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Algebra 1 Review. You will get off to a great start in Honors Algebra 2 if you take the time to practice and really understand these problems! . 1) Evaluate the following expression when m = 6: 8m + (2m – 9) 3. 1) Evaluate the following expression when m = 6: 8m + (2m – 9) 3.
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Algebra 1 Review You will get off to a great start in Honors Algebra 2 if you take the time to practice and really understand these problems!
1) Evaluate the following expression when m = 6: 8m + (2m – 9)3
1) Evaluate the following expression when m = 6: 8m + (2m – 9)3 Recall: Substitute 6 in for m and evaluate following order of operations: 8(6) + (2(6) – 9)3 8(6) + (12 – 9)3 8(6) + 33 8(6) + 27 48 + 27 75
2) Simplify: 7(m – 3) + 4(m + 5) Recall: Use the Distributive Property to eliminate parentheses, then combine like terms 7m – 21 + 4m + 20 7m + 4m – 21 + 20 11m - 1
3) Simplify: 9(x2 + 2y) - 3(x2 – y) Recall: Use the Distributive Property to eliminate parentheses, then combine like terms: 9x2 + 18y – 3x2 + 3y 9x2 – 3x2 + 18y + 3y 6x2 + 21y
4) Solve: Recall: There are always different approaches for solving equations. REMEMBER TO ALWAYS PERFORM THE INVERSE OPERATION TO UNDO THE CURRENT OPERATION: The simplest approach here is to subtract 8 from both sides, then multiply both sides by the reciprocal. x = 15
5) During one shift, a waiter earns wages of $30 and gets an additional 15% in tips on customers’ food bills. The waiter earns $105. What is the total of the customers’ food bills?
5) During one shift, a waiter earns wages of $30 and gets an additional 15% in tips on customers’ food bills. The waiter earns $105. What is the total of the customers’ food bills? Recall: The variable in an equation represents the unknown quantity –- Let b = total of customers’ food bill Translate from words to equation The waiter earns a total of $105 which is $30 plus 15% of total food bill 105 = 30 + .15b 75 = .15b b = $500
6) Solve: 3(5x – 8) = -2(-x + 7) – 12x Recall: Simplify each side of the equation before applying any inverse operations. 15x – 24 = 2x – 14 – 12x 15x – 24 = -10x – 14 add 10x to both sides, add 24 to both sides to obtain variable term on one side, constant on other 25x = 10 x =
7) Solve: Recall: The easiest way to solve equations involving fractions is to clear the fractions by multiplying both sides of the equation by the LCD: ( 3t + 2t = 60 5t = 60 t = 12
8) Solve for y: 2y + xy = 6 Recall: Solving literal equations – when the variable you are solving for appears in more than one variable term, reverse the Distributive Property to factor out the variable as one factor. Then, continue to solve. y(2 + x) = 6 Divide both sides by (2 + x) y =
9) Solve for b1: A = Recall: Use inverse operations to isolate b1: Multiply both sides by 2 2A = (b1 + b2)h Divide both sides by h Subtract b2 from both sides
10) Solve: 0.6g + 0.5 = 2.9 Recall: The easiest approach for solving equations containing decimals is to multiply both sides of the equation by a power of 10 that will clear all the decimals: Multiply both sides by 10 10(0.6g + 0.5) = (2.9)10 6g + 5 = 29 6g = 24 g = 4
Tell whether the lines are parallel, perpendicular, or neither: Line 1: through points (-2, 2) and (0, -1) Line 2: through points (-4, -1) and (2, 3)
Tell whether the lines are parallel, perpendicular, or neither: Line 1: through points (-2, 2) and (0, -1) Line 2: through points (-4, -1) and (2, 3) Recall: If the slopes of two lines are equal, the lines are parallel. If the slopes of two lines are opposite reciprocals, the lines are perpendicular. Otherwise, the lines are not parallel, nor perpendicular. Formula to calculate slope of a line given two points: m = Line 1: m = Line 2: m = Line 1: m = Line 2: m = Opposite reciprocals, therefore the lines are perpendicular
12) Graph the following equations: a) y = 2 b) x = -1 c) y = 3x - 2
12) Graph the following equations: a) y = 2 x = -1 c) y = 3x - 2 Recall: Slope-intercept form of an equation is y = mx + b, where m is the slope (rise over run) and b is the y-intercept. y = 2 has a slope of zero and a y-intercept of 2, therefore the line is horizontal line intersecting the y-axis at 2. x = -1 has an undefined slope, therefore the line is a vertical line passing through the x-axis at -1 y = 3x – 2 has a slope of 3 and a y-intercept of -2, therefore, the line crosses the y-axis at -2 and from that point the line will rise 3 and run 1 to create a positive slope.
13) Write the equation of the line that passes through (5, 4) and has a slope of -3.
13) Write the equation of the line that passes through (5, 4) and has a slope of -3. Recall: Since the given information is the slope of the line and a point it is passing through, you can easily write an equation in Point-Slope Form: y – y1 = m(x – x1) Substitute the given information y – 4 = -3(x – 5)
14) Write the equation of the line that passes through (5, -2) and (2, 10).
14) Write the equation of the line that passes through (5, -2) and (2, 10). Recall: Since the given information is two points that the line passes through, you can write an equation by first calculating the slope and then using that slope and choosing one of the points to write an equation in Point-Slope Form: y – y1 = m(x – x1) Calculate slope: y – (-2) = -4(x – 5) y + 2 = -4(x – 5)
