# 3.2 Solving Systems of Equations Algebraically Substitution Method Elimination Method.

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3.2 Solving Systems of Equations Algebraically Substitution Method Elimination Method

Substitution Method Here you replace one variable with an expression. x + 4y = 26 x – 5y = - 10 Solve for a variable, x = 26 – 4y Replace “x” in the other equation (26 – 4y) – 5y = -10 Solve for y

(26 – 4y) – 5y = -10 26 – 4y – 5y = - 10Remove parentheses by multiplying by 1 26 – 9y = - 10Add like terms -9y = - 36Subtract 26 from both sides y = 4Divide by - 9

Solve for x x = 26 – 4y x = 26 – 4(4)Substitute for y x = 26 – 16 x = 10 The order pair is (10, 4). This is where the lines cross.

The Elimination Method Here we add the equations together when the coefficients are different signs. x + 2y = 10 x + y = 6 Here both lead coefficients are 1. We can change the coefficient to – 1, by multiplying by – 1.

x + 2y = 10 x + y = 6 Multiply the bottom equation by – 1. x + 2y = 10 - x - y = - 6When adding the y = 4equations together, x go to zero. Find x by replace it back in either equation. x + 2(4) = 10;x + 8 = 10;x = 2

So the order pair (2, 4) works in both equations. 2 + 2(4) = 10 2 + 4 = 6 We have two way to solve the systems, Substitution and Elimination; which way is better depends on the problem.

What about this problem 2x + 3y = 12 5x – 2y = 11 Here we have to multiply both equations If we wanted to remove the “x”, then we have to find the Least common multiple (L.C.M.) of 2 and 5. If we wanted to remove the “y”, then we have to find the least common multiple of 3 and -2.

Lets get rid of the “y” The L.C.M of 2 and 3 is 6. Since we want the coefficients to be opposite, - 2 will help in the equation. we multiply the top equation by 2. 2x + 3y = 12 4x + 6y = 24 The bottom equation by 3 5x – 2y = 1115x – 6y = 33

Add the new equations together 4x + 6y = 24 15x – 6y = 33 19x = 57 Divide by 19 x = 3 Replace in original equation and solve for y 2(3) + 3y = 126 + 3y = 123y = 6 y= 2

What about inconsistent systems? y – x = 5 Multiply the top equation by – 2, 2y – 2x = 8 2y – 2x = -10 then add the bottom.2y – 2x = 8 0 = - 2 This shows no solutions.

What if it is dependent (Many solutions) 1.6y = 0.4x + 1 0.4y = 0.1x + 0.25 Multiply the top and bottom equation by 100 to remove decimals. 160y = 40x + 100 40y = 10x + 25 Then multiply the bottom equation by -4 -160y = -40x – 100

Add the new equations together 160y = 40x + 100 -160y = -40x – 100 0 = 0 This is a system with many solutions.

Solve this system a – b = 2 -2a + 3b = 3