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Linear Equations and Word Problems I. Return Quiz II. Homework III. Algebraic Expressions We will define the term variable which is analogous to a pronoun in grammar. Definition: A variable is a symbol that represents a number that may change depending on the circumstance. Examples: We use the variable t to represent time. W can use the variable h to represent height. We use the letter x for the variable when the variable stands for a general number without context. Definition: A constant is a symbol that represents a known fixed number. Examples: c = 3x108 pi = 3.14... Definition: An algebraic expression is an expression consisting of constants, variables, and symbols of operators. Examples: mc2 pi r2 3x2 +2x - 1 where x2 means xx. A term is an expression that is related by addition or subtraction. In 3x2 +2x - 1, 3x2, 2x, and 1 are the three terms. 3,2,1 are called the coefficients of x2, x, and 1 respectively. We say two terms are like terms if they differ only in their coefficient. For example 3x2 and 5x2 are like terms. We can add like terms. 3x2 + 5x2 = 8x2 Exercise: Simplify the expression 5x2 -2p + 3x2 + 5s2 + 2p - 3(2x2 -3s2) IV. Linear Equations Recall that an expression is a statement involving x's, constants, and operators. An expression is like the subject of a sentence. An equation is more like a complete sentence which as a subject, verb and predicate. The verb will typically be an "= " sign. Example: 3x + 1 = 4 When we have an equation, we can attempt to solve to find which x values satisfy the equation. An x value satisfies an equation if when we substitute the value in the equation for x, the equation is true. To solve for the variable x means that we have an equation of the form. The set of all x that satisfy the equation is called the solution set of the equation. x = a number. To solve we isolate x by 1) Using the distributive property 2) Combining like terms 3) Subtracting all the x terms from the right hand side- "Do unto the left as you would do unto the right" 4) Subtracting all the constant terms from the left hand side 5) dividing by the coefficient in front of x Example: 3(2x - 1) + 4x = 2(4x + 2) 1) 6x - 3 + 4x = 8x + 4 2) 10x - 3 = 8x + 4 3) 2x - 3 = 4 4) 2x = 7 5) x = 7/2 Definition: An equation that is true for some x but not for others is called a conditional equation. The value of x that makes the equation true is called the solution of the equation. In the above example we say that x = 7/2 is a solution to the equation.
Example: 3(2x - 1) - 2(3x + 1) = 0 6x - 3 - 6x - 2 = 0 0 - 5 = 0 -5 = 0 Since this can never happen we say that the equation is a contradiction. Example 2(x - 3) - 2x + 8 = 2 2x - 6 - 2x + 8 = 2 0 + 2 = 2 2 = 2 Since this is always true, we say that the equation is an identity. V. Problem Solving With Formulas A formula is an equation that relates real world quantities. P = 2l + 2w is the formula for the perimeter P of a rectangle given the length l and width w d = rt is the formula for the distance travelled d given the speed s and the time t A = P + Prt is the formula for the amount A in a bank account t years after P dollars is put in at an interest rate of r V = pi r2h is the formula for the volume V of a cylinder of radius r and height h A = h/2 (b1 + b2) is the formula for the area A of a trapazoid with height h and bases b1 and b2. We say that a formula is solved for a variable x if the equation becomes x = stuff where the left hand side does not include any x's. Example Solve C = 2 pi r for r then determine the radius of a circle with circumference 4 Solution Divide both sides by 2 pi: C/(2 pi) = r Use the reflexive property to get r = C/(2 pi) Now plug in 4 for C to obtain r = 4/(2 pi) = 2/pi Steps for solving a Problem 1) Read the problem, sketch the proper picture, and label variables. 2) Write down what the answer should look like. 3)Come up with the appropriate formula. 4) Solve for the needed variable. 5) Plug in the known numbers. 6) Answer the question. Example: A pile of sand has the shape of a right circular cone. Find the height of the pile if it contains 100 cc of sand and the radius is 5 cm. Solution: 1) We will draw the diagram in class 2) The height of the pile is ________ cm 3) We use the formula from appendix 2: V = 1/3 pi r2h 4) Multiply by 3 on both sides to get 3V = pi r2h Divide both sides by pi r2 to obtain 3v/(pi r2) = h or h = 3V/(pi r2) 5) h = 3(100)/(pi 52) = 12/pi 6) The height of the pile is 12/pi cm.
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