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Math 140 Quiz 5 - Summer 2006 Solution Review

Math 140 Quiz 5 - Summer 2006 Solution Review. (Small white numbers next to problem number represent its difficulty as per cent getting it wrong.). Problem 1 (05). Using the method of substitution, select an equation, say (b), to solve for a selected variable, say, y :

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Math 140 Quiz 5 - Summer 2006 Solution Review

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  1. Math 140Quiz 5 - Summer 2006 Solution Review (Small white numbers next to problem number represent its difficulty as per cent getting it wrong.)

  2. Problem 1 (05) Using the method of substitution, select an equation, say (b), to solve for a selected variable, say, y: y = (5/12)x - (21/2). (c) Then, substitute this for y in (a) & solve for x: (7/3)x + (5/4) [(5/12)x - (21/2)] = 4 Hence, x = (4+105/8)/[(7/3)+(5/4) (5/12)] = 6. Put this in (c) to get: y = (5/12)(6) - (21/2) = -8. (6, -8) Solve the system. (7/3)x + (5/4)y = 4 (a) (5/6)x - 2y = 21 (b)

  3. Problem 2 (33) Replace (b) by (b) minus 4 times (a): 0 = 0. This is indicates a consistent system with an infinite number of solutions. Check by noting (b) is a line of same slope (4/3) and y-intercept (–5/3) as (a). Thus, it is the same line. Determine the number of solutions for the given system without solving the system. 4x - 3y = 5 (a) 16x - 12y = 20 (b)

  4. Problem 3 (33) Replace (b) by: (b) minus 4 times (a) => 0 = 11 !!! This is indicates an inconsistent system with no solution. Note: (a) & (b) are lines of same slope (-4/3) but differing y-intercepts (2/3) & (-1/4). Thus, they are parallel lines and do not intersect. Determine the number of solutions for the given system without solving the system. 3x + 3y = -2 (a) 12x + 12y = 3 (b)

  5. Problem 4 (19) Substitute for y in (a) the y given in (b) & solve for x. Use a calculator to solve the system of equations. y = 2.12x - 31 (a) y = -0.7x + 24 (b) Then, from (b): y = -0.7(33.09153) + 24 = 0.83593

  6. Problem 5 (43) Let speeds be p for plane & w for wind. Then, 5(p + w) = 1190 (a) 7(p - w) = 1190 (b) Solve: A twin-engine aircraft can fly 1190 miles from city A to city B in 5 hours with the wind and make the return trip in 7 hours against the wind. What is the speed of the wind? After dividing (a) by 5 & (b) by 7, subtract them. 2w = 238 – 170 = 68 w = 34 miles per hour

  7. Problem 6 (24) Use augmented matrix for system & manipulate to row echelon form by row operations. Solve the system of equations. R3 = -r1 + r3, R1 = r1/2, R2 = -5r1 + r2 R2 = -r2/14, R3 = 3r2 + r3

  8. Problem 6 cont’d (24) Use augmented matrix for system & manipulate to row echelon form by row operations. Solve the system of equations. z = -2 y = -7/2 - z/4 = -7/2 - (-2)/4 = -3 R3 = r3/(14/19) x = -4 - z/2 - 2y = -4 - (-2)/2 -2(-3) = 3

  9. Problem 7 (0!) The augmented matrix for the system is obtained by just copying the coefficients in the standard equations. Write the augmented matrix for the system.

  10. Problem 8 (0!) The standard equations are obtained from the augmented matrix for the system by just copying the coefficients into their places. Write a system of equations associated with the augmented matrix. Do not try to solve.

  11. Problem 9 (29) Perform in order (a), (b), and (c) on the augmented matrix. (a) R2 = -2r1 + r2 (b) R3 = 4r1 + r3 (c) R3 = 3r2 + r3

  12. Problem 10 (19) Find the value of the determinant.

  13. Problem 11 (38) Use, e.g., (a) R1 = r1 + r3, (b) R3 = -4r2 + r3, & expand down column 2. Find the value of the determinant.

  14. Problem 12 (81) Note the matrix in D2 differs from that in D1 only by (a) a row swap R1 = r3, R3 = r1; & (b) by the factor of 3 in row 1 of D2. In the R1 row expansion of D2 these yield a (-1) overall & an overall factor of 3 compared to the R3 row expansion of D1. Details are on next slide. The result is D2 = -3D1 = 12. Use the properties of determinants to find the value of the second determinant, given the value of the first.

  15. Since & above is D1 ’s R3 row expansion. Problem 12 cont’d (81) The R1 row expansion of D2 is: Use the properties of determinants to find the value of the second determinant, given the value of the first.

  16. Problem 13 (05) Construct & evaluate the determinants: Use Cramer's rule to solve the linear system. Then, x = Dx/D = -0.41024/(-.05128) = 8 y = Dy/D = - 0.2564/(-.05128) = 5

  17. Problem 14 (10) Construct & evaluate the determinants: Use Cramer's rule to solve the linear system. Then, x = Dx/D = 3/3= 1, y = Dy/D = 15/3 = 5, ______________z = Dz/D = 3/3= 1.

  18. Problem 15 (10) Using the method of substitution, select an equation, preferably (b), to solve for a selected variable, say, y: y = -x + 2. (c) Then, substitute this for y in (a) & solve for x: x2 + (-x + 2)2 = 100 => 2x2 –4x – 96 = 0. Divide by 2 & factor: (x-8)(x+6)=0 => x = 8 or -6. Put this in (c): y = -6 or 8. => {(8, -6), (-6, 8)} Solve the system. x2 + y2 = 100 (a) x + y = 2 (b)

  19. Problem 16 (10) Using the method of substitution, select an equation, preferably (b), to solve for a selected variable, say, y: y = - x + 9. (c) Then, substitute this for y in (a) & solve for x: x(- x + 9) = 20 => x2- 9x + 20 = 0. Factor: (x – 4)(x – 5) = 0 => x = 4 or 5. Put this in (c): y = 5 or 4. => {(4, 5), (5, 4)} See graph. Solve the system. xy = 20 (a) x + y = 9 (b)

  20. Graph of (a) & (b) Zoomed in Problem 16 cont’d (10) Solution is: {(4, 5), (5, 4)}. Solve the system. xy = 20 (a) x + y = 9 (b)

  21. Graph of (a) & (b) Problem 17 (10) Using the method of elimination, add & subtract the equations: 2x2 = 50 and 2y2 = 0. Thus, solving this for x & y: x = +5 and y = 0. {(-5, 0), (5, 0)} Solve the system. x2 + y2 = 25 (a) x2 – y2 = 25 (b)

  22. Graph of (a) & (b) Problem 18 (33) Using the method of elimination, subtract the equations & back substitute result: x2 = 25 and 25 + y2 = 41. Thus, solving this for x & y: x = +5 and y = + 4. {(-5, 4), (5, 4), (-5, -4), (5, -4)} Solve the system. 2x2 + y2 = 66 (a) x2 + y2 = 41 (b)

  23. Graph of (a) & (b) Problem 19 (52) Using the method of elimination, add the equations, simplify, & back substitute result: x2+ y2 = 5 and xy = 2. Then, solving this for x & y: y = 2/x => x2+ (2/x)2 = 5, x4- 5x2 + 4= 0 => x2 = 1 or 4. {(-1, -2), (1, 2), (-2, -1), (2, 1)} Solve the system. x2 - xy + y2 = 3 (a) 2x2 + xy + 2y2 = 12 (b)

  24. 3 3 w h Problem 20 (33) Let sides of tin be h & w. Then, Solve: A rectangular piece of tin has area 736 in.2. A square of 3 in. is cut from each corner, and an open box is made by turning up the ends and sides. If the volume of the box is 1200 in.3, what were the original dimensions of the piece of tin? Tin area: hw = 736 (a) Box volume: 3(h - 6)(w - 6) = 1200 (b)

  25. hw = 736 (a) 3(h - 6)(w - 6) = 1200 (b) Problem 20 cont’d (29) After dividing (b) by 3 & expanding (b)’s product hw- 6w - 6h +36 = 400. Simplifying this with use of (a) we replace (b) with: h + w = 62 (c) Solve (c) for w = 62 - h & substitute in (a), h (62 -h) = 736 => h2 - 62 h +736 = (h -16) (h -46) =0 So h = 16 or 46 & w = 46 or 16. Tin is 16 in. by 46 in.

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