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Math 60 Test # 2 Fall 2014 Chapter 3-4 Instructor: Smith-Subbarao _______Solutions__________________ Name Score: _______________ Percent: ______________ Directions: • Show all work and circle/box your answers. • Partial credit may be given, even if the answer is incorrect, if your work is clear – attach additional scratch pages you wish to be considered. • If you do not show your work, you may not get credit. • Unless otherwise instructed, leave all answers as fraction; improper fractions are OK. • Not all problems are worth the same number of points NO: Telephones, Books, Notes; CALCULATORS ARE NOT ALLOWEED Suggestions: • Choose the problems you understand best to work first. • If you get stuck, write down what you do understand for partial credit and move on • Show your work clearly • Check your solutions • Evaluate your solutions for “reasonableness” 1. Find the slope, y intercept, and equation of the line that goes through the points (-3, 1) and (-1, 3) and graph it on the coordinate plane. (15 points) Slope __1_______ Intercept ____(0, 4)_______ Equation _____y = x + 4_______ 11 10 9 8 7 6 5 4 Line 1 Line 2 y 3 2 1 0 -7 -6 -5 -4 -3 -2 -1 -1 0 1 2 3 4 5 6 7 -2 -3 -4 -5 x Then, make a table of values and graph the following equation: y = -x + 2 Find a point that solves both equations. (-1, 3) Scoring: slope 2pts, intercept 2 pts, equation 2 pt, graph part 1 2 pts, table 2 pts, 2nd graph 2 pts, solution 3 pts. Note, the intersection of the two lines is the solution – you should have put them on the same graph. 2. Find the solution to the following system of equations by substitution: (10 pts) 3x – z = 4x 2z + 4 = x + 1 The first becomes –z = x, substituting the second is -2x + 4 = x + 1 -3x = -3 x=1 z=1 To get full credit you had to use the method of substitution. 3. Determine whether the given pairs of lines is parallel, perpendicular, or neither. (10 pts) a. x – 4y = 5 b. 8x – 5y = 10 y – 8 = 1 - 4x 3x = 35 – 3y y = x/4 – 5/4 y = 8/5 x – 2 y = -4x + 9 y = -x – 35/3 Slopes are negative reciprocals Slopes have no relationship Perpendicular Neither 4. Is the ordered pair (1, -2) a solution to the following systems of equations? (10 pts) a. 3x – y = 5x y + 8 = 2(x+ 2) b. 10x – 5 = -5(y+ 1) x=y–1 3(1) – (-2) = 5(1) 10(1) – 5 = -5(-2 + 1) 3 + 2 = 5 OK 10 – 5 = 5 OK (-2) + 8 = 2 (1 + 2) 1 = (-2) – 1 6 = 6 OK 1 = -3 Not a solution Is a solution for both, so is solution Is not a solution of the pair 5. Find the solution to the following system of equations by addition (elimination): (10 pts) 5x – 3y = 10 3x – 2y = 5 Multiply the first equation by 2 and the second by 3 10x – 6y = 20 9x – 6y = 15 Subtract the 2nd from the 1st x=5 substitute to find y into the 1st: 25 – 3y = 10, y = 5 Y = 5, y = 5 Find the solution to the systems of equations in problems 6 and 7 using any method, (5 pts each) 6. x – 2y = 10 4y = 2(x – 2) + 2 = 2x – 4 + 2 = 2x - 2 From the first, x = 10+2y, substituted in the second 4y = 2(10 + 2y) – 2 4y = 20 + 4y – 2 0 = 18 NO SOLUTION is possible 7. x/2 + y/3 + 1/2 = 2 x + 2y = 3 Multiply the first by 6 3x + 2y + 3 = 12 or 3x + 2y = 9 x = 3 – 2y from the second, substituted into the equation above 3(3 – 2y) + 2y = 9 9 – 6y + 2y = 9 y=0 x = 3 -2y = 3 x = 3, y = 0 8. Two angles are complimentary. One is four times larger than the other. What are the measures of the two angles? (5 pts) x = 4y x + y = 90 since they are complimentary angles 4y + y = 90 y = 18, x = 72 9. Solve for x: (5 pts) 5(x+1) – 3x < 4x – 1 5x + 5 – 3x < 4x – 1 2x + 5 < 4x – 1 6 < 2x x>3 10. A student solved |3x – 5| = 22 as follows: (5 pts) |3x – 5| = 22 3x - 5 = 22 3x = 27 x=9 What errors, if any, did he make? If none, write none. If he erred, give the correct solution Need to also solve 3x – 5 = -22 3x = -17, x = -17/3 x = 9 or x = -17/3 to get full credit you needed to find the second solution 11. Solve for x: (10 pts) x/3 < x/2 + 1< 2x Break into two inequalities: x/3 < x/2 + 1 and x/2 + 1 < 2x For the first, multiply by 6 2x < 3x + 6, or x > -6; if you got to –x < 6, you needed to change the orientation if you divided by a negative number. For the second, multiply by 2: x + 2 < 4x 3x > 2, x > 2/3 We have x > -6 and x > 2/3. The only values which satisfy all inequalities are x > 2/3. If you did not come to that conclusion, you lost a point 12. Karen invested part of her $10,000 in savings bonds at 5% simple interest and the rest in stocks at 7% simple interest. If she receives $650 a year in interest, how much did she invest in each account? (10 pts) Let S be the amount is stocks, B the amount in bonds S + B = 10,000 0.05 S + 0.07 B = 650 Multiply by 100: 5S + 7B = 65000 From S + B = 10,000, we have S = 10,000 – B; substitute 5(10,000 – B) + 7B = 65000 50,000 + 2B = 65000 2B = 15000, B = 7500 10,000 – 7500 = 2500 = S