Solving Systems of Equations

Slides:



Advertisements
Similar presentations
Solving Systems of Linear Equations in Three Variables.
Advertisements

8.3 Solving Systems of Linear Equations by Elimination
Table of Contents Recall that to solve the linear system of equations in two variables... we needed to find the values of x and y that satisfied both equations.
4.2 Systems of Linear Equations in Three Variables BobsMathClass.Com Copyright © 2010 All Rights Reserved. 1 The Graph of a Three Variable Equation Recall.
3.2 Solving Systems Algebraically 2. Solving Systems by Elimination.
Solve each with substitution. 2x+y = 6 y = -3x+5 3x+4y=4 y=-3x- 3
Solving a System of Equations by ELIMINATION. Elimination Solving systems by Elimination: 1.Line up like terms in standard form x + y = # (you may have.
Table of Contents Solving Linear Systems of Equations - Addition Method Recall that to solve the linear system of equations in two variables... we need.
Lesson 6-3 – Solving Systems Using Elimination
Solving Systems of Equations: Elimination Method.
Identifying Solutions
Solving Systems of Equations
9.2 Solving Systems of Linear Equations by Addition BobsMathClass.Com Copyright © 2010 All Rights Reserved. 1 Step 1.Write both equations in the form Ax.
6-3: Solving systems Using Elimination
Chapter 4 Section 3 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.
Solving Systems of Linear Equations by Elimination Solve linear systems by elimination. Multiply when using the elimination method. Use an alternative.
Goal: Solve a system of linear equations in two variables by the linear combination method.
Thinking Mathematically Systems of Linear Equations.
Solving Linear Systems of Equations - Addition Method Recall that to solve the linear system of equations in two variables... we need to find the values.
Bell Ringer 2x – 3y = 4 5x + 3y = 10. HW Check Check elimination part 1 practice.
SOLVING SYSTEMS ALGEBRAICALLY SECTION 3-2. SOLVING BY SUBSTITUTION 1) 3x + 4y = 12 STEP 1 : SOLVE ONE EQUATION FOR ONE OF THE VARIABLES 2) 2x + y = 10.
Chapter 4 Section 3. Objectives 1 Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solving Systems of Linear Equations by Elimination Solve linear.
3-2 Solving Linear Systems Algebraically Objective: CA 2.0: Students solve system of linear equations in two variables algebraically.
7.4. 5x + 2y = 16 5x + 2y = 16 3x – 4y = 20 3x – 4y = 20 In this linear system neither variable can be eliminated by adding the equations. In this linear.
Do Now (3x + y) – (2x + y) 4(2x + 3y) – (8x – y)
Solving Systems of Equations by Elimination Solve by Elimination Solve by Multiplying then Elimination.
6.2 Solve a System by Using Linear Combinations
Section 3.5 Solving Systems of Linear Equations in Two Variables by the Addition Method.
SOLVING SYSTEMS USING ELIMINATION 6-3. Solve the linear system using elimination. 5x – 6y = -32 3x + 6y = 48 (2, 7)
 Recall that when you wanted to solve a system of equations, you used to use two different methods.  Substitution Method  Addition Method.
7-3: Solving Systems of Equations using Elimination
Systems of Equations By Substitution and Elimination.
Elimination Method - Systems. Elimination Method  With the elimination method, you create like terms that add to zero.
Solving a System of Equations by ELIMINATION. Elimination Solving systems by Elimination: 1.Line up like terms in standard form x + y = # (you may have.
Warm Up Find the solution to linear system using the substitution method. 1) 2x = 82) x = 3y - 11 x + y = 2 2x – 5y = 33 x + y = 2 2x – 5y = 33.
7.5 Solving Systems of Linear Equations by Elimination.
6) x + 2y = 2 x – 4y = 14.
WARM UP Find each equation, determine whether the indicated pair (x, y) is a solution of the equation. 2x + y = 5; (1, 3) 4x – 3y = 14; (5, 2)
Objective I can solve systems of equations using elimination with addition and subtraction.
Chapter 2: Equations of Order One
Solving Systems of Equations using Elimination
Solve Systems of Equations by Elimination
3. Further algebraic skills and techniques
3.2 Solve Linear Systems Algebraically
Stand Quietly.
3-2: Solving Linear Systems
Solving Systems Using Elimination
12 Systems of Linear Equations and Inequalities.
Matrix Algebra.
3-2: Solving Systems of Equations using Elimination
3.3: Solving Systems of Equations using Elimination
REVIEW: Solving Linear Systems by Elimination
Systems of Equations and Inequalities
Simultaneous Equations
3-2: Solving Systems of Equations using Elimination
Solve Linear Equations by Elimination
Linear Algebra Lecture 3.
Objectives Solve systems of linear equations in two variables by elimination. Compare and choose an appropriate method for solving systems of linear equations.
3-2: Solving Linear Systems
3-2: Solving Systems of Equations using Elimination
Solving a System of Equations in Two Variables by the Addition Method
Warm Up 12/3/2018 Solve by substitution.
3-2: Solving Systems of Equations using Elimination
Solving Systems of Equations using Elimination
3-2: Solving Linear Systems
Example 2B: Solving Linear Systems by Elimination
3-2: Solving Linear Systems
The Substitution Method
Systems of three equations with three variables are often called 3-by-3 systems. In general, to find a single solution to any system of equations,
Solving Linear Equations
Presentation transcript:

Solving Systems of Equations 259 Lecture 12 Spring 2017 Solving Systems of Equations

A System of Equations Consider the system of two equations in two unknowns: x + 2y = 1 (1) 3x + 4y = -1 (2) Recall from algebra class that two ways to solve a problem like this are substitution and elimination.

Example 1 (Substitution) Solve (1) or (2) for one variable in terms of the other and substitute for that variable in the other equation. x + 2y = 1 (1) 3x + 4y = -1 (2) Solving (1) for x yields x = 1 – 2y (3)

Example 1 (cont.) Using (3) we can substitute 1 – 2y for x in (2): 3(1 – 2y) + 4y = -1 3 – 6y + 4y = -1 -2y = -4 y = 2 (4) (4) in (3) implies x = 1 – 2(2) = -3. Check (x, y) = (-3, 2) solves (1), (2).

Example 2 (Elimination) x + 2y = 1 (1) 3x + 4y = -1 (2) Multiply both sides of (1) or (2) by a non-zero constant to get coefficients in front of one variable that can be added to get zero. Then add to eliminate that variable and solve for the remaining variable.

Example 2 (cont.) Multiplying (1) by -2 on both sides, we get an equivalent system; -2x - 4y = -2 (3) 3x + 4y = -1 (2) Adding (3) to (2) gives an equation involving only x which can be solved for x. x = -3 (4)

Example 2 (cont.) Substituting x = -3 from (4) back into either (1) or (2) (why these?), we can solve for y. In this case, we choose (1): x + 2y = 1 -3 + 2y = 1 2y = 4 y = 2 Again, we find (x, y) = (-3, 2) and would need to check it works in (1),(2)!

Solving a System of Equations with Technology How could we use technology to help us solve system (1), (2). One way might be to use a calculator with this capability. How about Mathematica or Excel?

Example 3: Solving a System of Equations in Excel We can use the Solver to solve a system of equations! Let’s try with (1), (2). To do so, we need to choose one target cell (i.e. objective cell) and specify appropriate constraints. Think of the system as a*x + b*y = e c*x + d*y = f, with a=1, b=2, c=3, d=4, e=1, and f=-1. Then the objective can be to set ax+by = 1, subject to the constraints e = 1 and f = -1.

Example 3: Solving a System of Equations in Excel

Example 4: Solving a System of Equations in Mathematica For Mathematica, we use the Solve command: Solve[{x+2y==1, 3x+4y==-1},{x,y}]

Example 4: Solving a System of Equations in Mathematica We can also solve a system like this in general: To solve the system a*x + b*y = e c*x + d*y = f, use the command: Solve[{a*x+b*y==e, c*x+d*y==f},{x,y}]

Example 4: Solving a System of Equations in Mathematica We find that for the system a*x + b*y = e c*x + d*y = f, the solution is: x = (bf – de)/(ad – bc) y = (af – ce)/(ad – bc) Try with coefficients and right-hand side from (1), (2) to see if we get the same solution! For what choices of a, b, c, d, e and f, do we get a solution to this system?

Systems of Equations in General We can generalize the problem discussed above to a system of n equations in n unknowns! For example, here is a system of three equations in three unknowns: ax + by + cz = j dx + ey + fz = k gx + hy + iz = l How could we solve this? By hand? Calculator? With Mathematica? With Excel? Other? When are we guaranteed a solution?

Example 5: Mathematica Solution for 3 x 3 Case

Example 5 (cont.)

References The idea of how to use the Solver comes from a web supplement for Finite Mathematics (7th ed) by Margaret Lial et al. found at this link: http://web.archive.org/web/20141017004542/http://wps.aw.com/aw_lial_finitemath_7/0,1769,12520-,00.html