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Objective: Solve systems of equations by substitution.

y = x – 1. x + y = 7. Objective: Solve systems of equations by substitution. Solve systems of equations by elimination. Use substitution to solve the system of equations. Step 1 Solve one equation for one variable. The first equation is already solved for y : y = x – 1.

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Objective: Solve systems of equations by substitution.

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  1. y= x – 1 x + y = 7 Objective: Solve systems of equations by substitution. Solve systems of equations by elimination. Use substitution to solve the system of equations. Step 1 Solve one equation for one variable. The first equation is already solved for y: y = x – 1. Step 2 Substitute the expression into the other equation. x + y = 7 Substitute (x –1) for y in the other equation. x + (x – 1) = 7 2x– 1 = 7 Combine like terms. 2x = 8 x = 4

  2. Step 3 Substitute the x-value into one of the original equations to solve for y. y = x– 1 y = (4)– 1 Substitute x = 4. y = 3 The solution is the ordered pair (4, 3).

  3. 2y + x= 4 3x– 4y = 7 Use substitution to solve the system of equations. Since the x variable has a coefficient of 1 in the first equation, solve the first equation for x. Substitute x = 4 - 2y into the second equation. 2y + x = 4 x = 4 - 2y 3(4 - 2y)– 4y = 7 12 – 6y – 4y = 7 -10y + 12 = 7 -10y= -5 y=

  4. Substitute the value to solve for the other variable. 2y + x = 4 2 +x = 4 1+x=4 X=3 Second part of the solution The solution is

  5. 3x + 2y = 4 4x– 2y = –18 You can also solve systems of equations with the elimination method. With elimination, you eliminate of one of the variables by adding or subtracting equations. You may have to multiply one or both equations by a number to create variable terms that can be eliminated. Variable terms may be eliminated if their coefficients are additive inverses. The elimination method is sometimes called the addition method or linear combination. Use elimination to solve the system of equations. Step 1 Find the value of one variable. Select a variable to eliminate. In this case the y variables have coefficients that are additive inverses. 3x+ 2y = 4 The y-terms have opposite coefficients. + 4x– 2y = –18 Add the equations to eliminate y. 7x = –14 x = –2 First part of the solution

  6. Step 2 Substitute the x-value into one of the original equations to solve for y. 3(–2) + 2y = 4 2y = 10 Second part of the solution y = 5 The solution to the system is (–2, 5).

  7. 3x + y = 1 2y+ 6x = –18 Because isolating y is straightforward, use substitution. 3x + y = 1 y = 1 –3x 2(1 – 3x)+ 6x= –18 2– 6x+ 6x= –18 2= –18 Regardless of the method used, sometimes, all variables will disappear and a false statement will result. In this case, the system represents parallel lines and the number of solutions is zero. The solution is the empty set. IF the result is a true statement such as 4=4, then the system represents coincident lines and the number of solutions is infinte. The solution would be “All real numbers on the line y = ……” Be sure to state one of the lines!

  8. Three Scenarios

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