Dr Fowler CCM Solving Systems of Equations By

Dr. Fowler CCM Solving Systems of Equations By Elimination – Easier

Solving a system of equations by elimination using addition and subtraction. Step 1: Put the equations in Standard Form. Step 2: Determine which variable to eliminate. Standard Form: Ax + By = C Look for variables that have the same coefficient. Step 3: Add or subtract the equations. Solve for the variable. Step 4: Plug back in to find the other variable. Substitute the value of the variable into the equation. Step 5: Check your solution. Substitute your ordered pair into BOTH equations.

1) Solve the system using elimination. x+y=5 3 x – y = 7 Step 1: Put the equations in Standard Form. Step 2: Determine which variable to eliminate. Step 3: Add or subtract the equations. They already are! The y’s have the same coefficient. Add to eliminate y. x+ y=5 (+) 3 x – y = 7 4 x = 12 x=3

1) Solve the system using elimination. x+y=5 3 x – y = 7 Step 4: Plug back in to find the other variable. Step 5: Check your solution. x+y=5 (3) + y = 5 y=2 (3, 2) (3) + (2) = 5 3(3) - (2) = 7 The solution is (3, 2). What do you think the answer would be if you solved using substitution?

EXAMPLE #2: 5 x + 3 y = 11 5 x = 2 y + 1 STEP 1: Write both equations in Ax + By = C form. 5 x + 3 y =11 5 x - 2 y =1 STEP 2: Use subtraction to eliminate 5 x. 5 x + 3 y =11 5 x + 3 y = 11 -(5 x - 2 y =1) -5 x + 2 y = -1 Note: the (-) is distributed. STEP 3: Solve for the variable. 5 x + 3 y =11 -5 x + 2 y = -1 5 y =10 y=2

5 x + 3 y = 11 5 x = 2 y + 1 STEP 4: Solve for the other variable by substituting y = 2 into either equation. 5 x + 3 y =11 5 x + 3(2) =11 5 x + 6 =11 5 x = 5 x=1 The solution to the system is (1, 2).

3) Solve the system using elimination. 4 x + y = 7 4 x – 2 y = -2 Step 1: Put the equations in Standard Form. They already are! Step 2: Determine which variable to eliminate. The x’s have the same coefficient. Step 3: Add or subtract the equations. Subtract to eliminate x. 4 x + y = 7 (-) 4 x – 2 y = -2 3 y = 9 Remember to “keep-changey=3 change”

3) Solve the system using elimination. 4 x + y = 7 4 x – 2 y = -2 Step 4: Plug back in to find the other variable. Step 5: Check your solution. 4 x + y = 7 4 x + (3) = 7 4 x = 4 x=1 (1, 3) 4(1) + (3) = 7 4(1) - 2(3) = -2

4) Solve the system using elimination. y = 7 – 2 x 4 x + y = 5 Step 1: Put the equations in Standard Form. 2 x + y = 7 4 x + y = 5 Step 2: Determine which variable to eliminate. The y’s have the same coefficient. Step 3: Add or subtract the equations. Subtract to eliminate y. 2 x + y = 7 (-) 4 x + y = 5 -2 x = 2 x = -1

4) Solve the system using elimination. y = 7 – 2 x 4 x + y = 5 Step 4: Plug back in to find the other variable. Step 5: Check your solution. y = 7 – 2 x y = 7 – 2(-1) y=9 (-1, 9) (9) = 7 – 2(-1) 4(-1) + (9) = 5

Elimination using Addition 5) Solve the system x - 2 y = 5 2 x + 2 y = 7 Lets add both equations to each other

Elimination using Addition 5) Solve the system x - 2 y = 5 + 2 x + 2 y = 7 Lets add both equations to each other NOTE: We use the Elimination Method, if we can immediately cancel out two like terms.

Elimination using Addition 5) Solve the system x - 2 y = 5 + 2 x + 2 y = 7 = 12 3 x x=4 Lets add both equations to each other ANS: (4, y) NOTE: We use the Elimination Method, if we can immediately cancel out two like terms.

Elimination using Addition 5) Solve the system x - 2 y = 5 2 x + 2 y = 7 4 - 2 y = 5 - 2 y = 1 y= 1 2 Lets substitute x = 4 into this equation. Solve for y ANS: (4, y) NOTE: We use the Elimination Method, if we can immediately cancel out two like terms.

Elimination using Addition 5) Solve the system x - 2 y = 5 2 x + 2 y = 7 4 - 2 y = 5 - 2 y = 1 y= 1 2 Lets substitute x = 4 into this equation. Solve for y 1 Answer: (4, 2 ) NOTE: We use the Elimination Method, if we can immediately cancel out two like terms.

Elimination using Addition 5) Solve the system 3 x + y = 14 4 x - y = 7 NOTE: We use the Elimination Method, if we can immediately cancel out two like terms.

Elimination using Addition 5) Solve the system 3 x + y = 14 + 4 x - y = 7 7 x = 21 x=3 ANS: (3, y)

Elimination using Addition 5) Solve the system 3 x + y = 14 Substitute x = 3 into this equation 4 x - y = 7 3(3) + y = 14 9 + y = 14 y=5 Answer: (3, 5 ) NOTE: We use the Elimination Method, if we can immediately cancel out two like terms.

Excellent Job !!! Well Done

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