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1. Twice the cube of a number 2. The square of a number decreased by ten 3. The sum of three times a number and seven 4. Seven less than one third a number.

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Presentation on theme: "1. Twice the cube of a number 2. The square of a number decreased by ten 3. The sum of three times a number and seven 4. Seven less than one third a number."— Presentation transcript:

1 1. Twice the cube of a number 2. The square of a number decreased by ten 3. The sum of three times a number and seven 4. Seven less than one third a number 5. Three times the sum of a number and seven 6. If n represents an odd integer, what Is the next consecutive odd integer Algebra II 1

2 Literal Equations Algebra II

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6 1. 3x – 2y = 10 -2y = -3x + 10 y = 3/2x – 5 2. ½x + 3y = 5 3y = - ½x + 5 y = - 1/6x + 5/3 3. 10 = 3x – ¾ y -3x + 10 = - ¾y 4x – 40/3 = y 4. 4x + 8y = 17 8y = -4x + 17 y = -1/2x + 17/8 Algebra II 6

7 5. 6x – 5y = 13 y = 6/5x – 13/5 6. 2/3x + 3/2y = 10 y = -4/9x + 20/3 7. 4/3x – 2y = 12 y = 2/3 x – 6 8. 3x – 8y = 4 y = 3/8x – ½ Algebra II 7

8 Ax + By = C is solved for C in terms of A, x, B, and y ½bh = A Is solved for A in terms of b and h. Algebra II 8

9 1. What is P = 2L + 2w solved for? In terms of? Solved for P in terms of L and W 2. What is C = 2πr solved for? In terms of? Solved for C in terms of r 3. What are you finding in A = (b)(h)? In terms of? Finding A in terms of b and h 4. What are you finding in b = A/h ? In terms of? Finding b in terms A and h Algebra II 9

10 1. If A = c, 3t find t in terms of A and c. A = c 3t A = (3t)(c) A = 3tc A = t 3c 2. If 3L + 2W = 6, find w in terms of L. 3L + 2W = 6 2w = -3L + 6 w = -3/2 L + 3 Algebra II 10

11 3. If c = 4a 2, find a in terms of c. 4. If A = πrL + πr 2, Find L in terms of A and r. A - πr 2 = πrL A - πr 2 = L πr A – r = L πr Algebra II 11

12 5. If E=mc 2, find m in terms of c and E. 6. If V = 1/3πr 2 h, find r in terms of V and h. 12 Algebra II

13 7. If find p in terms of m, L, and q. 8. If A = 1/2bh, find h in terms of A and b. A = ½bh 2A = bh 2A = h b 13 Algebra II

14  Solve for y: 5x + 2/3 y = 10  Solve for x. 3x - 4y = 12  Solve for r in terms of V and h: V = 4 / 3 πr 2 h 14 Algebra II

15 Literal Equations Algebra II

16  What is P in terms of right now?  L and W  What should you replace? LL W = L + 3 W – 3 = L P = 2L + 2W P = 2(w – 3) + 2w P = 4w – 6 16 Algebra II

17  What is A in terms of right now?  b and h  What should you replace? BB b = 3h A = ½bh P = ½ (3h)(h) P = 3/2 h 2 17 Algebra II

18  What is V in terms of right now?  r and h  What should you replace? hh r = 1/2h 2r = h V = 1/3πr 2 h V = 1/3πr 2 (2r) V = 2/3πr 3 18 Algebra II

19  What is the Area in terms of now? rr  What should you replace? RR C = 2πr R = C/2π A = π(r) 2 A = π(C/2π) 2 A = C/(4π) 19 Algebra II

20  What is the Area in terms of now?  b and h  What should you replace? HH A = ½ bh A = ½ b(√3/2)b A = √3b 2 /4 20 Algebra II

21  h=d V=πr 2 h  What is V in terms of now?  Radius and height  What should we replace?  Height h=d h=2r V=πr 2 h V=πr 2 (2r) V=2πr 3 21 Algebra II

22  h=5 + 3r S=2πrh + 2πr 2  What is S in terms of now?  Radius and height  What should we replace?  Height h=5+3r  S= 2πrh + 2πr 2 S= 2πr(5 + 3r) + 2πr 2 S= 10πr + 6πr 2 +2πr 2 S= 10πr + 8πr 2 22 Algebra II

23  Solve for y: 1/3 x + 2/5 y = 10.  Given 4n + 1/2 m = 12, find m in terms of n.  If the area of a square can be found by A = s 2 and the perimeter of a square can be found by p = 4s, find area of a square (A) in terms of its perimeter (p). 23 Algebra II


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