# Solving Systems of Equations Using Elimination

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Solving Systems of Equations Using Elimination
Conncections - Thursday, November 11 Unit D.5 Solving Systems of Equations Using Elimination Students will be able to solve systems of equations using elimination Unit D.5 - Solving Systems Using Elimination

Solving Systems - Elimination
Another way to solve systems of equations is with the elimination method. With elimination, you get rid of (eliminate)one of the variables by adding or subtracting equations. The elimination method is sometimes called the addition and subtraction method. FHS Systems of Equations

Using Elimination - Example 1
Use elimination to solve the system of equations. Step 1: Find the value of one variable. 3x + 2y = 4 The y-terms have opposite coefficients. 4x – 2y = -18 7x = -14 Add the equations to eliminate y. x = -2 First part of the solution. FHS Systems of Equations

Example 1 (cont.) Step 2: Substitute the x-value into one of the original equations to solve for y. 3(-2) + 2y = 4 Substitute -2 in for x. 2y = 10 y = 5 Second part of the solution. The solution is the ordered pair (-2, 5). FHS Systems of Equations

What if Elimination Does Not Work?
When you cannot eliminate one of the variables by just adding or subtracting the two equations, it is still possible to solve the system. Sometimes you can multiply one or both of the equations by some number that would make elimination possible. FHS Systems of Equations

Conncections - Thursday, November 11
Unit D.5 Example 2 - Using Elimination Use elimination to solve the system of equations. Step 1: To eliminate x, multiply both sides of the first equation by 2 and both sides of the second equation by –3. 2(3x + 5y) = 2(–16) –3(2x + 3y) = –3(–9) 6x + 10y = –32 –6x – 9y = 27 Add the equations. First part of the solution: y = –5 FHS Systems of Equations Unit D.5 - Solving Systems Using Elimination

Conncections - Thursday, November 11
Unit D.5 Example 2 (Cont.) Step 2: Substitute the y-value into one of the original equations to solve for x. 3x + 5(–5) = –16 3x – 25 = –16 3x = 9 x = 3 Second part of the solution The solution for the system is (3, –5). FHS Systems of Equations Unit D.5 - Solving Systems Using Elimination