# Solving Systems Using Elimination

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Solving Systems Using Elimination
When both linear equations of a system are in the form Ax + By = C, you can solve the system using elimination. You can add or subtract equations to eliminate a variable.

Check Solve the system using elimination. 5x – 6y = -32 3x + 6y = 48
First, eliminate one variable. Then, find the value of the eliminated variable by plugging x into one of the equations. 5x – 6y = -32 3x + 6y = 48 5x – 6y = -32 5(2) – 6y = -32 10 – 6y = -32 8x + 0y = 16 - 6y = -42 x = 2 y = 7 Since x = 2 and y = 7, the solution is (2,7). See if (2,7) makes the other equation true. Check 3x + 6y = 48 = 48 3(2) + 6(7) = 48 48 = 48

Check Solve the system using elimination. x + y = 12 x – y = 2
First, eliminate one variable. Then, find the value of the eliminated variable by plugging x into one of the equations. x + y = 12 x - y = 2 x + y = 12 (7) + y = 12 y = 5 2x + 0y = 14 x = 7 Since x = 7 and y = 5, the solution is (7,5). See if (7,5) makes the other equation true. Check x - y = 2 2 = 2 (7) - (5) = 2

Check Solve the system using elimination. -x + 2y = -1 x – 3y = -1
First, eliminate one variable. Then, find the value of the eliminated variable by plugging y into one of the equations. -x + 2y = -1 x – 3y = -1 -x + 2(2) = -1 -x + 4 = -1 -x = -5 0x – y = -2 x = 5 y = 2 Since x = 5 and y = 2, the solution is (5,2). See if (5,2) makes the other equation true. Check x – 3y = -1 5 – 6 = -1 (5) – 3(2) = -1 -1 = -1

Check Solve the system using elimination. 3x + y = 8 x – y = -12
First, eliminate one variable. Then, find the value of the eliminated variable by plugging x into one of the equations. 3x + y = 8 x - y = -12 3x + y = 8 3(-1) + y = 8 -3 + y = 8 4x + 0y = -4 y = 11 x = -1 Since x = -1 and y = 11, the solution is (-1,11). See if (-1,11) makes the other equation true. Check x - y = -12 -12 = -12 (-1) – (11) = -12

Since 0 = 0 is a true statement, there are infinitely many solutions.
Solve the system using elimination. x + 4y = 1 3x + 12y = 3 First, eliminate one variable. -3( ) ( ) -3 Multiply by -3 x + 4y = 1 3x + 12y = 3 -3x – 12y = -3 3x + 12y = 3 Since 0 = 0 is a true statement, there are infinitely many solutions. 0x + 0y = 0 0 = 0 TRUE!

Check Solve the system using elimination. 3x + 4y = -10 5x – 2y = 18
First, eliminate one variable. Then, find the value of the eliminated variable by plugging x into one of the equations. 3x + 4y = -10 5x – 2y = 18 3x + 4y = -10 2( ) ( ) 2 Multiply by 2 3(2) + 4y = -10 3x + 4y = -10 10x – 4y = 36 6 + 4y = -10 4y = -16 13x + 0y = 26 y = -4 x = 2 Since x = 2 and y = -4, the solution is (2,-4). See if (2,-4) makes the other equation true. Check 5x – 2y = 18 = 18 5(2) – 2(-4) = 18 18 = 18

Check Solve the system using elimination. 2x – y = 6 -3x + 4y = 1
First, eliminate one variable. Then, find the value of the eliminated variable by plugging x into one of the equations. 4( ) ( ) 4 Multiply by 4 2x – y = 6 -3x + 4y = 1 2x – y = 6 2(5) - y = 6 8x – 4y = 24 -3x + 4y = 1 10 - y = 6 -y = -4 5x + 0y = 25 y = 4 x = 5 Since x = 5 and y = 4, the solution is (5,4). See if (5,4) makes the other equation true. Check -3x + 4y = 1 = 1 -3(5) + 4(4) = 1 1 = 1

Since 0 = 0 is a true statement, there are infinitely many solutions.
Solve the system using elimination. 4x – 2y = 6 -2x + y = -3 First, eliminate one variable. 4x – 2y = 6 -2x + y = -3 2( ) ( ) 2 Multiply by 2 4x – 2y = 6 -4x + 2y = -6 Since 0 = 0 is a true statement, there are infinitely many solutions. 0x + 0y = 0 0 = 0 TRUE!

Check Solve the system using elimination. 2x – 3y = 4 3x + 2y = 6
First, eliminate one variable. Then, find the value of the eliminated variable by plugging x into one of the equations. 2( ) ( ) 2 Multiply by 2 2x – 3y = 4 3x + 2y = 6 2x – 3y = 4 3( ) ( ) 3 Multiply by 3 2(2) + 3y = 4 4x – 6y = 8 9x + 6y = 18 4 + 3y = 4 3y = 0 13x + 0y = 26 y = 0 x = 2 Since x = 2 and y = 0, the solution is (2,0). See if (2,0) makes the other equation true. Check 3x + 2y = 6 6 + 0 = 6 3(2) + 2(0) = 6 6 = 6

no solution. Since 0 = 8 is a false statement, there is
Solve the system using elimination. x – 3y = 2 -2x + 6y = 4 First, eliminate one variable. 2( ) ( ) 2 Multiply by 2 x – 3y = 2 -2x + 6y = 4 2x – 6y = 4 -2x + 6y = 4 Since 0 = 8 is a false statement, there is no solution. 0x + 0y = 8 0 = 8 FALSE!

Check Solve the system using elimination. x + y = 6 x + 3y = 10
First, eliminate one variable. Then, find the value of the eliminated variable by plugging y into one of the equations. -1( ) ( ) Multiply by -1 x + y = 6 x + 3y = 10 x + y = 6 x + (2) = 6 -x - y = -6 x + 3y = 10 x = 4 0x + 2y = 4 y = 2 Since x = 4 and y = 2, the solution is (4,2). See if (4,2) makes the other equation true. Check x + 3y = 10 4 + 6 = 10 (4) + 3(2) = 10 10 = 10

Check Solve the system using elimination. 5x + 7y = -1 4x – 2y = 22
First, eliminate one variable. Then, find the value of the eliminated variable by plugging x into one of the equations. 2( ) ( ) 2 Multiply by 2 5x + 7y = -1 4x – 2y = 22 5x + 7y = -1 7( ) ( ) 7 Multiply by 7 5(4) + 7y = -1 10x + 14y = -2 28x – 14y = 154 20 + 7y = -1 7y = -21 38x + 0y = 152 y = -3 x = 4 Since x = 4 and y = -3, the solution is (4,-3). See if (4,-3) makes the other equation true. Check 4x - 2y = 22 = 22 4(4) – 2(-3) = 22 22 = 22

no solution. Since 0 = -9 is a false statement, there is
Solve the system using elimination. 2x – 3y = 6 6x – 9y = 9 First, eliminate one variable. -3( ) ( ) Multiply by -3 2x – 3y = 6 6x – 9y = 9 -6x + 9y = -18 6x – 9y = 9 Since 0 = -9 is a false statement, there is no solution. 0x + 0y = -9 0 = -9 FALSE!

Solving Systems Using Elimination
Time to Practice!

Solving Systems Using Elimination
2x – 3y = 5 x + 2y = -1 x + y = 10 x – y = 2 -x + 4y = 12 2x – 3y = 6 4x + 2y = 1 2x + y = 4 (1,-1) 2x + 5y = 20 3x – 10y = 37 3x + 2y = -19 x – 12y = 19 -2x + y = -1 6x – 3y = 3 4x + y = 8 -3x – y = 0 (11,-2/5) (6,4) (-5,-2) (12,6) infinitely many solutions no solution (8,-24)