true statement. 1 3𝑥 − 2𝑦 = 6 let x = 4 and y = 3 𝑦 = 𝑥 + 1 let x = 4 and y = 3 2 1 3 = 2𝑥 + 1 3=2+1 3=3 ✓ 3(4) − 2(3) = 6 12 − 6 = 6 6=6 ✓ For Example: Solve the system of equations using the substitution method. 4𝑥 − 8𝑦 = 12 16𝑦 = 8𝑥 – 27 Step 1: Pick an equation and solve for a variable. I will pick the bottom equation and solve for x. I started by solving the bottom equation for y and I got a fraction. So, I abandoned that idea. 4𝑥 − 8𝑦 = 12 +8𝑦 + 8𝑦 4𝑥 = 8𝑦 + 12 4𝑥 4 = 8𝑦 4 + 12 4 𝑥 = 2𝑦 + 3 Step 2: Substitute into the other equation. 16y = 8x – 24 (replace 𝑥 𝑤𝑖𝑡ℎ 2𝑦 + 3) 16𝑦 = 8(2𝑦 + 3) − 27 Step 3: Solve this equation 16𝑦 = 16𝑦 + 24 − 27 16𝑦 = 16𝑦 − 3 −16𝑦 − 16𝑦 0 = −3 Step 4: I must stop at step 3 as I do not have an answer for x nor y after finishing step 3. When all the variables drop out and a false statement remains, we write either “dependent” or “no solution” for our answer. Answer: dependent or no solution Step 5: I need to graph and confirm the lines are parallel to check. I used a program to graph and here is the result. You can see the lines are parallel. The lines are very close to each other. You may not see them both if this is printed in black and white. Solving a system of equations by the elimination method. Step 1: Write both equation in the 𝐴𝑥 + 𝐵𝑦 = 𝐶 𝑓𝑜𝑟𝑚 Step 2: Determine which variable to eliminate. Make that variable have the same coefficient with opposite signs in the equations. Step 3: Line up the new equations and combine to drop out the variable selected in step 2 and solve for the other variable. Step 4: • If step 3 gives an answer for a single variable, substitute that answer into either equation to solve for the remaining variable and write the solution. • If all the variables drop out and a false statement remains state that there is no solution or that the system is inconsistent • If all the variables drop out and a true statement remains, state that the system is dependent and write your answer using set builder notation. {(𝑥, 𝑦)|𝑤𝑟𝑖𝑡𝑒 𝑒𝑖𝑡ℎ𝑒𝑟 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 ℎ𝑒𝑟𝑒} Step 5: Check • • • If there is a single solution substitute the solution into both equations to check. If the system is inconsistent graph to confirm the lines are parallel If the system is dependent graph to confirm the lines are the same Solve the system of equations using the Elimination method: 3𝑥 − 2𝑦 = 13 2𝑥 + 6𝑦 = 16 Step 1: Write both equations in the form: 𝐴𝑥 + 𝐵𝑦 = 𝐶 This is done for me. Both equations are in the proper form. Step 2: Determine which variable to eliminate. I will eliminate the y’s by making the coefficient of the top 𝑦 𝑎 “ − 6”. In this way the coefficient of the 𝑦’𝑠 will be opposites and they will cancel. −6 𝑎𝑛𝑑 6 Change top equation to make elimination possible: 3(3𝑥 − 2𝑦 = 13) 9𝑥 − 6𝑦 = 39 Step 3: Replace the top equation and add the equations together to cancel the 𝑦’𝑠. 9𝑥 − 6𝑦 = 39 2𝑥 + 6𝑦 = 16 11𝑥 = 55 𝑥 = 5 Step 4: Substitute 𝑥 = 5 into either equation to solve for 𝑦. I will choose the bottom equation as it does not have any negative signs. 2𝑥 + 6𝑦 = 16 2(5) + 6𝑦 = 16 10 + 6𝑦 = 16 6𝑦 = 6 𝑦=1 Answer (5,1) Step 5: Check 3𝑥 − 2𝑦 = 13 let x = 5 and y = 1 3(5) − 2(1) = 13 15 − 2 = 13 13 = 13 ✓ 2𝑥 + 6𝑦 = 16 let x = 5 and y = 1 2(5) + 6(1) = 16 10 + 6 = 16 16 = 16 ✓