Section 6.1/6.2* 2x2 Linear Systems and some other Systems/Applications Solving 2x2 Linear Systems ax by c To solve a system of two linear equations means to find values for x and y dx ey f that satisfy both equations. You may use either the substitution or elimination/addition method. A 2x2 linear system of equations may have no solution, one solution or infinitely many solutions. Example 1: Solve each system of linear equations by the Substitution Method. a. 2x y = 5 5x + 2y = 8 b. x 2y = 3 2x 4y = 7 Section 6.1/6.2-2x2 Linear and Other Systems/Applications 1
Example 2: Solve the following systems by the Elimination Method. a. x + 4y = 10 1 x + 2y = 5 2 b. y 2x = -6 2y x = 0 c. 2x + 3y = -16 5x 10y = 30 Section 6.1/6.2-2x2 Linear and Other Systems/Applications 2
A system will have exactly one solution, no solution, or infinitely many solutions. 1. Exactly one solution, will look like: 2. No solution, will look like: Section 6.1/6.2-2x2 Linear and Other Systems/Applications 3
3. Infinitely many solutions, will look like: If you are interested in determining the number of solutions a 2X2 linear system of equations has, you can use the slope and y-intercept to determine just that. Parallel and Perpendicular Lines Two lines are parallel if and only if their slopes are the same. Two lines are perpendicular if and only if their slopes are negative reciprocals of each other. Example 3: Determine the number of solutions each of the following system has by using the slope and y-intercept. a. 3 y x 5 5 y x 1 3 b. y 3x 1 5y 15x 2 Section 6.1/6.2-2x2 Linear and Other Systems/Applications 4
c. x 2y 4 3x 6y 12 d. y 3x 5 x 2y 6 Example 4. Solve the following systems: a. 12 b. 5 3 Section 6.1/6.2-2x2 Linear and Other Systems/Applications 5
For each of the following problems: (a) Write a system of equations involving two variables to model the problem. (b) Solve your system of equations and answer the question. Example 1. Dillan is at a baseball game and is buying hot dogs and sodas for his family. Hot dogs cost $3 each and sodas cost $1.75 each. He purchases nine items and spends a total of $22.00. How many hot dogs did he buy? How many sodas did he buy? Example 2. Two numbers have a sum of 77 and a difference of 13. Find the two numbers Section 6.1/6.2-2x2 Linear and Other Systems/Applications 6
Example 3. A rectangle has a perimeter of 26 centimeters and an area of 36 square centimeters. Find the dimensions of the rectangle. Section 6.1/6.2-2x2 Linear and Other Systems/Applications 7
Math 1310 Section 2.4 Section 2.4: An Introduction to Complex Numbers In this section, you ll learn an introduction to complex numbers. Where a complex number has the form of. You will learn to add, subtract, multiply and divide these numbers Complex Numbers Definition: A complex number is a number that can be written in the form a + bi, where a is called the real part and bi is called the imaginary part. The a and b are real numbers and 1 Here are several properties of complex numbers: Addition of Complex Numbers: (a + bi) + (c + di) = (a + c) + (b + d)i Add the real parts together and add the imaginary parts together. Subtraction of Complex Numbers (a + bi) (c + di) = (a c) + (b d)i Subtract the real parts and subtract the imaginary parts. Multiplication of Complex Numbers: Multiply in the same manner as multiplying binomials and remember that i 2 = 1 Example 1: Simplify each. a. 16 b. 40 Example 2: Simplify each of the following and write the answer in form a + bi. a. (5 + 4i) + (2 i) b. ( 6 3i) ( 2 + 2i) c. i( 3 + 6i) d. ( 1 i)(2 + 5i)
Math 1310 Section 2.4 e. 16 36 9 10 Next, you ll need to be able to find various powers of i. You ll need to know these 4 powers: 1 1 1 1 1 1 For other powers of i, divide the exponent by 4 and find the remainder. Your answer will be i raised to the remainder power. If the remainder is zero, your answer will be i 4 or 1. Example 3: Simplify each. a. i 15 b. i 72 c. i 42 d. i 313
Math 1310 Section 2.4 Division of Complex Numbers The complex conjugate of the complex number a + bi is the complex number a bi. To simplify the quotient the denominator. multiply both the numerator and denominator by the complex conjugate of Example 4: Simplify each of the following and write the answer in form a + bi. a. 5 2 3 b. 1 c. 5 4 4
Math 1310 Section 2.4 Complex Roots of Quadratic Equations Using complex numbers, we can now find all solutions to quadratic equations. We can use any of the techniques from the previous section to solve, but usually, we will just take the square root of both sides of the equation, complete the square or use the quadratic formula. Example 5: Find all complex solutions of the following equations. Express your answer in form a + bi. a. 100 0 b. 49 36 0
Math 1310 Section 2.4 c. 6 13 d. 4 8 9 0