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3.1 Solving Linear Equations Part I. A linear equation in one variable can be written in the form: Ax + B = 0 Linear equations are solved by getting “x” by itself on one side of the equation Addition (Subtraction) Property of Equality:. 3.1 Solving Linear Equations Part I.
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3.1 Solving Linear Equations Part I • A linear equation in one variable can be written in the form: Ax + B = 0 • Linear equations are solved by getting “x” by itself on one side of the equation • Addition (Subtraction) Property of Equality:
3.1 Solving Linear Equations Part I • Multiplication Property of Equality: • Since division is the same as multiplying by the reciprocal, you can also divide each side by a number. • General rule: Whatever you do to one side of the equation, you have to do the same thing to the other side.
3.1 Solving Linear Equations Part I • Example: Solve by getting x by itself on one side of the equation.Subtract 7 from both sides:Divide both sides by 3:
3.2 Solving Linear Equations Part II- Fractions/Decimals • As with expressions, you need to combine like terms and use the distributive property in equations.Example:
3.2 Solving Linear Equations Part II- Fractions/Decimals • Fractions - Multiply each term on both sides by the Least Common Denominator (in this case the LCD = 4):Multiply by 4:Reduce Fractions:Subtract x:Subtract 5:
3.2 Solving Linear Equations Part II- Fractions/Decimals • Decimals - Multiply each term on both sides by the smallest power of 10 that gets rid of all the decimalsMultiply by 100:Cancel:Distribute:Subtract 5x:Subtract 50:Divide by 5:
3.2 Solving Linear Equations Part II- Fractions/Decimals • Eliminating fractions makes the calculation simpler:Multiply by 94:Cancel:Distribute:Subtract x:Subtract 10:
3.2 Solving Linear Equations Part II • 1 – Multiply on both sides to get rid of fractions/decimals • 2 – Use the distributive property • 3 – Combine like terms • 4 – Put variables on one side, numbers on the other by adding/subtracting on both sides • 5 – Get “x” by itself on one side by multiplying or dividing on both sides • 6 – Check your answers (if you have time)
3.2 Solving Linear Equations Part II • Example:Clear fractions:Combine like terms:Get variables on one side:Solve for x:
3.3 Applications of Linear Equations to General Problems • 1 – Decide what you are asked to find • 2 – Write down any other pertinent information (use other variables, draw figures or diagrams ) • 3 – Translate the problem into an equation. • 4 – Solve the equation. • 5 – Answer the question posed. • 6 – Check the solution.
Example: The sum of 3 consecutive integers is 126. What are the integers?x = first integer, x + 1 = second integer, x + 2 = third integer 3.3 Applications of Linear Equations to General Problems
Example: Renting a car for one day costs $20 plus $.25 per mile. How much would it cost to rent the car for one day if 68 miles are driven?$20 = fixed cost, $.25 68 = variable cost 3.3 Applications of Linear Equations to General Problems
3.4 Percent Increase/Decrease and Investment Problems • A number increases from 60 to 81. Find the percent increase in the number.
3.4 Percent Increase/Decrease and Investment Problems • A number decreases from 81 to 60. Find the percent increase in the number.Why is this percent different than the last slide?
3.4 Percent Increase/Decrease and Investment Problems • A flash drive is on sale for $12 after a 20% discount. What was the original price of the flash drive?
3.4 Percent Increase/Decrease and Investment Problems • Another Way: A flash drive is on sale for $12 after a 20% discount. What was the original price of the flash drive?Since $12 was on sale for 20% off, it is 100% - 20% = 80% of the original price set up as a proportion (see 3.6):
3.4 Percent Increase/Decrease and Investment Problems • Simple Interest Formula:I = interestP = principalR = rate of interest per yearT = time in years
3.4 Percent Increase/Decrease and Investment Problems • Example: Given an investment of $9500 invested at 12% interest for 1½ years, find the simple interest.
3.4 Percent Increase/Decrease and Investment Problems • Example: If money invested at 10% interest for 2 years yields $84, find the principal.
A = lw P = a + b + c Area of rectangle Area of a triangle Perimeter of triangle Sum of angles of a triangle Area of a circle Circumference of a circle 3.5 Geometry Applications and Solving for a Specific Variable
3.5 Geometry Applications and Solving for a Specific Variable • Complementary angles – add up to 90 • Supplementary angles – add up to 180 • Vertical angles – the angles opposite each other are congruent
3.5 Geometry Applications and Solving for a Specific Variable • Find the measure of an angle whose complement is 10 larger. • x = degree measure of the angle. • 90 – x = measure of its complement • 90 – x = 10 + x • Subtract 10: Add x: Divide by 2:
3.6 Applications of Linear Equations to Proportions, d=rt, and Mixture Problems • Ratio – quotient of two quantities with the same unitsNote: percents are ratios where the second number is always 100:
3.6 Applications of Linear Equations to Proportions, d=rt, and Mixture Problems • Percents : Example: If 70% of the marbles in a bag containing 40 marbles are red, how many of the marbles are red?:# of red marbles =
3.6 Applications of Linear Equations to Proportions, d=rt, and Mixture Problems • Proportion – statement that two ratios are equal:Solve using cross multiplication:
3.6 Applications of Linear Equations to Proportions, d=rt, and Mixture Problems • Solve for x: Solution:
Example: d=rt (distance = rate time)How long will it take to drive 420 miles at 50 miles per hour? 3.6 Applications of Linear Equations to Proportions, d=rt, and Mixture Problems
3.6 Applications of Linear Equations to Proportions, d=rt, and Mixture Problems • General form of a mixture problem: x units of an a% solution are mixed with y units of a b% solution to get z units of a c% solutionEquations will always be:
3.6 Applications of Linear Equations to Proportions, d=rt, and Mixture Problems • Example: How many gallons of a 10% indicator solution must be mixed with a 20% indicator solution to get 10 gallons of a 14% solution? Let x = # gallons of 10% solution,then 10 - x = # gallons of 20% solution :
3.7 Solving Linear Inequalities in One Variable • < means “is less than” • means “is less than or equal to” • > means “is greater than” • means “is greater than or equal to”note: the symbol always points to the smaller number
3.7 Solving Linear Inequalities in One Variable • A linear inequality in one variable can be written in the form: ax < b (a0) • Addition property of inequality:if a < b then a + c < b + c
3.7 Solving Linear Inequalities in One Variable • Multiplication property of inequality: • If c > 0 thena < b and ac < bc are equivalent • If c < 0 thena < b and ac > bc are equivalentnote: the sign of the inequality is reversed when multiplying both sides by a negative number
3.7 Solving Linear Inequalities in One Variable • Example: -9
3.8 Solving Compound Inequalities • For any 2 sets A and B, the intersection of A and B is defined as follows:AB = {x x is an element of A and x is an element of B} • For any 2 sets A and B, the union of A and B is defined as follows:AB = {x x is an element of A or x is an element of B}
3.8 Solving Compound Inequalities • Example: 1 3
3.8 Solving Compound Inequalities • Example: 1 3