Free Math Worksheets

- Use the inverse operation to isolate the placeholder. - A denominator cannot equal zero. - Identify which variable enters a denominator after rearranging.

1. a) Divide by \(h\): \(\square=\frac Ah\), where \(h\ne0\). 2. b) The placeholder is already isolated: \(\square=\frac st\), where \(t\ne0\). 3. c) Divide by \(r\): \(\square=\frac Cr\), where \(r\ne0\).

a) \(\square=\frac Ah\), where \(h\ne0\) b) \(\square=\frac st\), where \(t\ne0\) c) \(\square=\frac Cr\), where \(r\ne0\)

- Identify the known quantities and the variable being solved for. - Clear the denominator before isolating \(t\). - Compare how much algebra is repeated when several work values are used.

1. Solve \(P=\frac{W}{t}\) for \(t\): multiply by \(t\) to get \(Pt=W\), then divide by \(P\) to get \(t=\frac{W}{P}\). 2. Substitute the first values: \(t=\frac{2500\,\text{J}}{500\,\text{W}}=5\,\text{s}\). 3. Solving the formula once creates a general expression. Each additional work value can be substituted directly into \(t=\frac{W}{P}\) without repeating the algebra.

a) \(t=5\,\text{s}\) b) Solving for \(t\) once gives \(t=\frac{W}{P}\), so each additional value of \(W\) can be substituted directly.

- Use inverse operations to isolate the requested variable. - Clear a fraction by multiplying by its denominator. - Treat all other variables as constants.

1. a) Multiply by \(2\): \(2A = (a + c)h\). Divide by \(h\): \(\frac{2A}{h} = a + c\). Subtract \(a\): \(c = \frac{2A}{h} - a\). 2. b) Multiply by \(t\): \(vt = s_2 - s_1\). Rearranging gives \(s_1 = s_2 - vt\). 3. c) Multiply by \(A\): \(pA = F\). Divide by \(p\): \(A = \frac{F}{p}\). 4. The required conditions are \(h \ne 0\) in a), \(t \ne 0\) in b), and \(p \ne 0\) and \(F \ne 0\) in c). In c), these conditions ensure that the resulting value of \(A\) is nonzero, as required by the original denominator.

a) \(c = \frac{2A}{h} - a\), for \(h \ne 0\) b) \(s_1 = s_2 - vt\), for \(t \ne 0\) c) \(A = \frac{F}{p}\), for \(p \ne 0\) and \(F \ne 0\)

- Clear numerical fractions before isolating the placeholder. - When the placeholder is a denominator, its final value must be nonzero. - Treat a grouped expression such as \(a+c\) as one factor.

1. a) Multiply by 3 and divide by \(G\): \(\square=\frac{3V}{G}\). For a unique result, \(G\ne0\). 2. b) Multiply by the placeholder and divide by \(\rho\): \(\square=\frac m\rho\). A unique, defined result requires \(\rho\ne0\) and \(m\ne0\), because the placeholder is an original denominator. 3. c) Multiply by 2 and divide by \(a+c\): \(\square=\frac{2A}{a+c}\). For a unique result, \(a+c\ne0\). 4. d) Multiply by the placeholder and divide by \(T\): \(\square=\frac{pV}{T}\). A unique, defined result requires \(T\ne0\), \(p\ne0\), and \(V\ne0\).

a) \(\square=\frac{3V}{G}\), where \(G\ne0\) b) \(\square=\frac m\rho\), where \(\rho\ne0\) and \(m\ne0\) c) \(\square=\frac{2A}{a+c}\), where \(a+c\ne0\) d) \(\square=\frac{pV}{T}\), where \(T\ne0\), \(p\ne0\), and \(V\ne0\)

- A milliliter has the same volume as a cubic centimeter. - Rearrange the density formula for mass in part b. - Convert grams to kilograms at the end.

1. a) Since \(1\,\text{mL}=1\,\text{cm}^3\), the volume is \(250\,\text{cm}^3\). 2. Compute the density: \(\rho=\frac{210\,\text{g}}{250\,\text{cm}^3}=0.84\,\text{g/cm}^3\). 3. b) Convert the new volume: \(1.2\,\text{L}=1200\,\text{cm}^3\). 4. Solve the formula for mass: \(m=\rho V\). 5. Substitute: \(m=0.84\cdot1200=1008\,\text{g}=1.008\,\text{kg}\).

a) \(0.84\,\text{g/cm}^3\) b) \(1.008\,\text{kg}\)

- Clear the denominator before isolating \(I\). - Use \(U=RI\) to analyze part c. - Replace \(R\) with \(2R\) while keeping \(I\) fixed.

1. a) Multiply by \(I\): \(RI=U\). Divide by \(R\): \(I=\frac UR\). 2. b) Substitute: \(I=\frac{12}{5}=2.4\,\text{A}\). 3. c) From \(U=RI\), replacing \(R\) with \(2R\) gives \(U_{\text{new}}=(2R)I=2(RI)=2U\). 4. Therefore, the voltage doubles.

a) \(I=\frac UR\) b) \(I=2.4\,\text{A}\) c) The voltage doubles.

- Division undoes multiplication. - Treat the other variables as constants. - State the restriction on every quantity used as a divisor.

1. a) Divide by \(I\): \(U=\frac PI\), where \(I\ne0\). 2. b) Divide by \(v\): \(t=\frac sv\), where \(v\ne0\). 3. c) Divide by \(m\): \(a=\frac Fm\), where \(m\ne0\).

a) \(U=\frac PI\), where \(I\ne0\) b) \(t=\frac sv\), where \(v\ne0\) c) \(a=\frac Fm\), where \(m\ne0\)

- Undo addition or subtraction before undoing multiplication. - Clear the fraction in part b first. - Treat every variable except the target as a constant.

1. a) Add 12: \(y+12=3x\). Divide by 3: \(x=\frac{y+12}{3}\). 2. b) Multiply by 3: \(3V=Gh\). Divide by \(h\): \(G=\frac{3V}{h}\), where \(h\ne0\). 3. c) Subtract \(a\) and \(c\): \(b=u-a-c\).

a) \(x=\frac{y+12}{3}\) b) \(G=\frac{3V}{h}\), where \(h\ne0\) c) \(b=u-a-c\)

- Distribute before collecting terms. - Move all terms not containing \(x\) to the other side. - Divide by the coefficient of \(x\).

1. Distribute 8: \(8x-8a=16b+24a\). 2. Add \(8a\): \(8x=16b+32a\). 3. Divide by 8: \(x=2b+4a\).

\(x=4a+2b\)

- Clear the denominator first. - Undo the subtraction of \(12a\). - Divide by the coefficient of \(y\).

1. Multiply by 3: \(4y-12a=24b+12a\). 2. Add \(12a\): \(4y=24b+24a\). 3. Divide by 4: \(y=6b+6a\).

\(y=6a+6b\)

- Compare how changing the numerator or denominator changes a quotient. - Substitute the changed quantity into the original formula. - Clear the denominator before isolating a variable.

1. Replacing \(V\) with \(4V\) gives \(\rho_{\text{new}}=\frac{m}{4V}=\frac14\rho\). The density becomes one-fourth as large. 2. Replacing \(m\) with \(\frac12m\) gives \(\rho_{\text{new}}=\frac{\frac12m}{V}=\frac12\rho\). The density is halved. 3. Multiply by \(V\): \(m=\rho V\). 4. Divide \(m=\rho V\) by \(\rho\): \(V=\frac m\rho\). For a unique valid value in the original formula, \(\rho\ne0\) and \(m\ne0\).

1) \(\rho_{\text{new}}=\frac14\rho\) 2) \(\rho_{\text{new}}=\frac12\rho\) 3) \(m=\rho V\) 4) \(V=\frac m\rho\), where \(\rho\ne0\) and \(m\ne0\)

- Clear the denominator first. - Treat variables other than the target as constants. - Reverse the operations one at a time.

1. Multiply by \(c\): \(Pc=(a+b)d\). 2. a) Divide by \(d\): \(\frac{Pc}{d}=a+b\). Subtract \(b\): \(a=\frac{Pc}{d}-b\). 3. b) Divide \(Pc=(a+b)d\) by \(a+b\): \(d=\frac{Pc}{a+b}\). 4. c) Divide \(Pc=(a+b)d\) by \(P\): \(c=\frac{d(a+b)}{P}\).

a) \(a=\frac{Pc}{d}-b\) b) \(d=\frac{Pc}{a+b}\) c) \(c=\frac{d(a+b)}{P}\)

- Treat \(k\) as a fixed quantity while isolating \(z\). - Move the \(k\)-term away from the \(z\)-term first. - Divide every term by the coefficient of \(z\).

1. Treat \(k\) as a constant. Subtract \(4k\) from both sides: \(0.8z = 8k - 1.6\). 2. Divide every term by \(0.8\): \(z = \frac{8k - 1.6}{0.8}\). 3. Simplify: \(z = 10k - 2\).

\(z = 10k - 2\)

- Clear the denominator first. - Isolate the expression \(x-b\). - Add \(b\) after dividing by \(c\).

1. Multiply both sides by \(x-b\): \(a=c(x-b)\). 2. Divide by \(c\): \(\frac ac=x-b\). 3. Add \(b\): \(x=\frac ac+b\). 4. Since \(a>0\) and \(c>0\), \(\frac ac>0\), so the result satisfies \(x>b\).

\(x=\frac ac+b\), equivalently \(x=\frac{a+bc}{c}\)

- Combine the fractions before taking a reciprocal. - When the requested variable is in a denominator, clear the denominator first. - Check every original denominator after solving.

1. a) Subtract \(\frac{1}{b}\): \(\frac{1}{g} = \frac{1}{f} - \frac{1}{b} = \frac{b - f}{fb}\). Taking reciprocals gives \(g = \frac{fb}{b - f}\). 2. The restrictions for a) are \(f \ne 0\), \(b \ne 0\), and \(b - f \ne 0\). 3. b) Multiply by \(R_i + R_a\): \(I(R_i + R_a) = U\). Divide by \(I\), then subtract \(R_a\): \(R_i = \frac{U}{I} - R_a\). 4. The restrictions for b) are \(I \ne 0\) and \(U \ne 0\). These ensure that the resulting denominator \(R_i + R_a = \frac{U}{I}\) is nonzero in the original formula.

a) \(g = \frac{fb}{b - f}\), for \(f \ne 0\), \(b \ne 0\), and \(b - f \ne 0\) b) \(R_i = \frac{U}{I} - R_a\), for \(I \ne 0\) and \(U \ne 0\)

- Isolate \(\frac1y\) before taking a reciprocal. - Combine \(1-\frac1x\) into one fraction. - A contradiction means no value satisfies the condition.

1. a) If \(x=2\), then \(\frac12+\frac1y=1\), so \(\frac1y=\frac12\) and \(y=2\). 2. If \(x=5\), then \(\frac15+\frac1y=1\), so \(\frac1y=\frac45\) and \(y=\frac54\). 3. b) Isolate the reciprocal: \(\frac1y=1-\frac1x=\frac{x-1}{x}\). Taking reciprocals gives \(y=\frac{x}{x-1}\), where \(x\ne0,1\). 4. c) Setting \(y=1\) gives \(1=\frac{x}{x-1}\). Multiplying by \(x-1\) produces \(x-1=x\), or \(-1=0\), a contradiction.

a) If \(x=2\), \(y=2\); if \(x=5\), \(y=\frac54\). b) \(y=\frac{x}{x-1}\), where \(x\ne0,1\) c) The condition \(y=1\) leads to the contradiction \(x-1=x\), so no solution exists.

- Translate “twice their sum” into an algebraic expression. - Solve the equation for one variable, then choose convenient values for the other. - Substitute \(a=4\) directly for part c.

1. a) The relationship is \(ab=2(a+b)\). 2. To express \(b\) in terms of \(a\), rearrange: \(ab-2b=2a\), so \(b(a-2)=2a\). For \(a\ne2\), \(b=\frac{2a}{a-2}\). 3. b) If \(a=3\), then \(b=\frac6{1}=6\), giving \((3, 6)\). If \(a=0\), then \(b=0\), giving \((0, 0)\). 4. c) Substitute \(a=4\): \(4b=2(4+b)\). Thus, \(4b=8+2b\), so \(2b=8\) and \(b=4\).

a) \(ab=2(a+b)\) b) For example, \((3, 6)\) and \((0, 0)\) c) Substituting \(a=4\) gives \(b=4\).

- Cross-multiply when two ratios are equal. - Give the left side a common denominator in part c. - Track both the original denominators and any new denominator created while solving.

1. a) Cross-multiply: \(a\cdot\square=bc\), so \(\square=\frac{bc}{a}\). The conditions are \(a\ne0\), \(b\ne0\), and \(c\ne0\). 2. b) Multiply by \(x\): \(\square=\frac{xy}{z}\). The conditions are \(x\ne0\) and \(z\ne0\). 3. c) Write \(1=\frac yy\): \(\frac xy+1=\frac{x+y}{y}\). Thus, \(\square=x+y\), where \(y\ne0\). 4. d) Cross-multiply: \(kn=m\cdot\square\), so \(\square=\frac{kn}{m}\). The conditions are \(k\ne0\), \(m\ne0\), and \(n\ne0\).

a) \(\square=\frac{bc}{a}\), where \(a,b,c\ne0\) b) \(\square=\frac{xy}{z}\), where \(x,z\ne0\) c) \(\square=x+y\), where \(y\ne0\) d) \(\square=\frac{kn}{m}\), where \(k,m,n\ne0\)

- Translate the verbal relationship into an equation. - Collect the terms containing \(y\) and factor. - Choose rational values of \(x\) other than 1.

1. The condition is \(x+y=xy\). 2. Rearrange: \(x=xy-y=y(x-1)\). Therefore, \(y=\frac{x}{x-1}\), where \(x\ne1\). 3. If \(x=2\), then \(y=2\), giving \((2, 2)\). 4. If \(x=0\), then \(y=0\), giving \((0, 0)\). 5. If \(x=-1\), then \(y=\frac12\), giving \(\left(-1, \frac12\right)\).

1) \(y=\frac{x}{x-1}\) 2) \(x\ne1\) 3) For example, \((2, 2)\), \((0, 0)\), and \(\left(-1, \frac12\right)\)

- Clear the denominator first. - Collect and factor the terms containing \(a\). - Track both the original denominator and the new factor used to divide.

1. The equation is \(2(a+b)=\frac ab\). 2. Since \(b\ne0\), multiply by \(b\): \(2ab+2b^2=a\). 3. Collect the terms containing \(a\): \(2b^2=a(1-2b)\). For \(b\ne\frac12\), \(a=\frac{2b^2}{1-2b}\). 4. The restrictions are \(b\ne0\) and \(b\ne\frac12\). 5. Choosing \(b=1\), \(b=-1\), and \(b=2\) gives \((-2, 1)\), \(\left(\frac23, -1\right)\), and \(\left(-\frac83, 2\right)\).

1) \(2(a+b)=\frac ab\) 2) \(a=\frac{2b^2}{1-2b}\) 3) \(b\ne0,\frac12\) 4) For example, \((-2, 1)\), \(\left(\frac23, -1\right)\), and \(\left(-\frac83, 2\right)\)

- Isolate \(\frac1z\) first. - Combine the fractions on the other side using a common denominator. - Take the reciprocal only after the side is written as one fraction.

1. Isolate the term containing \(z\): \(\frac1z=\frac1y-\frac1x\). 2. Combine the right side: \(\frac1z=\frac{x-y}{xy}\). 3. Take reciprocals: \(z=\frac{xy}{x-y}\). 4. The restrictions are \(x\ne0\), \(y\ne0\), and \(x-y\ne0\). Under these conditions, the resulting \(z\) is also nonzero.

\(z=\frac{xy}{x-y}\), where \(x\ne0\), \(y\ne0\), and \(x\ne y\)

- Clear the factor \(\frac13\) before isolating \(A\). - Substitute \(2h\) for \(h\) in the rearranged formula. - Compare the new expression with the original value of \(A\).

1. a) Multiply by 3: \(3V=Ah\). Divide by \(h\), where \(h\ne0\): \(A=\frac{3V}{h}\). 2. b) Replace \(h\) with \(2h\): \(A_{\text{new}}=\frac{3V}{2h}=\frac12\left(\frac{3V}{h}\right)\). 3. Therefore, \(A_{\text{new}}=\frac12A\), so the base area is halved.

a) \(A=\frac{3V}{h}\) b) The base area is halved: \(A_{\text{new}}=\frac12A\).

- Isolate the variable that is in the denominator. - Lower density means greater volume when mass is fixed. - Divide the larger volume by the smaller volume to find the factor.

1. a) Multiply by \(V\) and divide by \(\rho\): \(V=\frac m\rho\). 2. b) \(V_1=\frac{540}{9.0}=60\,\text{cm}^3\). 3. \(V_2=\frac{540}{2.7}=200\,\text{cm}^3\). 4. c) The volume factor is \(\frac{V_2}{V_1}=\frac{200}{60}=\frac{10}{3}\approx3.33\).

a) \(V=\frac m\rho\) b) \(V_1=60\,\text{cm}^3\) and \(V_2=200\,\text{cm}^3\) c) \(\frac{10}{3}\), or approximately \(3.33\)

- Multiply length, width, and thickness to find volume. - Rearrange the density formula for mass. - Round the number of trips up to the next whole number.

1. a) Compute the rectangular-prism volume: \(V=15\cdot12\cdot0.5=90\,\text{ft}^3\). 2. b) Solve the density formula for mass: \(m=\rho V\). 3. Substitute: \(m=150\cdot90=13{,}500\,\text{lb}\). 4. Since \(2000\,\text{lb}=1\) short ton, \(13{,}500\,\text{lb}=6.75\) tons. 5. c) Divide by the truck capacity: \(6.75\div1.5=4.5\). A partial trip is impossible, so at least 5 trips are needed.

a) \(90\,\text{ft}^3\) b) \(6.75\) short tons c) \(5\) trips

- Treat \(x\) like any positive number while rearranging the formula. - Set the given circumference equal to \(2\pi r\). - When squaring \(5x\), square both the coefficient and the variable.

1. Set the circumference formulas equal: \(2\pi r=10\pi x\). 2. Divide by \(2\pi\): \(r=5x\). 3. The diameter is \(d=2r=10x\). 4. Substitute the radius into the area formula: \(A=\pi(5x)^2=25\pi x^2\).

a) \(r=5x\); \(d=10x\) b) \(A=25\pi x^2\)

- Clear the denominator and isolate \(A\). - Substitute the given force and pressure. - Use the rearranged formula to compare \(A\) and \(3F\).

1. a) Multiply by \(A\): \(pA=F\). Divide by \(p\): \(A=\frac Fp\). 2. b) Substitute: \(A=\frac{450}{1500}=0.3\,\text{m}^2\). 3. c) If \(F\) becomes \(3F\), then \(A_{\text{new}}=\frac{3F}{p}=3A\). 4. Therefore, the area must triple.

a) \(A=\frac Fp\) b) \(A=0.3\,\text{m}^2\) c) The area must triple.

- Clear the denominator before isolating \(t\). - Use time equals distance divided by speed. - Substitute \(2s\) and \(4t\) into the speed formula and simplify.

1. a) Multiply by \(t\): \(vt=s\). Divide by \(v\): \(t=\frac sv\). 2. b) Substitute: \(t=\frac{90}{60}=1.5\) hours. 3. c) The new speed is \(v_{\text{new}}=\frac{2s}{4t}=\frac12\cdot\frac st\). 4. Since \(v=\frac st\), \(v_{\text{new}}=\frac12v\).

a) \(t=\frac sv\) b) \(1.5\) hours c) \(v_{\text{new}}=\frac12v\)

- Clear fractions before isolating the target variable. - Treat a grouped expression as one factor. - Divide by the complete product attached to the target variable.

1. a) Multiply by 2: \(2A=(a+c)h\). Divide by \(h\): \(\frac{2A}{h}=a+c\). Subtract \(c\): \(a=\frac{2A}{h}-c\), where \(h\ne0\). 2. b) Divide by \(Pt\): \(r=\frac{I}{Pt}\), where \(P\ne0\) and \(t\ne0\). 3. c) Subtract \(L\): \(S-L=2B\). Divide by 2: \(B=\frac{S-L}{2}\).

a) \(a=\frac{2A}{h}-c\), where \(h\ne0\) b) \(r=\frac{I}{Pt}\), where \(P\ne0\) and \(t\ne0\) c) \(B=\frac{S-L}{2}\)

- Undo addition or subtraction before division. - Isolate a grouped factor before separating its terms. - Track any expression used as a divisor.

1. a) Subtract \(n\): \(y-n=mx\). Divide by \(m\): \(x=\frac{y-n}{m}\), where \(m\ne0\). 2. b) Multiply by 2: \(2A=(a+c)h\). Divide by \(h\): \(\frac{2A}{h}=a+c\). Subtract \(a\): \(c=\frac{2A}{h}-a\), where \(h\ne0\). 3. c) Divide by \(A_0\): \(\frac{A_f}{A_0}=1+r\). Subtract 1: \(r=\frac{A_f}{A_0}-1\), where \(A_0\ne0\).

a) \(x=\frac{y-n}{m}\), where \(m\ne0\) b) \(c=\frac{2A}{h}-a\), where \(h\ne0\) c) \(r=\frac{A_f}{A_0}-1\), where \(A_0\ne0\)

- Find the amount by which Account B gains on Account A each month. - Set the two balance expressions equal. - Isolate \(n\) by factoring it from the variable terms. - Examine how increasing a positive denominator affects a fraction with fixed numerator.

1. After \(n\) months, Account A has \(x+an\), and Account B has \(bn\). 2. Set the balances equal: \(x+an=bn\). 3. Subtract \(an\): \(x=n(b-a)\). Divide by \(b-a\): \(n=\frac{x}{b-a}\). 4. Equality occurs after a whole number of months only when \(\frac{x}{b-a}\) is a nonnegative integer. 5. If \(b\) increases, the positive denominator \(b-a\) increases while \(x\) stays fixed, so \(n\) decreases.

a) \(n=\frac{x}{b-a}\). For equality after a whole number of months, \(\frac{x}{b-a}\) must be a nonnegative integer. b) Increasing \(b\) decreases \(n\), so Account B catches up sooner.

- Distribute before combining the terms containing \(a\). - Isolate \(x\) while treating \(a\) as a parameter. - Substitute \(x = 15\) into the general solution.

1. Distribute \(3\): \(3x - 6a + 5a = 14a\). 2. Combine like terms: \(3x - a = 14a\). 3. Add \(a\): \(3x = 15a\). 4. Divide by \(3\): \(x = 5a\). 5. Use the condition \(x = 15\): \(15 = 5a\). 6. Divide by \(5\): \(a = 3\).

\(x = 5a\). For \(x = 15\), the parameter must be \(a = 3\).

- Isolate \(\frac1{R_1}\) first. - Combine the other reciprocals using a common denominator. - Take the reciprocal only after the other side is one fraction.

1. Subtract \(\frac1{R_2}\): \(\frac1{R_1}=\frac1R-\frac1{R_2}\). 2. Combine the right side: \(\frac1{R_1}=\frac{R_2-R}{RR_2}\). 3. Take reciprocals: \(R_1=\frac{RR_2}{R_2-R}\). 4. The formula requires \(R\ne0\), \(R_2\ne0\), and \(R_2-R\ne0\). These conditions also make the resulting \(R_1\) nonzero.

\(R_1=\frac{RR_2}{R_2-R}\), where \(R\ne0\), \(R_2\ne0\), and \(R_2\ne R\)

- Treat \(a+c\) as one factor when analyzing proportional changes. - Clear the factor \(\frac12\) before isolating a variable. - Undo multiplication before undoing addition.

1. Replacing \(h\) with \(2h\) gives \(A_{\text{new}}=\frac{a+c}{2}(2h)=2A\). The area doubles. 2. Replacing \(a+c\) with \(\frac12(a+c)\) gives \(A_{\text{new}}=\frac12A\). The area is halved. 3. Multiply by 2 and divide by \(a+c\): \(h=\frac{2A}{a+c}\), where \(a+c\ne0\). 4. From \(2A=(a+c)h\), divide by \(h\) and subtract \(c\): \(a=\frac{2A}{h}-c\), where \(h\ne0\).

1) The area doubles. 2) The area is halved. 3) \(h=\frac{2A}{a+c}\) 4) \(a=\frac{2A}{h}-c\)

- Clear the denominator before isolating a variable. - Divide by the complete product attached to the target variable. - Convert the decimal rate to a percentage at the end.

1. a) Multiply by 360: \(360I=Prd\). Divide by \(rd\): \(P=\frac{360I}{rd}\). 2. b) From \(360I=Prd\), divide by \(Pr\): \(d=\frac{360I}{Pr}\). 3. c) Solve for the rate: \(r=\frac{360I}{Pd}\). 4. Substitute: \(r=\frac{360\cdot100}{5000\cdot180}=0.04\). 5. As a percentage, the annual rate is \(4\%\).

a) \(P=\frac{360I}{rd}\) b) \(d=\frac{360I}{Pr}\) c) \(r=0.04=4\%\)

- Substitute the known solution into the equation in part a). - In part b), replace the parameter with its given value. - Distribute before moving variable terms. - Treat whichever symbol is unknown as the variable to isolate.

1. For a), substitute \(x = 2\): \(5 \cdot 2 - k = 3(2 + 4)\). 2. Simplify: \(10 - k = 18\). Subtract \(10\): \(-k = 8\). Multiply by \(-1\): \(k = -8\). 3. For b), substitute \(k = 2\): \(5x - 2 = 3(x + 4)\). 4. Distribute: \(5x - 2 = 3x + 12\). 5. Subtract \(3x\) and add \(2\): \(2x = 14\). Divide by \(2\): \(x = 7\).

a) \(k = -8\) b) \(x = 7\)

- Expand both sides first. - Collect every term containing \(x\). - The factor used as a divisor cannot be zero.

1. Expand: \(kx-km=5x+5n\). 2. Collect the \(x\)-terms: \(kx-5x=5n+km\). 3. Factor: \(x(k-5)=5n+km\). 4. Divide by \(k-5\): \(x=\frac{5n+km}{k-5}\). 5. A unique solution requires \(k-5\ne0\), so \(k\ne5\).

\(x=\frac{5n+km}{k-5}\), where \(k\ne5\)

- Expand both sides first. - Collect the terms containing \(x\) on one side. - Combine like parameter terms before dividing.

1. Expand: \(4x-8a+3b=2b-2x-6a\). 2. Add \(2x\): \(6x-8a+3b=2b-6a\). 3. Add \(8a\) and subtract \(3b\): \(6x=2a-b\). 4. Divide by 6: \(x=\frac{2a-b}{6}\).

\(x=\frac{2a-b}{6}\)

- How long has Alex been running when Blake has run for \(t\) seconds? - At the catch-up point, what must be true about the runners' distances? - After distributing, collect all terms containing \(t\) on one side. - Factor out \(t\) before dividing.

1. When Blake has run for \(t\) seconds, Alex has run for \(t + 90\) seconds. Therefore, \(s_A = v_A(t + 90)\). 2. Blake's distance is \(s_B = v_Bt\). 3. At the catch-up point, the distances are equal: \(v_A(t + 90) = v_Bt\). 4. Distribute: \(v_At + 90v_A = v_Bt\). 5. Move the terms containing \(t\) to one side: \(90v_A = v_Bt - v_At\). 6. Factor out \(t\): \(90v_A = t(v_B - v_A)\). 7. Divide by \(v_B - v_A\): \(t = \frac{90v_A}{v_B - v_A}\).

a) \(s_A = v_A(t + 90)\) b) \(s_B = v_Bt\) c) \(t = \frac{90v_A}{v_B - v_A}\)

- Start by distributing and combining like terms. - What operations will isolate \(y\) on one side? - Once you have \(y\) in terms of \(x\), choose convenient x-values to generate ordered pairs. - Think about how many points lie on a line.

1. Distribute: \(3x + 6y - 5 = 4x + 7y - 2\). 2. Rearrange the terms: \(6y - 7y = 4x - 3x - 2 + 5\). 3. Simplify: \(-y = x + 3\), so \(y = -x - 3\). 4. For example, when \(x = 0\), \(y = -3\), giving \((0, -3)\). When \(x = 1\), \(y = -4\), giving \((1, -4)\). 5. Every real value of \(x\) produces exactly one corresponding value of \(y\). Therefore, there are infinitely many solutions, represented by all points on the line \(y = -x - 3\).

a) \(y = -x - 3\) b) For example, \((0, -3)\) and \((1, -4)\) c) Infinitely many solutions, because every real value of \(x\) gives a corresponding point on the line.

- What common denominator can eliminate both fractions? - Be careful with the subtraction when you distribute through the second expression. - For an integer solution, choose an x-value that makes the numerator divisible by \(11\).

1. Multiply the entire equation by the least common denominator, \(12\): \(4(x + 2y) - 3(2x - y) = 12\). 2. Distribute and combine like terms: \(4x + 8y - 6x + 3y = 12\), so \(-2x + 11y = 12\). 3. Solve for \(y\): \(11y = 2x + 12\), so \(y = \frac{2x + 12}{11}\). 4. Choose \(x = 5\). Then \(y = \frac{2 \cdot 5 + 12}{11} = 2\), so \((5, 2)\) is one integer solution.

a) \(-2x + 11y = 12\) b) \(y = \frac{2x + 12}{11}\) c) For example, \((5, 2)\)

- Multiply by the common denominator. - Collect all terms containing \(x\) on one side. - Factor out \(x\) before dividing.

1. Multiply by \(ab\): \(a(x-a)=b(x-b)+ab\). 2. Expand: \(ax-a^2=bx-b^2+ab\). 3. Collect the \(x\)-terms: \(ax-bx=a^2-b^2+ab\). 4. Factor: \(x(a-b)=a^2-b^2+ab\). 5. Since \(a-b\ne0\), divide to obtain \(x=\frac{a^2-b^2+ab}{a-b}\).

\(x=\frac{a^2-b^2+ab}{a-b}\)

- First solve the formula for \(t\). - To find a scale factor between the times, divide the larger time by the smaller time. - Consider how the height ratio appears under the square root.

1. Solve the formula for time: \(t = \sqrt{\frac{2h}{g}}\). 2. For \(h_1 = 45\,\text{m}\), \(t_1 = \sqrt{\frac{2 \cdot 45}{9.81}}\,\text{s} \approx 3.03\,\text{s}\). 3. For \(h_2 = 180\,\text{m}\), \(t_2 = \sqrt{\frac{2 \cdot 180}{9.81}}\,\text{s} \approx 6.06\,\text{s}\). 4. Exactly, \(\frac{t_2}{t_1} = \sqrt{\frac{180}{45}} = \sqrt{4} = 2\). Thus, multiplying the height by \(4\) doubles the fall time.

a) \(t_1 \approx 3.03\,\text{s}\); \(t_2 \approx 6.06\,\text{s}\) b) The fall time is multiplied by \(2\).

- A solution must make the equation true when substituted. - Replace every \(x\) with \(6\). - Simplify both sides before isolating \(k\). - Check the resulting value in the original equation.

1. Substitute \(x = 6\): \(5(6 - 2) + k = 2(6 + 1)\). 2. Simplify: \(20 + k = 14\). 3. Subtract \(20\): \(k = -6\).

\(k = -6\)

- Start with the area formula for a trapezoid. - Combine the expressions for the parallel sides. - Reverse multiplication by dividing to isolate the requested variable. - Check the numerical result in the simplified area formula.

1. Use the trapezoid area formula: \(A = \frac{a + b}{2}h\). 2. Substitute the parallel side lengths: \(A = \frac{3x + x}{2}h = \frac{4x}{2}h = 2xh\). 3. Divide by \(2x\) to solve for the height: \(h = \frac{A}{2x}\). 4. Substitute \(x = 5\,\text{in.}\) and \(A = 70\,\text{in.}^2\): \(h = \frac{70}{2 \cdot 5}\,\text{in.} = 7\,\text{in.}\). 5. Divide \(A = 2xh\) by \(2h\) to solve for \(x\): \(x = \frac{A}{2h}\).

a) \(A = 2xh\) b) \(h = \frac{A}{2x}\), and \(h = 7\,\text{in.}\) for the given values. c) \(x = \frac{A}{2h}\)

- Undo the multiplication by \(h\) before isolating \(c\). - Substitute the measurements only after solving the formula. - In part c, replace \(A\) with \(2A\) in the rearranged formula.

1. a) Multiply by 2: \(2A=(a+c)h\). Divide by \(h\): \(\frac{2A}{h}=a+c\). Subtract \(a\): \(c=\frac{2A}{h}-a\). 2. b) Substitute: \(c=\frac{2\cdot40}{5}-10=16-10=6\,\text{cm}\). 3. c) If \(A\) is replaced by \(2A\), then \(c_{\text{new}}=\frac{4A}{h}-a\). 4. Since \(c+a=\frac{2A}{h}\), \(c_{\text{new}}=2(c+a)-a=2c+a\). Thus, \(c\) does not simply double.

a) \(c=\frac{2A}{h}-a\) b) \(c=6\,\text{cm}\) c) \(c_{\text{new}}=2c+a\)

- Clear the denominator and collect all terms containing \(x\). - Factor out \(x\). - Before dividing by \(k-y\), consider when that expression is zero and check \(x=-1\).

1. The original expression requires \(x\ne-1\). 2. Multiply by \(x+1\): \(y(x+1)=kx\). 3. Expand and collect the \(x\)-terms: \(yx+y=kx\), so \(y=x(k-y)\). 4. If \(k-y\ne0\), then \(x=\frac{y}{k-y}\). 5. This value equals the excluded input \(-1\) exactly when \(k=0\). Therefore, the solution is unique when \(k\ne0\) and \(k\ne y\). 6. If \(k=y\ne0\), the equation has no solution. If \(k=0\) and \(y\ne0\), the only algebraic candidate is \(-1\), so there is no solution. If \(k=y=0\), every real \(x\ne-1\) is a solution.

For \(k\ne0\) and \(k\ne y\), \(x=\frac{y}{k-y}\). If \(k=y\ne0\), or if \(k=0\) and \(y\ne0\), there is no solution. If \(k=y=0\), every \(x\ne-1\) is a solution.

- Translate “2 greater than \(u\)” as \(u+2\). - Clear the denominator and factor out \(u\). - Check separately the values that make a denominator or a division factor zero.

1. For \((2, 0.5)\), \(\frac{2}{0.5}=4\) and \(2+2=4\), so the pair satisfies the condition. 2. The equation is \(\frac uv=u+2\). Multiply by \(v\): \(u=uv+2v\). 3. Rearrange and factor: \(u(1-v)=2v\). For \(v\ne1\), \(u=\frac{2v}{1-v}\). 4. The quotient requires \(v\ne0\). When \(v=1\), the original equation becomes \(u=u+2\), so it has no solution. 5. If \(v=-1\), then \(u=\frac{-2}{2}=-1\), giving \((-1, -1)\).

1) Yes. 2) \(u=\frac{2v}{1-v}\), for \(v\ne1\) 3) \(v=0\) is excluded; \(v=1\) gives no solution. 4) For example, \((-1, -1)\)

- Write a time expression for each leg using distance divided by speed. - Add the two time expressions. - Use total distance divided by total time for average speed. - Simplify the complex fraction before solving part c).

1. The outbound time is \(t_1 = \frac{12}{v_1}\). 2. The return time is \(t_2 = \frac{12}{1.5v_1} = \frac{8}{v_1}\). 3. Therefore, \(t_{\text{total}} = \frac{12}{v_1} + \frac{8}{v_1} = \frac{20}{v_1}\). 4. The total distance is \(24\) miles, so \(v_g = \frac{24}{20/v_1} = \frac{24v_1}{20} = 1.2v_1\). 5. Set the required average speed equal to the formula: \(54 = 1.2v_1\). 6. Divide by \(1.2\): \(v_1 = 45\,\text{mph}\).

a) \(t_{\text{total}} = \frac{20}{v_1}\) b) \(v_g = \frac{24}{20/v_1} = 1.2v_1\) c) \(v_1 = 45\,\text{mph}\)

- Divide by every factor multiplying the target variable. - Solve the same formula for \(P\) before analyzing the change. - Dividing by half of a quantity doubles the quotient.

1. a) Divide by \(Pr\): \(t=\frac{I}{Pr}\), where \(P\ne0\) and \(r\ne0\). 2. Solve for principal: \(P=\frac{I}{rt}\). 3. Replace \(t\) with \(\frac12t\): \(P_{\text{new}}=\frac{I}{r(\frac12t)}=2\frac{I}{rt}=2P\). 4. Therefore, the principal must double.

a) \(t=\frac{I}{Pr}\) b) The principal must double: \(P_{\text{new}}=2P\).

- Treat \(k\) as a fixed value while solving each equation for \(x\). - What equation represents the condition that the two solutions for \(x\) are equal? - Set the two expressions from part a) equal and solve for \(k\).

1. For equation (1), distribute: \(4x + 8k = 20k\). Subtract \(8k\): \(4x = 12k\). Divide by \(4\): \(x = 3k\). 2. For equation (2), add \(5k\): \(2x = 16k\). Divide by \(2\): \(x = 8k\). 3. For the solutions to be equal, set \(3k = 8k\). 4. Subtract \(3k\): \(0 = 5k\). 5. Divide by \(5\): \(k = 0\).

a) (1) \(x = 3k\); (2) \(x = 8k\) b) The solutions are equal when \(k = 0\).

- Isolate \(\frac1g\) and combine the other fractions. - Use the numerical example to test the proposed rule. - Analyze what happens to \(\frac1b\) as \(b\) grows.

1. a) Subtract \(\frac1b\): \(\frac1g=\frac1f-\frac1b=\frac{b-f}{fb}\). 2. Take reciprocals: \(g=\frac{fb}{b-f}\). 3. b) The correct formula gives \(g=\frac{6\cdot10}{10-6}=15\). The claim gives \(g=6-10=-4\), so the claim is false. 4. c) In \(\frac1g=\frac1f-\frac1b\), the value \(\frac1b\) decreases as \(b\) increases. Therefore, \(\frac1g\) increases, so \(g\) decreases. 5. As \(b\) becomes very large, \(\frac1b\) approaches 0, so \(g\) approaches \(f\) from above.

a) \(g=\frac{fb}{b-f}\) b) The claim is false; the correct value is \(g=15\), not \(-4\). c) \(g\) decreases and approaches \(f\) from above.

- Multiply by the common denominator. - Collect and factor the terms containing \(x\). - Track both the original denominators and the final divisor.

1. The original denominators require \(a\ne0\) and \(c\ne0\). 2. Multiply by \(ac\): \(cx-abc=ax\). 3. Collect and factor the \(x\)-terms: \(x(c-a)=abc\). 4. Divide by \(c-a\): \(x=\frac{abc}{c-a}\). 5. A unique solution also requires \(c-a\ne0\), so \(c\ne a\).

\(x=\frac{abc}{c-a}\), where \(a\ne0\), \(c\ne0\), and \(c\ne a\)

- Represent the smaller and larger numbers with variables. - Translate the ratio into an equation relating the two numbers. - Translate the difference into a second equation. - Substitute one relationship into the other and solve symbolically.

1. Let \(x\) be the smaller number and \(y\) the larger number. 2. The ratio gives \(y = nx\), and the difference gives \(y - x = d\). 3. Substitute \(y = nx\) into the difference equation: \(nx - x = d\). 4. Factor: \(x(n - 1) = d\). 5. Divide by \(n - 1\): \(x = \frac{d}{n - 1}\). 6. Substitute into \(y = nx\): \(y = \frac{nd}{n - 1}\).

The smaller number is \(\frac{d}{n - 1}\), and the larger number is \(\frac{nd}{n - 1}\).

- Expand the products and simplify the quadratic terms. - Collect all terms containing \(x\). - The coefficient used to divide must be nonzero.

1. Expand both sides: \(x^2+ax+bx+ab-x^2=cx+c\). 2. Simplify: \(ax+bx+ab=cx+c\). 3. Collect the \(x\)-terms: \(ax+bx-cx=c-ab\). 4. Factor: \(x(a+b-c)=c-ab\). 5. Divide: \(x=\frac{c-ab}{a+b-c}\). 6. Exactly one solution requires \(a+b-c\ne0\).

\(x=\frac{c-ab}{a+b-c}\), where \(a+b-c\ne0\)

- Write each number of revolutions as distance divided by circumference. - Set the difference between the revolution counts equal to \(n\). - Rewrite \(\frac{1}{1.25}\) in a simpler form. - Isolate \(s\) in the resulting equation.

1. Wheel A makes \(\frac{s}{u}\) revolutions. 2. Wheel B makes \(\frac{s}{1.25u}\) revolutions. 3. Their difference is \(n\), so \(\frac{s}{u} - \frac{s}{1.25u} = n\). 4. Because \(\frac{1}{1.25} = 0.8\), the equation becomes \(\frac{s}{u} - 0.8\frac{s}{u} = n\). 5. Combine like terms: \(0.2\frac{s}{u} = n\). 6. Multiply by \(5u\): \(s = 5nu\).

\(s = 5nu\)

- Which amount stays unchanged when only water is added? - Write the original fruit-juice volume in terms of \(V\) and \(p\). - Express the final total volume in terms of \(V\) and \(w\). - Use the target concentration to form an equation and isolate \(w\).

1. The concentrate contains \(V\frac{p}{100}\) liters of fruit juice. 2. After adding \(w\) liters of water, the total volume is \(V + w\) liters, while the fruit-juice volume is unchanged. 3. The target concentration gives \(\frac{V\frac{p}{100}}{V+w} = \frac{q}{100}\). 4. Multiply and simplify: \(Vp = q(V+w)\). 5. Solve for \(w\): \(Vp = qV + qw\), so \(qw = V(p-q)\) and \(w = \frac{V(p-q)}{q}\).

\(w = \frac{V(p-q)}{q}\)

- Write expressions for the original salt mass and the new total mass. - Adding pure salt changes both the salt mass and the total mass. - Set the new salt portion equal to the target percent. - Rearrange the resulting literal equation to isolate \(z\).

1. The original solution contains \(m\frac{s}{100}\) grams of salt. 2. After adding \(z\) grams of pure salt, the salt mass is \(m\frac{s}{100} + z\), and the total mass is \(m + z\). 3. The target concentration gives \(\frac{m\frac{s}{100}+z}{m+z} = \frac{k}{100}\). 4. Multiply by \(100(m+z)\): \(ms + 100z = k(m+z)\). 5. Rearrange: \(ms + 100z = km + kz\), so \(z(100-k) = m(k-s)\). 6. Therefore, \(z = \frac{m(k-s)}{100-k}\).

\(z = \frac{m(k-s)}{100-k}\)

- Clear the original denominator first. - Collect all terms containing the target variable. - Factor the target variable before dividing.

1. a) Multiply by \(x+m\): \(y(x+m)=kx\). Divide by \(x\): \(k=\frac{y(x+m)}{x}\). 2. b) Expand \(y(x+m)=kx\): \(yx+ym=kx\). 3. Collect and factor the \(x\)-terms: \(ym=x(k-y)\). 4. Divide by \(k-y\): \(x=\frac{ym}{k-y}\). 5. The stated conditions also ensure that the resulting \(x\) does not make \(x+m=0\).

a) \(k=\frac{y(x+m)}{x}\), where \(x\ne0\) b) \(x=\frac{ym}{k-y}\), where \(m\ne0\), \(k\ne0\), and \(k\ne y\)

- Factor \(x\) from the left side. - Use a common denominator for the expressions in parentheses. - Recognize the difference-of-squares product.

1. Factor out \(x\): \(x\left(\frac1{a-b}-\frac1{a+b}\right)=\frac{2b}{a}\). 2. Combine the fractions: \(\frac1{a-b}-\frac1{a+b}=\frac{(a+b)-(a-b)}{(a-b)(a+b)}=\frac{2b}{a^2-b^2}\). 3. Thus, \(x\cdot\frac{2b}{a^2-b^2}=\frac{2b}{a}\). 4. Since \(b\ne0\), cancel \(2b\) and solve: \(x=\frac{a^2-b^2}{a}=a-\frac{b^2}{a}\).

\(x=\frac{a^2-b^2}{a}\), equivalently \(x=a-\frac{b^2}{a}\)

- Multiply by the common denominator. - Expand carefully and combine the terms containing \(x\). - Factor the right side after isolating the \(x\)-term.

1. Multiply by \((a+b)(a-b)=a^2-b^2\): \(a(x+b)(a-b)+b(x-a)(a+b)=a(a^2-b^2)\). 2. Expand and combine like terms: \((a^2+b^2)x-2ab^2=a^3-ab^2\). 3. Add \(2ab^2\): \((a^2+b^2)x=a^3+ab^2\). 4. Factor the right side: \((a^2+b^2)x=a(a^2+b^2)\). 5. Since \(a\ne0\), \(a^2+b^2>0\). Divide to obtain \(x=a\).

\(x=a\)

- Multiply by the least common denominator. - Distribute the subtraction carefully. - Use the formula from part a to apply the given value of \(x\).

1. a) Multiply every term by 6: \(2(2x-a)-3(x+a)=6a\). 2. Expand: \(4x-2a-3x-3a=6a\). 3. Simplify: \(x-5a=6a\), so \(x=11a\). 4. b) Set \(x=22\): \(22=11a\). 5. Divide by 11: \(a=2\).

a) \(x=11a\) b) \(a=2\)

- Write the surface area as the sum of two square bases and four rectangular faces. - Isolate \(h\) in the surface-area equation. - Multiply the expression for height by the base area \(a^2\). - For the domain, require both \(a\) and \(h(a)\) to be positive.

1. The surface area is \(1000 = 2a^2 + 4ah\). Solving for \(h\) gives \(4ah = 1000 - 2a^2\), so \(h(a) = \frac{250}{a} - 0.5a\). 2. A physical prism requires \(a > 0\) and \(h(a) > 0\). Thus, \(\frac{250}{a} - 0.5a > 0\), which gives \(a^2 < 500\). The domain is \(0 < a < 10\sqrt{5}\), with \(a\) in centimeters. 3. Since \(V = a^2h\), \(V(a) = a^2\left(\frac{250}{a} - 0.5a\right) = 250a - 0.5a^3\). 4. \(V(5) = 250 \cdot 5 - 0.5(5^3) = 1187.5\,\text{cm}^3\). 5. \(V(15) = 250 \cdot 15 - 0.5(15^3) = 2062.5\,\text{cm}^3\). The prism with \(a = 15\,\text{cm}\) has the greater volume.

a) \(h(a) = \frac{250}{a} - 0.5a\), for \(0 < a < 10\sqrt{5}\) b) \(V(a) = a^2\left(\frac{250}{a} - 0.5a\right) = 250a - 0.5a^3\) c) \(V(5) = 1187.5\,\text{cm}^3\); \(V(15) = 2062.5\,\text{cm}^3\). The prism with \(a = 15\,\text{cm}\) has the greater volume.