# Math lecture 6 (System of Linear Equations)

Trainer at Training Pakistan Private Limited um Training Pakistan Private Limited
8. Feb 2014
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### Math lecture 6 (System of Linear Equations)

• 1. System of Linear Equations A System of Equations is when we have two or more equations working together  Solve by Elimination  Solve by Substitution Solve by Elimination: Two equations in Two Variables:  x+y=6  -3x + y = 2 The two equations are shown on this graph: Our task is to find where the two lines cross. OK, we can see where they cross, but let's solve it using Algebra! Hmmm ... how should we solve this? There can be many ways! In this case both equations have "y" so I will try subtracting the second equation from the first: x + y - (-3x + y) = 6 - 2 Which simplifies to: x + y + 3x - y = 6 - 2 4x = 4 x=1 So now we know that x=1 is on both lines. And we can find the matching value of y using either of the two original equations (because we know they have the same value at x=1). Let's use the first one (you can try the second one yourself): x+y=6 1+y=6 y=5 And the solution is: x = 1 and y = 5 And the graph shows us we are right!
• 2. Solving By Substitution: 3 equations in 3 variables OK! Let's move to a longer example: 3 equations in 3 variables. This is not hard to do... it just takes a long time to do it! Example:    x+z=6 z - 3y = 7 2x + y + 2z = 11 I will try to line up my variables neatly, or I may lose track of what I am doing: x + z = 6 - + z = 7 + 2x 3y y + 2z = 11 One again, I can start with any equation and any variable. I will use the first equation and the variable "x". Write one of the equations so it is in the style "variable = ...": x = 6-z - + z = 7 + 2x 3y y + 2z = 11 Now replace "x" with "6 - z" in the other equations: (Luckily there is only one other equation with x in it) x = 6-z - + z = 7 + 2(6-z) 3y y + 3z = 15 Solve using the usual algebra methods: 2(6-z) + y + 3z = 15 simplifies to y + z = 3: x = - 6-z 3y + z = 7 y + z = 3 Good. We have made some progress, but not there yet. Now repeat the process, but just for the last 2 equations.
• 3. Write one of the equations so it is in the style "variable = ...": I will choose the last equation and the variable z: x = - 3y + z = z 6-z 7 = 3-y Now replace "z" with "3 - y" in the other equation: x = - 3y + 3-y = z 6-z 7 = 3-y Solve using the usual algebra methods: -3y + (3-y) = 7 simplifies to -4y = 4, or in other words y = -1 x = y = z 6-z -1 = 3-y Almost Done! Knowing that y = -1 we can calculate that z = 3-y = 4: x = y 6-z = z -1 = 4 And knowing that z = 4 we can calculate that x = 6-z = 2: x = = y z And the answer is: 2 -1 = 4 x = 2, y = -1 and z = 4
• 4. Cramer's Rule Given a system of linear equations, Cramer's Rule is a handy way to solve for just one of the variables without having to solve the whole system of equations. They don't usually teach Cramer's Rule this way, but this is supposed to be the point of the Rule: instead of solving the entire system of equations, you can use Cramer's to solve for just one single variable. Let's use the following system of equations: 2x + y + z = 3 x– y–z=0 x + 2y + z = 0 We have the left-hand side of the system with the variables (the "coefficient matrix") and the right-hand side with the answer values. Let D be the determinant of the coefficient matrix of the above system, and let Dx be the determinant formed by replacing the x-column values with the answer-column values: system of equations coefficient matrix's determinant answer column Dx: coefficient determinant with answer-column values in x-column 2x + 1y + 1z = 3 1x – 1y – 1z = 0 1x + 2y + 1z = 0 Similarly, Dy and Dz would then be: Evaluating each determinant, we get: Copyright © Elizabeth Stapel 2004-2011 All Rights Reserved
• 5. Cramer's Rule says that x = Dx ÷ D, y = Dy ÷ D, and z = Dz ÷ D. That is: x = 3/3 = 1, y = –6/3 = –2, and z = 9/3 = 3 That's all there is to Cramer's Rule. To find whichever variable you want (call it "ß" or "beta"), just evaluate the determinant quotient Dß ÷ D. (Please don't ask me to explain why this works. Just trust me that determinants can work many kinds of magic.)  Given the following system of equations, find the value of z. 2x + y + z = 1 x – y + 4z = 0 x + 2y – 2z = 3 To solve only for z, I first find the coefficient determinant. Then I form Dz by replacing the third column of values with the answer column: Then I form the quotient and simplify: z=2 The point of Cramer's Rule is that you don't have to solve the whole system to get the one value you need. This saved me a fair amount of time on some physics tests. I forget what we were working on (something with wires and currents, I think), but Cramer's Rule was so much faster than any other
• 6. solution method (and God knows I needed the extra time). Don't let all the subscripts and stuff confuse you; the Rule is really pretty simple. You just pick the variable you want to solve for, replace that variable's column of values in the coefficient determinant with the answer-column's values, evaluate that determinant, and divide by the coefficient determinant. That's all there is to it. Almost. What if the coefficient determinant is zero? You can't divide by zero, so what does this mean? I can't go into the technicalities here, but "D = 0" means that the system of equations has no unique solution. The system may be inconsistent (no solution at all) or dependent (an infinite solution, which may be expressed as a parametric solution such as "(a, a + 3, a – 4)"). In terms of Cramer's Rule, "D = 0" means that you'll have to use some other method (such as matrix row operations) to solve the system. If D = 0, you can't use Cramer's Rule. Quadratic Equations 1. Direct Method In disguise In standard form a, b and c x2 = 3x -1 x2 - 3x + 1 = 0 a=1, b=-3, c=1 2(x2 - 2x) = 5 2x2 - 4x - 5 = 0 a=2, b=-4, c=-5 x(x-1) = 3 x -x-3=0 a=1, b=-1, c=-3 5 + 1/x - 1/x2 = 0 5x2 + x - 1 = 0 a=5, b=1, c=-1 2 2. By Using Quadratic Formula
• 7. 3. By Using Factorizing Method Example The factors of x2 + 3x - 4 are: (x+4) and (x-1) Why? Well, let us multiply them to see: (x+4)(x-1) = x(x-1) + 4(x-1) = x2 - x + 4x - 4 = x2 + 3x - 4 4. By Completing the Square Steps Now we can solve a Quadratic Equation in 5 steps:  Step 1 Divide all terms by a (the coefficient of x2).  Step 2 Move the number term (c/a) to the right side of the equation.  Step 3 Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation. We now have something that looks like (x + p) 2 = q, which can be solved rather easily:  Step 4 Take the square root on both sides of the equation.  Step 5 Subtract the number that remains on the left side of the equation to find x.
• 8. Example 1: Solve x2 + 4x + 1 = 0 Step 1 can be skipped in this example since the coefficient of x 2 is 1 Step 2 Move the number term to the right side of the equation: x2 + 4x = -1 Step 3 Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation. (b/2)2 = (4/2)2 = 22 = 4 x2 + 4x + 4 = -1 + 4 (x + 2)2 = 3 Step 4 Take the square root on both sides of the equation: x + 2 = ±√3 = ±1.73 (to 2 decimals) Step 5 Subtract 2 from both sides: x = ±1.73 – 2 = -3.73 or -0.27 Example 2: Solve 5x2 – 4x – 2 = 0 Step 1 Divide all terms by 5 x2 – 0.8x – 0.4 = 0 Step 2 Move the number term to the right side of the equation: x2 – 0.8x = 0.4 Step 3 Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation: (b/2)2 = (0.8/2)2 = 0.42 = 0.16 x2 – 0.8x + 0.16 = 0.4 + 0.16 (x – 0.4)2 = 0.56 Step 4 Take the square root on both sides of the equation: x – 0.4 = ±√0.56 = ±0.748 (to 3 decimals) Step 5 Subtract (-0.4) from both sides (in other words, add 0.4): x = ±0.748 + 0.4 = -0.348 or 1.148