Systems of Equations In Electronics and with Matrices Back to Activities page This site explores the solution of two simultaneous equations for two unknowns. There are three different manners in which you may want to explore this site depending on your interest and needs. They can be interchanged or used by themselves. Electronics Application of Simultaneous Linear Equations and a solver Systems of Equations in terms of x and y and a SOLVER Using Matrices and Cramer's Rule for Solving a System of Equations Just the Solver Illinois Learning Standards Addressed

Electronics Application of Simultaneous Linear Equations Linear Simultaneous Equations are frequently encountered in electronics and technology. For example, when using Kirchhoff's Laws (KCL & KVL) for analyzing a circuit you eventually arrive at a system of loop equations that must be solved for CURRENTS I1 and I2. These two unknowns are solution to the network that is established by Kirchhoff's Laws. Links to Kirchhoff's Laws Kirchhoff's Voltage Law Active Experiment 1 Active Experiment 2 Active Experiment 3 Active Experiment 4 Active Experiment 5 As per the diagrams to the right the SUM of the Voltage Terms in the loops should be ZERO. The three different currents are I1 , I2 and I3 with I3 = I1 + I2. Steps to solving Kirchhoff Law Equations. 1. Consider each loop as a seperate equation with I1 &  I2 as variables to solve for. 2. Solve simultaneous equations for the unknown quantities. I1 & I2 Solving for the variables is described below.   Loop 1 - Loop 2 = 0         A System of Equations has one solution. This solution is a set of ordered pairs of the form (x,y) that satisfies two or more equations. A System of Equations is generally of the (standard) form. It is helpful to arrange (rewrite the equations of the system) to look like the standard form: AX + BY = C DX + EY = F and one ordered pair (x,y) satisfies these two equations. Or in electronics AI1 + BI2 = C DI1 + EI2 = F and one ordered pair (I1,I2) satisfies these two equations. Example: Eq 1. ) 3I1 + I2 = 14 Eq 2. ) 2I1 + 3I2 = 7 The next step is to perform operations on one equation in the system to eliminate one variable and solve for the other. Using the example, we want to find a way to make the coefficient of either I1 or I2 equal so that we can isolate the other variable and solve for it. Try multiplying the entire Eq 1. by "3" so that we have Eq 1. ) (3)3I1+ (3)I2 = (3)14 Eq 2. ) 2I1 + 3I2 = 7 Now we have Eq 1. ) 9I1+ 3I2 = 42 Eq 2. ) 2I1 + 3I2 = 7 We can then subtract Eq. 2 from Eq. 1 and solve for I1. 7I1 = 35 and so, I1 = 5 Insert this I1 = 5 back into either Eq. 1 or Eq. 2 and then you will be able to solve for X. Let's Insert it into Eq. 2. Eq 2. ) 2I1 + 3I2 = 7 Eq 2. ) 2(5) + 3I2 = 7 Eq 2. ) 10 + 3I2 = 7 Eq 2. ) 3I2 = 7 - 10 Eq 2. ) 3I2 = -3 Eq 2. ) I2 = -1 top   Interactive Solver for I1 and I2 Eq 1. ) 3I1 + I2 = 14 Eq 2. ) 2I1 + 3I2 = 7 Try these Equations in the Solver below and hit the "Solve" Button. I1+ I2 = I1+ I2 = I1 = I2 = top Systems of Equations in terms of x and y and a SOLVER A System of Equations has one solution. This solution is a set of ordered pairs of the form (x,y) that satisfies two or more equations. A System of Equations is generally of the (standard) form. It is helpful to arrange (rewrite the equations of the system) to look like the standard form: AX + BY = C DX + EY = F and one ordered pair (x,y) satisfies these two equations. One way to solve this system of equations is to graph both equations (make sure to set y = equation) and observe the intersection of the lines that are plotted. This intersection point is the ordered pair that represents the solution. You can try this by hand on paper or by using a graphing utility like a graphing calculator, or even a tool available on the web. Try this example by graphing it: Eq 1. ) X+ 2Y = 10 Eq 2. ) 2X - Y = 10 Another method is to perform operations on one equation in the system to eliminate one variable and solve for the other. Using the same example: Eq 1. ) X+ 2Y = 10 Eq 2. ) 2X - Y = 10 We want to find a way to make the coefficient of either x or y equal so that we can isolate the other variable and solve for it. Try multiplying the entire Eq 1. by "2" so that we have Eq 1. ) (2)X+ (2)2Y = (2)10 = Eq 1. ) 2X+ 4Y = 20 Now we have Eq 1. ) 2X+ 4Y = 20 Eq 2. ) 2X - Y = 10 We can then subtract them and solve for Y. 5Y = 10 and so, Y = 2 Insert this Y = 2 into either Equation and then you will be able to solve for X. Eq 2. ) 2X - Y = 10 Eq 2. ) 2X - (2) = 10 Eq 2. ) 2X = 12 Eq 2. ) X = 6 Does this match with what you have graphed? Try it out numerically by entering values here (remember to put "-" for a negative sign.) and then hit the "Solve" button. X + Y = X + Y = X = Y = top  Solver: Just Enter your Coefficient Values and hit solve X + Y = X + Y = X = Y = Matrices and Cramer's Rule for Solving a System of Equations Solutions to systems of equations of the form AX + BY = C DX + EY = F Can be found using the following: (CE - BF) X = (AE - BD) (AF - CD) Y = (AE - BD) This method is derived from the determinants of matrices that can be arranged from a system of equations Recall that a matrix A is of the form [A] = | a b | | c d | The value of the determinant of the 2 X 2 matrix is the difference of the diagonal products. a*d - c * b = determinant A To solve a system of the form AX + BY = C DX + EY = F 1.) Find the determinants of the following matrices. [A] = | a b | | d e | a*e - d * b = determinant A [Ax] = | c b | | f e | c * e - f * b = determinant Ax [Ay] = | a c | | d f | a*f - d * c = determinant Ay 2.) Use the determinants to find the solution if determinant A is not zero. X = determinant Ax determinant A Y = determinant Ay determinant A

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Email Jim Dildine
Date Last Modified: 8/31/99
James P. Dildine