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Foundation Lesson III Literal Equations Manipulating Variables and Constants A literal equation is one which is expressed in terms of variable symbols (such as d, v, and a) and constants (such as R, g, and π). Often in science and mathematics you are given an equation and asked to solve it for a particular variable symbol or letter called the unknown. The symbols which are not the particular variable we are interested in solving for are called literals, and may represent variables or constants. Literal equations are solved by isolating the unknown variable on one side of the equation, and all of the remaining literal variables on the other side of the equation. Sometimes the unknown variable is part of another term. A term is a combination of symbols such as the products ma or πr2. In this case the unknown (such as r in πr2) must factored out of the term before we can isolate it. The following rules, examples, and exercises will help you review and practice solving literal equations from physics and chemistry. PROCEDURE In general, we solve a literal equation for a particular variable by following the basic procedure below. 1. Recall the conventional order of operations, that is, the order in which we perform the operations of multiplication, division, addition, subtraction, etc.: a. Parenthesis b. Exponents c. Multiplication and Division d. Addition and Subtraction This means that you should do what is possible within parentheses first, then exponents, then multiplication and division from left to right, then addition and subtraction from left to right. If some parentheses are enclosed within other parentheses, work from the inside out. 2. If the unknown is a part of a grouped expression (such as a sum inside parentheses), use the distributive property to expand the expression. 3. By adding, subtracting, multiplying, or dividing appropriately, (a) move all terms containing the unknown variable to one side of the equation, and (b) move all other variables and constants to the other side of the equation. Combine like terms when possible. 4. Factor the unknown variable out of its term by appropriately multiplying or dividing both sides of the equation by the other literals in the term. Laying the Foundation in Biology 69 Foundation Lesson III 5. If the unknown variable is raised to an exponent (such as 2, 3, or ½), perform the appropriate operation to raise the unknown variable to the first power, that is, so that it has an exponent of one. EXAMPLES 1. F = ma . Solve for a. F = ma Divide both sides by m: F =a m Since the unknown variable (in this case a) is usually placed on the left side of the equation, we can switch the two sides: a= F m 2. PV 1 1 = P2 V2 . Solve for V2. PV 1 1 = P2 V2 Divide both sides by P2: PV 1 1 = V2 P2 V2 = PV 1 1 P2 d . Solve for t. t Multiply each side by t: 3. v = tv = d Divide both sides by v: t= 70 d v Laying the Foundation in Biology Foundation Lesson III 4. PV = nRT . Solve for R. PV = nRT Divide both sides by n: PV = RT n Divide both sides by T: PV =R nT PV R= nT 5. R = ρL A ρL R= A . Solve for L . Multiply both sides by A: RA = ρ L Divide both sides by ρ: RA ρ L= =L RA ρ 6. A = h ( a + b ) . Solve for b. Distribute the h: A = ha + hb Subtract ha from both sides: A − ha = hb Laying the Foundation in Biology 71 Foundation Lesson III Divide both sides by h: A − ha =b h A − ha b= h 7. P = P0 + ρ gh . Solve for g. Subtract P0 from both sides: P − P0 = ρ gh Divide both sides by ρh: P − P0 =g ρh P − P0 g= ρh 1 8. U = QV . Solve for Q. 2 Multiply both sides by 2: 2U = QV Divide both sides by V: 2U =Q V 2U Q= V 1 2 kx . Solve for x. 2 Multiply both sides by 2: 9. U = 2U = kx 2 Divide both sides by k: 2U = x2 k 72 Laying the Foundation in Biology Foundation Lesson III Take the square root of both sides: 2U =x k x= 2U k L . Solve for L . g Divide both sides by 2π: 10. T = 2π T L = 2π g Square both sides: T2 L = 2 4π g Multiply both sides by g: gT 2 =L 4π 2 gT 2 L= 4π 2 Gm1m2 . Solve for r. r2 Multiply both sides by r2: 11. F = Fr 2 = Gm1m2 Divide both sides by F: r2 = Gm1m2 F Take the square root of both sides: r= Gm1m2 F Laying the Foundation in Biology 73 Foundation Lesson III 12. hi s = − i . Solve for so. ho so Cross-multiply: hi s o = − ho si Divide both sides by hi: s0 = − 13. ho si hi 1 1 1 1 = + + . Solve for R3. REQ R1 R2 R 3 Subtract 1 1 from both sides: + R1 R2 1 1 1 1 − − = REQ R1 R2 R 3 Take the reciprocal of both sides: 1 = R3 1 1 1 − − REQ R1 R2 R3 = 1 1 1 1 − − REQ R1 R2 This equation could be solved further with several more algebraic steps. 14. F = qvB sin θ . Solve for θ. Divide both sides by qvB: F = sin θ qvB Take the inverse sine of both sides: F qvB θ = sin −1 74 Laying the Foundation in Biology Foundation Lesson III 15. µ mg cos θ = mg sin θ . Solve for µ. Divide both sides by mgcos θ: µ= mg sin θ sin θ = = tan θ mg cos θ cos θ Laying the Foundation in Biology 75 Foundation Lesson III Name _____________________________________ Period _____________________________________ Literal Equations Manipulating Variables and Constants EXERCISES Directions: For each of the following equations, solve for the variable in bold print. Be sure to show each step you take to solve the equation for the bold variable. 1. v = at 2. P = F A 3. λ = h p 4. F ( ∆t ) = m∆v 5. U = Gm1m2 r 6. C = 5 ( F − 32 ) 9 7. v 2 = v0 2 + 2a∆x 8. K avg = 9. K = 76 3 kBT 2 1 2 mv 2 Laying the Foundation in Biology Foundation Lesson III 3RT M 10. vrms = 3k BT 11. vrms = 12. F = µ Kq1q2 4πε 0 r 2 1 13. 1 1 1 + = si so f 14. 1 1 1 = + CEQ C1 C2 4 15. V = π r 3 3 1 16. P + Dgy + Dv 2 = C 2 1 17. P + Dgy + Dv 2 = C 2 1 18. x = x0 + v0t + at 2 2 19. n1 sin θ1 = n2 sin θ2 M +m 20. mg sin θ = µ mg cos θ m Laying the Foundation in Biology 77