algebra 1 review

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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!

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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. - PowerPoint PPT Presentation

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Page 1: Algebra 1 Review

ALGEBRA 1 REVIEWYou will get off to a great start in

Honors Algebra 2 if you take the time to practice and really understand these problems!

Page 2: Algebra 1 Review

1) Evaluate the following expression when m = 6:

8m + (2m – 9)3

Page 3: Algebra 1 Review

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) + 2748 + 27

75

Page 4: Algebra 1 Review

2) Simplify: 7(m – 3) + 4(m + 5)

Page 5: Algebra 1 Review

2) Simplify: 7(m – 3) + 4(m + 5)

Recall:Use the Distributive Property to eliminate parentheses, then combine like terms

7m – 21 + 4m + 207m + 4m – 21 + 20

11m - 1

Page 6: Algebra 1 Review

3) Simplify: 9(x2 + 2y) - 3(x2 – y)

Page 7: Algebra 1 Review

3) Simplify: 9(x2 + 2y) - 3(x2 – y)

Recall:Use the Distributive Property to eliminate parentheses, then combine like terms:

9x2 + 18y – 3x2 + 3y9x2 – 3x2 + 18y + 3y

6x2 + 21y

Page 8: Algebra 1 Review

4) Solve: 20854

x

Page 9: Algebra 1 Review

4) Solve: 20854

x

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

Page 10: Algebra 1 Review

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?

Page 11: Algebra 1 Review

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 equationThe waiter earns a total of $105 which is $30

plus 15% of total food bill

105 = 30 + .15b75 = .15bb = $500

Page 12: Algebra 1 Review

6) Solve: 3(5x – 8) = -2(-x + 7) – 12x

Page 13: Algebra 1 Review

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 – 12x15x – 24 = -10x – 14

add 10x to both sides, add 24 to both sides to obtain variable term on one side, constant on

other

25x = 10x =

Page 14: Algebra 1 Review

7) Solve: 1031

21

tt

Page 15: Algebra 1 Review

7) Solve: 1031

21

tt

Recall:The easiest way to solve equations involving fractions is to clear the fractions by multiplying both sides of the equation by the LCD of the fractions:

(3t + 2t = 60

5t = 60t = 12

Page 16: Algebra 1 Review

8) Solve for y: 2y + xy = 6

Page 17: Algebra 1 Review

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) = 6Divide both sides by (2 + x)

y =

Page 18: Algebra 1 Review

9) Solve for b1: A =

Page 19: Algebra 1 Review

9) Solve for b1: A =

Recall:Use inverse operations to isolate b1:

Multiply both sides by 22A = (b1 + b2)h

Divide both sides by h

Subtract b2 from both sides

Page 20: Algebra 1 Review

10) Solve: 0.6g + 0.5 = 2.9

Page 21: Algebra 1 Review

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 1010(0.6g + 0.5) = (2.9)10

6g + 5 = 296g = 24

g = 4

Page 22: Algebra 1 Review

11)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)

Page 23: Algebra 1 Review

11)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

Page 24: Algebra 1 Review

12) Graph the following equations:a) y = 2b) x = -1c) y = 3x - 2

Page 25: Algebra 1 Review

12) Graph the following equations: a) y = 2

b) 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.a) 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.

b) x = -1 has an undefined slope, therefore the line is a vertical line passing through the x-axis at -1

c) 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.

Page 26: Algebra 1 Review

13) Write the equation of the line in slope-intercept form that passes through (5, 4) and has a slope of -3.

Page 27: Algebra 1 Review

13) Write the equation of the line in slope-intercept form 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 start by writing an equation in Point-Slope Form: y – y1 = m(x – x1)

Substitute the given informationy – 4 = -3(x – 5)

Then just solve for y (to put in y = mx + b form)y – 4 = -3x + 15y = -3x + 19

Page 28: Algebra 1 Review

14) Write the equation of the line in slope-intercept form that passes through (5, -2) and (2, 10).

Page 29: Algebra 1 Review

14) Write the equation of the line in slope-intercept form 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)y + 2 = -4x + 20

y = -4x + 18

Page 30: Algebra 1 Review

15) Find the slope of the line given the equation 2x – 3y = 5

Page 31: Algebra 1 Review

15) Find the slope of the line given the equation 2x – 3y = 5

Recall:When an equation is in Slope-intercept form y = mx + b, the slope is always m (the coefficient of x).

Rewrite the given equation in slope-intercept form to find the slope:2x – 3y = 5 (subtract 2x)

-3y = -2x + 5 (divide by -3)

The slope, m, of this line is 2/3.

(y-intercept is (0, -5/3)

35

32

xy

Page 32: Algebra 1 Review

16) Find the x- and y-intercepts of the line with the equation 4x – 8y = 12.

Page 33: Algebra 1 Review

16) Find the x- and y-intercepts of the line with the equation 4x – 8y = 12.

Recall:To find the x-intercept of a line, substitute 0 for y and solve for x:

4x – 8(0) = 124x = 12

X = 3 x-intercept (3, 0)To find the y-intercept of a line, substitute 0 for x and solve for y:

4x – 8y = 124(0) – 8y = 12

-8y = 12Y = -3/2 y-intercept (0, -3/2)

Page 34: Algebra 1 Review