Solving systems using substitution Three methods to solving

Solving systems using substitution

Three methods to solving systems 1) Graphing 2) Elimination 3) Substitution

The purpose of substitution substitute To for one of the variables, so that you only have 1 variable to solve for.

Substitution Basics 4 Y + 2 Y=3 4(3) + 2 = 14 Y = X 4(x) + 2 = 4 x + 2 Y = (x + 2) 4(x+2) + 2 = 4 x + 10

y = 4 x 3 x + y = -21 Notice that one of the equations is solved for Y. So we can Replace Y in the other equation with _____ 4 x 3 x + y = -21 3 x + 4 x = -21 Now we can solve for x. 7 x = -21 X = -3

y = 4 x 3 x + y = -21 Now that we know X = -3. We can solve for Y. Y = 4(-3) Y = -12 Solution to the system is (-3, -12).

y = 4 x 3 x + y = -21 Check the solution in both equations. Solution to the system is (-3, -12). y = 4 x -12 = 4(-3) -12 = -12 3 x + y = -21 3(-3) + (-12) = -21 -9 + (-12) = -21= -21

x + y = 10 5 x – y = 2 Neither of the equations is solved for x or y, so you will Need to solve one of the equations for x or y. X = 10 – Y or Y = 10 - X You choose which one you want. For right now we are Going to choose to use x = 10 -Y

x + y = 10 5 x – y = 2 Substitute x = 10 – y in the second equation 5(10 -y) –y = 2 50 -5 y – y = 2 50 -6 y = 2 -6 y = -48 Y=8

x + y = 10 5 x – y = 2 Now that we know that Y = 8 we can solve for X X + 8 = 10 X=2 Solution to the system is (2, 8).

x + y = 10 5 x – y = 2 Check the solution in both equations. Solution to the system is (2, 8). x + y =10 2 + 8 =10 10 =10 5 x – y = 2 5(2) - (8) = 2 10 – 8 = 2 2=2

1. 2. Solve by substitution: