Solving Systems of Linear Equations

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Directions: Solve the linear systems of equations by graphing. Use the graph paper from the table. Tell whether you think the problems have one solution,
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Presentation transcript:

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

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

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.

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!

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).

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.

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!

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).

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!

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)

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!

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).

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.

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