1. Math 140 Quiz 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) Solve the system. (7/3)x + (5/4)y = 4 (a) (5/6)x - 2y = 21 (b) 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)

3. Problem 2 (33) Determine the number of solutions for the given system without solving the system. 4x - 3y = 5 (a) 16x - 12y = 20 (b) 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.

4. Problem 3 (33) Determine the number of solutions for the given system without solving the system. 3x + 3y = -2 (a) 12x + 12y = 3 (b) 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.

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

6. Problem 5 (43) 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? Let speeds be p for plane & w for wind. Then, 5(p + w) = 1190 (a) 7(p - w) = 1190 (b) After dividing (a) by 5 & (b) by 7, subtract them. 2w = 238 – 170 = 68 w = 34 miles per hour

7. Problem 6 (24) Solve the system of equations. Use augmented matrix for system & manipulate to row echelon form by row operations. R 3 = -r 1 + r 3, R 1 = r 1 /2, R 2 = -5r 1 + r 2 R 2 = -r 2 /14, R 3 = 3r 2 + r 3

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

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

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

11. Problem 9 (29) Perform in order (a), (b), and (c) on the augmented matrix. (a) R 2 = -2r 1 + r 2 (b) R 3 = 4r 1 + r 3 (c) R 3 = 3r 2 + r 3

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

13. Problem 11 (38) Find the value of the determinant. Use, e.g., (a) R 1 = r 1 + r 3, (b) R 3 = -4r 2 + r 3, & expand down column 2.

14. Problem 12 (81) Use the properties of determinants to find the value of the second determinant, given the value of the first. Note the matrix in D2 D2 differs from that in D1 D1 only by (a) a row swap R1 R1 = r 3, R3 R3 = r 1 ; & (b) by the factor of 3 in row 1 of D 2. In the R1 R1 row expansion of D2 D2 these yield a (-1) overall & an overall factor of 3 compared to the R 3 row expansion of D 1. Details are on next slide. The result is D 2 = -3D 1 = 12.

15. Problem 12 cont’d (81) Use the properties of determinants to find the value of the second determinant, given the value of the first. The R 1 row expansion of D 2 is: Since & above is D 1 ’s R 3 row expansion.

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

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

18. Problem 15 (10) Solve the system. x 2 + y 2 = 100 (a) x + y = 2 (b) 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: x 2 + (-x + 2) 2 = 100 => 2x 2 –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)}

19. Problem 16 (10) Solve the system. xy = 20 (a) x + y = 9 (b) 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 => x 2 - 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.

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

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

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

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

24. Problem 20 (33) 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) w h 3 3 Let sides of tin be h & w. Then,

25. Problem 20 cont’d (29) hw = 736 (a) 3(h - 6)(w - 6) = 1200 (b) Solve (c) for w = 62 - h & substitute in (a), h (62 -h) = 736 => h 2 - 62 h +736 = (h -16) (h -46) =0 So h = 16 or 46 & w = 46 or 16. Tin is 16 in. by 46 in. 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)