Answers to the problems will be posted on the school website, go to Academics tab, then select Mathematics and select Summer Packets.

Transcription

1 Name Geometry SUMMER PACKET This packet contains Algebra I topics that you have learned before and should be familiar with coming into Geometry. We will use these concepts on a regular basis throughout Geometry. There are a few example problems at the beginning of each section, followed by practice problems for you to complete. Answers to the problems will be posted on the school website, go to Academics tab, then select Mathematics and select Summer Packets. You are expected to have this packet done before returning to school in the fall. You will take a series of quizzes during the 1 st Marking Period on these topics. Have a great summer!!

2 Combining Like Terms Like terms are algebraic expressions with the same variables and the same exponents for each variable. Like terms may be combined by performing addition and/or subtraction of the coefficients of the terms. Example 1 3x x 2 Example 2 3x 2 5x 3y + 10x + 7y 8x 2-5x 2 + 5x + 4y Combine like terms for each expression below. 1. 4x 2 + 8x + 9-6x 2 - x x + 9-5x 2-6x y 2 + 3x y 2 + x y 2 + 3x 2 + 5y x 2-2y 2 + y 5. 3x 3 + 4y 2 - y - y 2 + 7x x x

3 Solving Algebraic Equations Solving equations involves undoing what has been done to create the equation. In this sense, solving an equation can be described as working backward, generally known as using inverse (or opposite) operations. For example, to undo x + 2 = 5, that is, adding 2 to x, subtract 2 from both sides of the equation. The result is x = 3, which makes x + 2 = 5 true. For = 17, x is multiplied by 2, so divide both sides by 2 and the result is x = 8.5. For equations like those in the examples and exercises below, apply the idea of inverse (opposite) operations several times. Always follow the correct order of operations. Example Solve for x: 2( - 1) - 6 = -x + 2 First distribute to remove the parentheses, then combine like terms. 4x = -x + 2 4x - 8 = -x + 2 Next, move variables and constants by addition of opposites to get the variable term on one side of the equation. 4x - 8 = -x + 2 +x + x 5x - 8 = 2 5x - 8 = x = 10 Now, divide by 5 to get the value of x. 5x = x = 2 Finally, check that your answer is correct. 2(2(2) -1) - 6 = -(2) + 2 2(4-1) - 6 = 0 2(3) - 6 = = 0 checks 3.

4 Solve each equation below. 7. x + 7 = x 8 = y 9 = x = x = x = x + 8 = = x 6 = x + 2 = -x x - 2 = x x + 4x - 2 = x - 3x + 2 = (m - 2) = -2(m - 7) 21. 3( + 2) + 2(x - 7) = x

5 Slope Slope, or rise over run, measures a line s steepness. The variable m is used for slope, where y y2 y1 m. An increasing line has positive slope, and a decreasing line has negative slope. x x x 2 1 Examples Find the slope of each line. Example 1 To find slope, first choose two points that have ordered pairs made of integers. The points chosen at left are (0, -2) and (3, 0). You can find slope by counting or by using the slope formula. To count: Create a slope triangle. Then count the vertical change (rise), divide it by the horizontal change (run). This line has a slope of 2 3. y2 y1 Slope formula: m x x ( 2) Example 2 The points chosen at left are (0, 3) and (1, 1). To count: Create a slope triangle. Then count the vertical change (rise), divide it by the horizontal change (run). This 2 line has a slope of, or y2 y Slope formula: m 2 x x Determine the slope of each line. 22. m = 23. m = 24. m = 5.

6 Graphing One of the most useful ways to write the equation of a line is slope-intercept form, or y mx b, where m is the slope and b is the y-intercept. Examples Write the equation of the line. Example 1 The line at left has a slope of 2. Its y-intercept, where it 3 crosses the y-axis, is (0, -2). The slope of the line in slope-intercept form is 2 y x 3 2. Example 2 The line at left has a slope of 2. Its y-intercept, where it crosses the y-axis, is (0, 3). The slope of the line in slope-intercept form is y 3. Given the graph, write the equation of the line in slope-intercept form

7 y x Graph each of the following equations. y 3x y y x x 2y 8 7.

8 Solving Systems of Linear Equations in Two Variables Solving by the substitution method Consider this system: 10y 3x 14 4y 4 Look for the equation that is easiest to solve for x or y. In this case, we chose to solve the second equation for x by dividing both sides by two. Hint: Pick the equation that will avoid getting fractions. 4y 4 4 4y x 2 2y Now substitute (replace) the x in the other equation with (-2-2y). This will allow you to solve for y first. 10y 3( 2 2 y) 14 10y 6 6y 14 16y y y Next, find x by substituting 0.5 for y into either original equation. 4y 4 4(0.5) x 3 The solution to this system of equations is the coordinate pair (-3, 0.5). You can check your solution by substituting the coordinate pair you found into both original equations. If both of the original equations balance, then you have found the correct coordinate pair. NOTE: When both variables drop out of the equation, you either have no solution (false statement) ex. 5 = 0, or infinitely many solutions (true statement) ex. 3=3. 8.

9 Solving by the Elimination Method First, rewrite the equation so that the x s and y s are lined up vertically on the same side of the equal sign. Constants should be on the opposite side of the equal sign. Next, decide which variable you would like to eliminate, it does not matter. Hint: try to pick the variable that might already have a constant that is a multiple of the other or pick the variable that has opposite signs already. Consider the same system as above: 10y 3x 14 4y 4 In this case, we chose to elimintate the x s since their coefficients are already opposite signs. Next, find the LCM for the x s coefficients. In this case, it is 6. Decide what each equation needs to get multiplied by to make the coefficients of either the x s or the y s the same numbers with opposite signs. For our example above, we will multiply the top equation by two and the bottom equation by three to get: 20y 6x 28 12y 6x 12 If we add the two equations together, we eliminate the x. This will allow us to solve for y. 32y 16 y 0.5 Finally, go back and substitute 0.5 for y in either original equation: 10(0.5) 3x x 14 3x 9 x 3 The solution to this system of equations is the coordinate pair (-3, 0.5). You can check your solution by substituting the coordinate pair you found into both original equations. If both of the original equations balance, then you have found the correct ordered pair. NOTE: When both variables drop out of the equation, you either have no solution (false statement) ex. 5 = 0, or infinitely many solutions (true statement) ex. 3=3. 9.

10 Solve the following systems of equations by substitution: 33. y = 3x x + 2y = 10 y = -x + 3 3x 2y = -2 Solve the following systems of equations by elimination: 35. 7x 3y = x 7y = 4 y = 12 3x + y = -10 Examples Use either substitution or elimination. Determine which one would be easiest before beginning. 37. y = x + 2y = 4 x + y = 15 x + 2y = x + 3y = x + 2y = 14 9y = 35 3y =

11 41. x + y = x y = 1 -x + 2y = 13 y = x = 4y y = 10 x 4y = 2 5x 4y = y = -4x x 2y = -2 3x + 5y = -19 6x 4y =

12 SOLVING QUADRATIC EQUATIONS At this point, we have learned two ways to solve quadratic equations: ZERO PRODUCT PROPERTY QUADRATIC FORMULA Must be factored & set equal to zero. Can be used to solve any quadratic equation. ZERO PRODUCT PROPERTY: 1. Set equation equal to zero. 2. Factor completely. 3. Set each factor equal to zero and solve. Solve for x. Example: x 2 4x = x 2 4x 21= 0 2. (x 7)(x +3) = 0 3. x 7 = 0 and x + 3 = 0 x = 7 x =

13 QUADRATIC FORMULA: 1. Set equation equal to zero. QUADRATIC FORMULA: 2. Apply quadratic formula. ax 2 +bx + c = 0 EXAMPLE: 1. 3x 2 + 4x 7 = 0 Solve for x

14 Area and Perimeter Find the area and perimeter of each of the following figures. All measurements are in feet. Show all work. Refer to the formula sheet for formulas

15 Find the area and circumference of each of the following circles. All measurements are in feet. Show all work. Refer to the formula sheet for formulas