Graphing is just one way to solve a system of equations. What are the potential problems with using this method of graphing to find the intersection point of two lines?

Solving by graphing is not always accurate. Unless you have a graphing calculator or other software, it may be hard to determine exactly what this intersection point is just by looking at the graph.

Chapter 3.2 Solving Systems of Equations Algebraically

Substitution is one method of solving a system of equations Substitution is one method of solving a system of equations. See Page 142, problem 1 Solve one of the equations for one of its variables. Try to find a variable with a coefficient of 1 or -1. Substitute the expression from step 1 into the other equation and solve for the other variable. Use this value from step 2 into the first equation, and solve for the other variable.

Substitution Example 3x + 6y = 12 y = -2x - 1 . y = -2x - 1 SUBSTITUTE this for y into the 1st equation. 3x + 6y = 12 3x + 6(-2x-1) = 12 3x – 12x – 6 = 12 -9x – 6 = 12 -9x = 18 x = -2 Now, plug this back into the original equation….. y = -2x – 1 y = -2(-2) – 1 y = 4-1 y = 3 The solution for this system of equations is (-2,3)

Another substitution example y = 3x – 9 4x + 5y = -7 Substitute “3x – 9” in place of Y in the 2nd equation… 4x + 5(3x-9) = -7 4x + 15x – 45 = -7 19x – 45 = -7 19x = 38 x = 2 Plug this back into one of the original equations…. y = 3(2) – 9 y = 6-9 y = -3

2x + 3y = 5 y = 4x – 1 Since y = 4x – 1, substitute the expression “4x-1” into the other equation in place of y 2x + 3(4x-1) = 5 then solve for x 2x + 12x – 3 = 5 14x – 3 = 5 14x = 8 X = 4/7 Now that you know x, solve for y. Use either of the original equations: y = 4x – 1 y = 4 (4/7) – 1 y = 16/7 – 1 y = 16/7 - 7/7 y = 9/7 The solution for this system of equations is (4/7, 9/7)

Solve the system of equations: y = 3x + 7 -3x + y = 11

are good examples. It’s easy to solve these for either x or y. Substitution is great when either one of the equations has a variable with a coefficient of 1 or -1. 2x + y = 5; 2y – x = 7; x + 5y = 12 are good examples. It’s easy to solve these for either x or y. Very useful when one or both equations are in slope-intercept form, such as y = 3x – 7; y = -2x + 4

Classwork Substitution Page 146, #10-18

Sometimes, the ELIMINATION method may be better for solving a System of Equations. This method is also called the ADDITION method. We are eliminating one of the variables by adding the like terms together 3x + 2y = 13 5x – 2y = 11 Look for opposites in the two equations. 2y and -2y are opposites. Combine the two equations by adding like terms, which eliminates one of the variables (y in this case).

________________ Opposites? Add the columns 8x = 24 x = 3 3x + 2y = 13 5x – 2y = 11 ________________ Opposites? Add the columns 8x = 24 x = 3 Plug this value for x into either one of the equations. 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 The solution is (3,2)

2x – 5y = 8 4x + 5y = 7

-6x + 2y = 16 6x + 5y = 5

4x + 3y = 14 2x – y = 2 Since there are no opposites in these two equations, we can multiply one (or both) of the equations by a constant to obtain a pair of opposites. 4x + 3y = 14 2x – y = 2 multiply the 2nd equation by 3 4x + 3y = 14 6x – 3y = 6 Now we have opposites of 3 and -3, so we can eliminate the y’s _____________ 10x = 20 x = 2 Now plug this into one of the original equations. 2(2) – y = 2 4 – y = 2 y = 2 The solution is (2,2)

Classwork Elimination Method Page 146, #22-30

3x – 6y = -30 4x + 5y = 12 Sometimes, you may need to multiply BOTH equations by a number in order to get a pair of opposites. You don’t have to multiply both equations by the same number. For example, we can multiply the 1st equation by 5 and the 2nd equation by -6. Or we can multiply the 1st equation by -4 and the 2nd equation by 3.

2x + 3y = 18 3x + 5y = 29 Since there are no opposites in these two equations, we can multiply one (or both) of the equations by a constant to obtain a pair of opposites. 2x + 3y = 18 3x + 5y = 29 multiply the 1st equation by -3, and the 2nd equation by 2 -6x – 9y = -54 6x + 10y = 58 ____________ Now we have opposites of -6 and 6, so we can eliminate the x’s y = 4 Now plug this into one of the original equations. 2x + 3(4) = 18 2x + 12 = 18 2x = 6 x = 3 The solution is (3,4)

If you have decimals, you have a couple of options: Use a calculator and solve by substitution or elimination Multiply everything by a power of 10 (10,100,1000, whatever is needed) to get rid of all of the decimals… Then solve 5x + 2.75y = 4 2.5x – 0.05y = 2.25

0.02x – 1.5y = 4 0.5y – 0.02x = 1.8

Classwork Page 146, #31-42

BELLRINGER – 10/4/2012 Thursday You will operate a concession stand, selling hamburgers and drinks. The burgers will sell for $3.00, and the drinks will sell for $2.00. The sales total for the day is $480. If you sold a total of 210 items combined, how many each of burgers and drinks did you sell? HINT: Let x = the number of burgers sold Let y = the number of drinks sold Answer: 60 burgers, 150 drinks