 Download Download Presentation Solving Systems of Linear Equations

# Solving Systems of Linear Equations

Download Presentation ## Solving Systems of Linear Equations

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1. Solving Systems of Linear Equations Addition (Elimination) Method Tutorial 14c

2. 3 Methods to Solve • There are 3 methods that we can use to solve systems of linear equations. • Solve by the Graphing Method • Solve by the Substitution Method • Solve by the Addition (Elimination) Method

3. Elimination Method:Using Addition & Subtraction • In systems of equations where the coefficient of the x or y terms are opposites of each other, you can solve the system by adding the equations together. • Adding them together causes one of the variables to be eliminated.

4. Elimination Method:Using Addition Use elimination to solve the system of equations: x – 3y = 7 and 3x + 3y = 9. • Check to see if 2 like-terms have opposite coefficients: The 1st equation has a –3y and the 2nd has a +3y. • Therefore, add the two equations. x – 3y = 7 4 – 3y = 7 – 3y = 3 y = -1 x – 3y = 7 • Substitute 4 for x in either original equation and then solve for the y coordinate. + 3x + 3y = 9 4x = 16 x = 4 • Step 4 is to check the answer. Click here!

5. Elimination Method:Using Addition Use elimination to solve the system of equations: x – 3y = 7 and 3x + 3y = 9. • The last step is to check your solution by substituting both values, x = 4 & y = -1, in both equations. x – 3y = 7 4 –3(-1) = 7 4 – (-3) = 7 7 = 7  3x + 3y = 9 3(4) +3(-1) = 9 12 + (-3) = 9 9 = 9  The solution of the system is (4, -1).

6. Elimination Method:Using Multiplication with Addition • Some systems of equations cannot be solved by simply adding the equations. • In many cases one or both equations must first be multiplied by a number before the system can be solved by elimination. • Consider the following examples.

7. Elimination Method:Using Multiplication with Addition Use elimination to solve the system of equations: -x + 5y = -2 and 2x + 5y = 4. • Check to see if 2 like-terms have opposite coefficients: No terms have opposite coefficients. However, each equation has a 5y! Therefore, Multiply one of the equations by –1 to get a 5y and a –5y. -1( ) ( )-1 -x + 5y = -2 x – 5y = 2 • Then add the equations. -x + 5y = -2 -2 + 5y = -2 5y = 0 y = 0 x – 5y = 2 • Substitute 2 for x in either original equation and then solve for the y coordinate. + 2x + 5y = 4 3x = 6 x = 2 • Step 4 is to check the answer. Click here!

8. Elimination Method:Using Multiplication with Addition Use elimination to solve the system of equations: -x + 5y = -2 and 2x + 5y = 4. • The last step is to check your solution by substituting both values, x = 2 & y = 0, in both equations. -x + 5y = -2 -2 +5(0) = -2 -2 + 0 = -2 -2 = -2  2x + 5y = 4 2(2) +5(0) = 4 4 + 0 = 4 4 = 4  The solution of the system is (2, 0).

9. Elimination Method:Using Multiplication with Addition Use elimination to solve the system of equations: x + 10y = 3 and 4x + 5y = 5. • Check to see if 2 like-terms have opposite coefficients: No terms have opposite coefficients Therefore, multiply the first equation by -4 -4( ) ( )-4 x + 10y = 3 -4x – 40y = -12 • Then add the equations. • Substitute 1/5 for y in either original equation and then solve for the x coordinate. x + 10y = 3 x + 10(1/5) = 3 x + 2 = 3 x = 1 -4x – 40y = -12 +4x + 5y = 5 35y = 7 y =1/5 • Step 4 is to check the answer. Click here!

10. Elimination Method:Using Multiplication with Addition Use elimination to solve the system of equations: x + 10y = 3 and 4x + 5y = 5. • The last step is to check your solution by substituting both values, x = 1 & y = 1/5, in both equations. x + 10y = 3 1 +10(1/5) = 3 1 + 2 = 3 3 = 3  4x + 5y = 5 4(1) +5(1/5) = 5 4 + 1 = 5 5 = 5  The solution of the system is (1, 1/5)

11. Elimination Method:Using Multiplication with Addition Use elimination to solve the system of equations: 2x – 3y = 8 and 3x – 7y = 7. -3( ) ( )-3 2x – 3y = 8 • Check to see if 2 like-terms have opposite coefficients:No terms have opposite coefficients. Therefore, multiply the first equation by –3 and the second equation by 2. -6x + 9y = -24 2( ) ( )2 3x – 7y = 7 6x – 14y = 14 • Then add the equations. 2x – 3y = 8 2x – 3(2) = 8 2x – 6 = 8 2x = 14 x = 7 -6x + 9y = -24 • Substitute 2 for y in either original equation and then solve for the x coordinate. + 6x – 14y = 14 -5y = -10 y = 2 • Step 4 is to check the answer. Click here!

12. Elimination Method:Using Multiplication with Addition Use elimination to solve the system of equations: 2x – 3y = 8 and 3x – 7y = 7. • The last step is to check your solution by substituting both values, x = 7 & y = 2, in both equations. 2x – 3y = 8 2(7) –3(2) = 8 14 – 6 = 8 8 = 8  3x – 7y = 7 3(7) –7(2) = 7 21 – 14 = 7 7 = 7  The solution of the system is (7, 2).

13. Elimination Method:Using Multiplication with Addition – In Review Steps for the Multiplication with Addition Method • Check to see if any like-terms have opposite coefficients. If necessary, multiply each side of either or both equations by numbers that will make opposite coefficients for one variable. • Add to eliminate on of the variables. Solve the resulting equation. • Substitute the known value of one variable in either of the original equations of the system. Solve for the other variable. • Check the answer by substituting both values in both equations of the system.

14. The End Time to move on to the assignment or the next lesson