* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Evaluating algebraic expressions:
Path integral formulation wikipedia , lookup
Two-body problem in general relativity wikipedia , lookup
Two-body Dirac equations wikipedia , lookup
Debye–Hückel equation wikipedia , lookup
Schrödinger equation wikipedia , lookup
Maxwell's equations wikipedia , lookup
Unification (computer science) wikipedia , lookup
BKL singularity wikipedia , lookup
Van der Waals equation wikipedia , lookup
Calculus of variations wikipedia , lookup
Itô diffusion wikipedia , lookup
Navier–Stokes equations wikipedia , lookup
Euler equations (fluid dynamics) wikipedia , lookup
Equations of motion wikipedia , lookup
Equation of state wikipedia , lookup
Derivation of the Navier–Stokes equations wikipedia , lookup
Computational electromagnetics wikipedia , lookup
Schwarzschild geodesics wikipedia , lookup
Differential equation wikipedia , lookup
Evaluating algebraic expressions: PEMDAS Parenthesis Exponents Multiplication Division Add Subtract Inverse Operations: Original Inverse + - x, (opposite) 2 sin Sin-1 cos cos-1 tan tan-1 Solving equations: Single variable 1. Remove any parentheses and combine like terms, on each side of the equation. 2. Clear any fractions by multiplying all terms on both sides by the denominator 3. Group like terms on each side of the equal sign. 4. Isolate the variable on one side of the equation by using inverse operations 5. Divide or multiply both sides of the equation by the coefficient of the variable to solve 6. Check the solution by substituting your answer into the original equation wherever you have a variable. If the solution is correct, both sides of the equation will be equal. Solve for the given variable: 1) 5c - 4 - 2c + 1 = 8c + 2 2) 8(m + 5) = 16 Literal Equations Can be solved the same way as all other equations. Just remember you can only combine like terms and it is ok to have an answer with multiple parts. Solve for x: 3) xy - d = m x z y 4) y Solving Quadratic Equations Any equations with an x2 Always keep your squared variable positive 1-Set your equation equal to 0 2-Identify the type of equation you have 3-Pick a method of factoring 1. x2 – 5x = 24 2. x2 + 5x = 0 Solving Systems of Equations Algebraically - Using Substitution The substitution method is used to eliminate one of the variables by replacement when solving a system of equations. 1. Solve one of the equations for either "x =" or "y =". This example solves the second equation for "y =". 3y - 2x = 11 y = 9 - 2x 2. Replace the "y" value in the first equation by what "y" now equals. Grab the "y" value and plug it into the other equation. 3(9 - 2x) - 2x = 11 3. Solve this new equation for "x". (27 - 6x) - 2x = 11 27 - 6x - 2x = 11 27 - 8x = 11 -8x = -16 x=2 4. Place this new "x" value into either of the ORIGINAL equations in order to solve for "y". Pick the easier one to work with! y = 9 - 2(2) y=9-4 y=5 Solve: x = y - 5 2x – y = 7 Solving Systems of Equations Algebraically - Using Addition or Subtraction Think of the adding or subtracting method as simply "eliminating" one of the variables to make your life easier. 1. Solve this system of equations: x - 2y = 14 x + 3y = 9 a. First, be sure that the variables are "lined up" under one another. In this problem, they are already "lined up". x - 2y = 14 x + 3y = 9 b. Decide which variable ("x" or "y") will be easier to eliminate. In order to eliminate a variable, the numbers in front of them (the coefficients) must be the same or negatives of one another. Looks like "x" is the easier variable to eliminate in this problem. x - 2y = 14 x + 3y = 9 c. Now, subtract to eliminate the "x" variable. (Remember: when you subtract signed numbers, you change the signs and add) x - 2y = 14 -x - 3y = - 9 - 5y = 5 d. Solve this simple equation. -5y = 5 y = -1 e. Plug "y = -1" into either of the ORIGINAL equations to get the value for "x". x - 2y = 14 x - 2(-1) = 14 x + 2 = 14 x = 12 Ex: The sum of two numbers is 19. Their difference is 9. Find each number. Solving a Linear Quadratic System Algebraically Through substitution We'll use the following example. Solve algebraically: y = x2 - x - 6 y = 2x - 2 First we solve for one y = 2x - 2 Since this is already of the variables in the done for us in this linear equation. example, we can go to the next step. 2 Next we substitute for y=x -x-6 that variable in the 2x-2 = x2 - x - 6 Subtract 2 from both quadratic equation, and 2x = x2 - x - 4 sides. solve the resulting 2 Subtract 2x from both equation. 0 = x - 3x - 4 sides. 0 =(x-4)(x+1) Factor. x-4=0 x+1=0 Set each factor =0 and x=4 x = -1 solve. We now have two values for x, but we still need to find the corresponding values for y. We find the yvalues by substituting each value of x into the linear y = 2x - 2 y = 2(4) - 2 y=8-2 y=6 (4,6) equation. {(4,6),(-1,-4)} y = 2x - 2 y = 2(-1) - 2 y = -2 - 2 y = -4 (-1,-4)