- Treat \(F\) and \(m\) as known quantities and isolate \(a\). - Which inverse operation undoes multiplication by \(m\)? - The assumption \(m\ne0\) tells you the required division is defined.

1. Divide both sides by \(m\): \(a=\frac{F}{m}\).

\(a=\frac{F}{m}\)

- Try a value for \(a\) and work out the matching value of \(k\). - Your new formula should give back the same \(a\) from that value of \(k\).

1. Subtract \(7\) from both sides: \(k-7=a\). 2. Write the subject first: \(a=k-7\).

\(a=k-7\)

- Choose simple values of \(a\) and \(b\), then calculate \(P\). - Check that your rearrangement recovers the original \(b\) from \(P\) and \(a\).

1. Subtract \(a\) from both sides: \(P-a=-b\). 2. Multiply by \(-1\): \(b=a-P\).

\(b=a-P\). Change the signs on both sides when removing the negative sign.

- Try choosing values for \(n\) and \(p\) and working out the matching \(C\). - An equivalent equation must still work for those same values. - In rearranging, you do not have to assign numerical values to every letter.

1. The target is \(n\); \(C\) and \(p\) are treated as fixed numbers during rearrangement. 2. Subtract \(p\) from both sides to get \(C-p=5n\).

Treat \(C\) and \(p\) as fixed; \(C-p=5n\).

- Try \(x=4\) and find \(y\). Would your new equation accept the same pair? - Think about an operation that clears the fraction without changing the solution pairs.

1. Multiply both sides by \(4\): \(4y=4\cdot\frac{3x}{4}\). 2. The factor \(4\) cancels the denominator, giving \(4y=3x\).

\(4y=3x\). Multiplying both sides by \(4\) keeps the equation equivalent because \(4\) is nonzero.

- 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: \(z = 10k - 2\).

\(z = 10k - 2\)

- Isolate the variable that currently appears in the denominator before using any numbers. - After obtaining one symbolic formula for \(t\), use that same expression for all three calculations. - Check how changing \(W\) affects \(t\) when \(P\) is fixed.

1. Start with \(P=\frac{W}{t}\). Multiply both sides by \(t\): \(Pt=W\). 2. Divide by \(P\): \(t=\frac{W}{P}\), for \(P\ne0\). 3. For \(W=2500\,\text{J}\), \(t=\frac{2500}{500}\,\text{s}=5\,\text{s}\). 4. For \(W=5000\,\text{J}\), \(t=\frac{5000}{500}\,\text{s}=10\,\text{s}\). 5. For \(W=10{,}000\,\text{J}\), \(t=\frac{10000}{500}\,\text{s}=20\,\text{s}\). 6. Once \(t=\frac{W}{P}\) has been derived, each new work value can be substituted directly without rearranging the original formula again.

a) \(t=\frac{W}{P}\), for \(P\ne0\) b) \(5\,\text{s}\), \(10\,\text{s}\), and \(20\,\text{s}\), respectively c) The solved formula can be reused by substituting each new value of \(W\).

- Treat the letters other than \(M\) as fixed quantities. - Ask which inverse operation isolates \(M\) in each equation. - Check whether any division used in the original or rearranged equation could have a zero denominator.

1. In (a), divide by \(h\): \(M=\frac{A}{h}\), which requires \(h\ne0\). 2. In (b), \(M\) is already isolated: \(M=\frac{s}{t}\), and the original fraction requires \(t\ne0\). 3. In (c), divide by \(r\): \(M=\frac{C}{r}\), which requires \(r\ne0\).

a) \(M=\frac{A}{h}\), for \(h\ne0\) b) \(M=\frac{s}{t}\), for \(t\ne0\) c) \(M=\frac{C}{r}\), for \(r\ne0\)

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

a) Divide by \(I\): \(U=\frac{P}{I}\), where \(I\ne0\). b) Divide by \(v\): \(t=\frac{s}{v}\), where \(v\ne0\). c) Divide by \(m\): \(a=\frac{F}{m}\), where \(m\ne0\).

a) \(U=\frac{P}{I}\), where \(I\ne0\) b) \(t=\frac{s}{v}\), where \(v\ne0\) c) \(a=\frac{F}{m}\), where \(m\ne0\)

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

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

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

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

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

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

- 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. - When isolating a variable, keep track of the original denominator as well as any new divisor.

1) Replacing \(V\) with \(4V\) gives \(\rho_{\text{new}}=\frac{m}{4V}=\frac{1}{4}\rho\). The density becomes one-fourth as large. 2) Replacing \(m\) with \(\frac{1}{2}m\) gives \(\rho_{\text{new}}=\frac{\frac{1}{2}m}{V}=\frac{1}{2}\rho\). The density is halved. 3) Multiply by \(V\): \(m=\rho V\). The original formula requires \(V\ne0\). 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) \(\rho_{\text{new}}=\frac{1}{4}\rho\) 2) \(\rho_{\text{new}}=\frac{1}{2}\rho\) 3) \(m=\rho V\), with \(V\ne0\) in the original formula 4) \(V=\frac{m}{\rho}\), where \(\rho\ne0\) and \(m\ne0\)

- Clear the original denominator first. - Treat variables other than the target as constants. - Use the stated nonzero assumptions before dividing by a factor.

a) Multiply by \(c\): \(Pc=(a+b)d\). Divide by \(d\): \(\frac{Pc}{d}=a+b\). Subtract \(b\): \(a=\frac{Pc}{d}-b\). b) From \(Pc=(a+b)d\), divide by \(a+b\): \(d=\frac{Pc}{a+b}\). c) From \(Pc=(a+b)d\), divide by \(P\): \(c=\frac{d(a+b)}{P}\).

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

- Which cost depends on the number of students and which does not? - What factor multiplies the target variable in each rearrangement? - Check the number of students using the original cost formula.

1. Subtract \(G\): \(K-G=np\). 2. Divide by \(p\): \(n=\frac{K-G}{p}\). 3. Substitute: \(n=\frac{468-120}{14.50}=24\) students. 4. To solve for \(p\), start again with \(K-G=np\) and divide by \(n\): \(p=\frac{K-G}{n}\).

a) \(n=\frac{K-G}{p}\) b) \(n=24\) c) \(p=\frac{K-G}{n}\)

- Expand both sides before isolating the parameter. - Keep \(x\) symbolic until \(k\) has been isolated. - Substitute the given x-value only after obtaining the general formula.

1. Distribute: \(5x-10+k=2x+2\). 2. Subtract \(5x-10\) from both sides: \(k=2x+2-5x+10\). 3. Combine like terms: \(k=12-3x\). 4. Substitute \(x=6\): \(k=12-18=-6\).

\(k=12-3x\); when \(x=6\), \(k=-6\)

- The target variable appears in both terms on the right side. - Look for a common factor before trying to divide. - Use the stated assumption when performing the final division.

1. Both terms on the right contain \(y\), so factor: \(A=y(x+z)\). 2. Divide by \(x+z\): \(y=\frac{A}{x+z}\).

\(y=\frac{A}{x+z}\)

- Which number tells you how many copies of the same price are added? - When \(n=0\), what does the original equation say about \(C\)? - Could two different prices produce the same total in that situation?

1. If \(n>0\), divide by \(n\): \(p=\frac{C}{n}\). 2. If \(n=0\), then \(C=0\) for every price \(p\). The cost does not determine one notebook's price.

a) \(p=\frac{C}{n}\), for \(n>0\) b) No. Buying zero notebooks costs \(\$0\) whatever \(p\) is.

- Which part of the right side is grouped together? - How can you keep the equation equivalent while getting \(u\) alone? - Does your expression produce the given value of \(T\) after substitution?

1. Divide by \(3\): \(\frac{T}{3}=u+4\). 2. Subtract \(4\): \(u=\frac{T}{3}-4\). 3. At \(T=30\), \(u=10-4=6\); indeed, \(3(6+4)=30\).

\(u=\frac{T}{3}-4\); when \(T=30\), \(u=6\)

- In the original formula, what Fahrenheit value corresponds to \(C=0\)? - Be careful: the \(32\) is not multiplied by \(C\). - A correct reverse formula must return the same Celsius value when checked forward.

1. Subtract \(32\): \(F-32=\frac{9}{5}C\). 2. Multiply by \(\frac{5}{9}\): \(C=\frac{5}{9}(F-32)\). 3. With \(F=68\), \(C=\frac{5}{9}\cdot36=20\) degrees Celsius.

\(C=\frac{5}{9}(F-32)\); \(20\,\text{°C}\)

- Try a simple value of \(m\) in the original equation and see how \(y\) changes. - The vertical starting value and the part that changes with \(x\) play different roles. - Check your rearranged formula against the supplied numbers.

1. Subtract \(b\): \(y-b=mx\). 2. Divide by \(x\): \(m=\frac{y-b}{x}\). 3. Substitute the values: \(m=\frac{11-3}{2}=4\). Check: \(4\cdot2+3=11\).

\(m=\frac{y-b}{x}\); with the given values, \(m=\frac{11-3}{2}=4\). Check: \(4\cdot2+3=11\).

- 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 \(c\): \(\frac{a}{c}=x-b\). 3. Add \(b\): \(x=\frac{a}{c}+b\). 4. Since \(a>0\) and \(c>0\), \(\frac{a}{c}>0\), so the result satisfies \(x>b\).

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

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

a) Multiply by \(2\): \(2A = (a + c)h\). Divide by \(h\): \(\frac{2A}{h}=a+c\). Subtract \(a\): \(c=\frac{2A}{h}-a\), with \(h\ne0\). b) Multiply by \(t\): \(vt=s_2-s_1\). Rearranging gives \(s_1=s_2-vt\), with \(t\ne0\) in the original formula. c) Multiply by \(A\): \(pA=F\). Divide by \(p\): \(A=\frac{F}{p}\). A unique valid value requires \(p\ne0\) and \(F\ne0\), which keeps the resulting \(A\) 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 \(F \ne 0\)

- Isolate \(\frac{1}{y}\) before taking a reciprocal. - Combine \(1-\frac{1}{x}\) into one fraction. - A contradiction means no value satisfies the condition.

a) If \(x=2\), then \(\frac{1}{2}+\frac{1}{y}=1\), so \(\frac{1}{y}=\frac{1}{2}\) and \(y=2\). If \(x=5\), then \(\frac{1}{5}+\frac{1}{y}=1\), so \(\frac{1}{y}=\frac{4}{5}\) and \(y=\frac{5}{4}\). b) Isolate the reciprocal: \(\frac{1}{y}=1-\frac{1}{x}=\frac{x-1}{x}\). Taking reciprocals gives \(y=\frac{x}{x-1}\), where \(x\ne0,1\). c) Setting \(y=1\) gives \(1=\frac{x}{x-1}\). Multiplying by \(x-1\) gives \(x-1=x\), or \(-1=0\), a contradiction. Therefore, no value of \(x\) works.

a) If \(x=2\), \(y=2\); if \(x=5\), \(y=\frac{5}{4}\). 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.

- Gather every term containing \(b\) on one side before trying to isolate it. - Once the \(b\)-terms are together, look for a common factor. - Check separately the value that would make the factor multiplying \(b\) equal to zero.

1. Expand the right side: \(ab=2a+2b\). 2. Collect the terms containing \(b\): \(ab-2b=2a\). 3. Factor: \(b(a-2)=2a\). For \(a\ne2\), divide by \(a-2\): \(b=\frac{2a}{a-2}\). 4. If \(a=2\), the original equation becomes \(2b=4+2b\), which is impossible. Thus, no ordered pair has \(a=2\). 5. For example, \(a=3\) gives \(b=6\), so \((3,6)\) works. Also, \(a=0\) gives \(b=0\), so \((0,0)\) works. 6. For \(a=4\), \(b=\frac{2\cdot4}{4-2}=4\).

a) \(b=\frac{2a}{a-2}\) for \(a\ne2\); when \(a=2\), the original equation has no solution for \(b\). b) For example, \((3,6)\) and \((0,0)\) c) \(b=4\)

- Treat \(M\) as the quantity to isolate in each formula. - When \(M\) appears in a denominator, remember that its final value cannot be zero. - For a product such as \((a+c)M\), keep the grouped factor together while isolating \(M\).

1. In (a), multiply by \(3\) and divide by \(G\): \(M=\frac{3V}{G}\), requiring \(G\ne0\). 2. In (b), multiply by \(M\) and divide by \(\rho\): \(M=\frac{m}{\rho}\). A unique defined value requires \(\rho\ne0\) and \(m\ne0\), because \(M\) is an original denominator. 3. In (c), multiply by \(2\) and divide by \(a+c\): \(M=\frac{2A}{a+c}\), requiring \(a+c\ne0\). 4. In (d), multiply by \(M\) and divide by \(T\): \(M=\frac{pV}{T}\). A unique defined value requires \(T\ne0\), \(p\ne0\), and \(V\ne0\).

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

- In the proportion-style equations, think about how cross-multiplication removes the fractions. - In part (c), first rewrite the whole number as a fraction with denominator \(y\). - Track both original denominators and any quantity you divide by while isolating \(M\).

1. In (a), cross-multiply: \(aM=bc\), so \(M=\frac{bc}{a}\). A unique defined value requires \(a\ne0\), \(b\ne0\), and \(c\ne0\). 2. In (b), multiply by \(x\): \(M=\frac{xy}{z}\). The original fractions require \(x\ne0\) and \(z\ne0\). 3. In (c), write \(1=\frac{y}{y}\): \(\frac{x}{y}+1=\frac{x+y}{y}\). Thus \(M=x+y\), with \(y\ne0\). 4. In (d), cross-multiply: \(kn=mM\), so \(M=\frac{kn}{m}\). A unique defined value requires \(k\ne0\), \(m\ne0\), and \(n\ne0\).

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

- 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 \(x+y=xy\). Rearrange: \(x=xy-y=y(x-1)\). Therefore, \(y=\frac{x}{x-1}\). 2) The restriction is \(x\ne1\), because solving for \(y\) requires division by \(x-1\). 3) If \(x=2\), then \(y=2\), giving \((2, 2)\). If \(x=0\), then \(y=0\), giving \((0, 0)\). If \(x=-1\), then \(y=\frac{1}{2}\), giving \(\left(-1, \frac{1}{2}\right)\).

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

- 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 \(2(a+b)=\frac{a}{b}\). 2) Since \(b\ne0\), multiply by \(b\): \(2ab+2b^2=a\). Collect the terms containing \(a\): \(2b^2=a(1-2b)\). For \(b\ne\frac{1}{2}\), \(a=\frac{2b^2}{1-2b}\). 3) The restrictions are \(b\ne0\) and \(b\ne\frac{1}{2}\). 4) Choosing \(b=1\), \(b=-1\), and \(b=2\) gives \((-2, 1)\), \(\left(\frac{2}{3}, -1\right)\), and \(\left(-\frac{8}{3}, 2\right)\).

1) \(2(a+b)=\frac{a}{b}\) 2) \(a=\frac{2b^2}{1-2b}\) 3) \(b\ne0,\frac{1}{2}\) 4) For example, \((-2, 1)\), \(\left(\frac{2}{3}, -1\right)\), and \(\left(-\frac{8}{3}, 2\right)\)

- Treat the speed symbols other than the one being isolated as fixed quantities. - After solving the second relation for \(v_1\), substitute that entire expression into the first relation. - Delay the numerical substitution until both symbolic relationships have been obtained.

1. From \(v_g=1.2v_1\), divide by \(1.2\): \(v_1=\frac{v_g}{1.2}=\frac{5}{6}v_g\). 2. Substitute into \(v_2=1.5v_1\): \(v_2=1.5\left(\frac{5}{6}v_g\right)=\frac{5}{4}v_g=1.25v_g\). 3. If \(v_g=54\,\text{mph}\), then \(v_1=\frac{5}{6}\cdot54=45\,\text{mph}\). 4. Then \(v_2=1.25\cdot54=67.5\,\text{mph}\).

a) \(v_1=\frac{5}{6}v_g\) b) \(v_2=\frac{5}{4}v_g\) c) \(v_1=45\,\text{mph}\) and \(v_2=67.5\,\text{mph}\)

- Isolate the reciprocal containing \(z\) 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 \(z\): \(\frac{1}{z}=\frac{1}{y}-\frac{1}{x}\). 2. Combine the right side: \(\frac{1}{z}=\frac{x-y}{xy}\). 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 numerical fraction before isolating \(A\). - Substitute \(2h\) for \(h\) in the rearranged formula. - Compare the new expression with the original value of \(A\).

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

a) \(A=\frac{3V}{h}\) b) The base area is halved: \(A_{\text{new}}=\frac{1}{2}A\).

- Clear the denominator before isolating either requested variable. - For part b), note which quantity you divide by. - Convert the volume to cubic centimeters before using the density in \(\text{g}/\text{cm}^3\).

1. Multiply \(\rho=\frac{m}{V}\) by \(V\): \(\rho V=m\). Thus, \(m=\rho V\). 2. To solve for \(V\), multiply by \(V\) and divide by \(\rho\): \(V=\frac{m}{\rho}\), for \(\rho\ne0\). 3. Convert \(1.2\,\text{L}\) to \(1200\,\text{cm}^3\). 4. Use \(m=\rho V\): \(m=0.84\cdot1200=1008\,\text{g}=1.008\,\text{kg}\).

a) \(m=\rho V\) b) \(V=\frac{m}{\rho}\), for \(\rho\ne0\) c) \(1.008\,\text{kg}\)

- 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.

a) Multiply by \(V\) and divide by \(\rho\): \(V=\frac{m}{\rho}\). b) \(V_1=\frac{540}{9.0}=60\,\text{cm}^3\), and \(V_2=\frac{540}{2.7}=200\,\text{cm}^3\). 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 \(3.33\)

- 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.

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

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

- Track the variable that appears in the original denominator before rearranging. - After isolating \(I\), check what makes the solved value zero or undefined. - For part c), compare the product \(RI\) before and after changing \(R\).

a) The original formula requires \(I\ne0\). Multiply by \(I\): \(RI=U\). For \(R\ne0\), divide by \(R\): \(I=\frac{U}{R}\). For this value of \(I\) to remain nonzero, \(U\ne0\). Thus a unique solution valid in the original formula requires \(R\ne0\) and \(U\ne0\). b) Substitute: \(I=\frac{12}{5}=2.4\,\text{A}\). c) From \(U=RI\), replacing \(R\) with \(2R\) gives \(U_{\text{new}}=(2R)I=2(RI)=2U\). Therefore, the voltage doubles.

a) \(I=\frac{U}{R}\), for \(R\ne0\) and \(U\ne0\) b) \(I=2.4\,\text{A}\) c) The voltage doubles.

- Track the area restriction from the original denominator before rearranging. - Check both the new denominator and whether the solved area could equal zero. - For part c), compare the rearranged formula before and after the force changes.

a) The original formula requires \(A\ne0\). Multiply by \(A\): \(pA=F\). For \(p\ne0\), divide by \(p\): \(A=\frac{F}{p}\). For this value of \(A\) to remain nonzero, \(F\ne0\). Thus a unique solution valid in the original formula requires \(p\ne0\) and \(F\ne0\). b) Substitute: \(A=\frac{450}{1500}=0.3\,\text{m}^2\). c) If \(F\) becomes \(3F\), then \(A_{\text{new}}=\frac{3F}{p}=3A\). Therefore, the area must triple.

a) \(A=\frac{F}{p}\), for \(p\ne0\) and \(F\ne0\) b) \(A=0.3\,\text{m}^2\) c) The area must triple.

- Track the time restriction from the original denominator before rearranging. - Check both the new denominator and whether the solved time could equal zero. - For part c), compare the factors multiplying distance and time before simplifying.

a) The original formula requires \(t\ne0\). Multiply by \(t\): \(vt=s\). For \(v\ne0\), divide by \(v\): \(t=\frac{s}{v}\). For this value of \(t\) to remain nonzero, \(s\ne0\). Thus a unique solution valid in the original formula requires \(v\ne0\) and \(s\ne0\). b) Substitute: \(t=\frac{90}{60}=1.5\) hours. c) The new speed is \(v_{\text{new}}=\frac{2s}{4t}=\frac{1}{2}\cdot\frac{s}{t}\). Since \(v=\frac{s}{t}\), \(v_{\text{new}}=\frac{1}{2}v\).

a) \(t=\frac{s}{v}\), for \(v\ne0\) and \(s\ne0\) b) \(1.5\) hours c) \(v_{\text{new}}=\frac{1}{2}v\)

- How much does the amount of water change in one minute? - Check your formula for \(m\) by putting it back into the original equation. - Think about both an empty tank and a full tank.

1. Subtract \(B\): \(W-B=m(a+b-c)\). 2. Divide by the net rate: \(m=\frac{W-B}{a+b-c}\), provided \(a+b-c\ne0\). 3. Substitute the expression for \(W\): \(V-W=V-[B+m(a+b-c)]\). This is the unused capacity after \(m\) minutes. 4. Time must be nonnegative, and the water level must stay between empty and full. Thus \(m\ge0\) and \(0\le B+m(a+b-c)\le V\). These conditions also allow a zero or negative net flow rate.

a) \(m=\frac{W-B}{a+b-c}\), for \(a+b-c\ne0\) b) \(V-[B+m(a+b-c)]\); the unused capacity c) \(m\ge0\) and \(0\le B+m(a+b-c)\le V\)

- Find each balance after the same number of monthly deposits. - Which values of \(n\) represent times just after a monthly deposit? - Try two different amounts for Devin's deposit and compare the results.

1. After \(n\) months, Rosa has \(x+an\) dollars and Devin has \(bn\) dollars. Equal balances require \(x+an=bn\). 2. Rearrange: \(x=n(b-a)\), so \(n=\frac{x}{b-a}\), since \(b>a\). 3. The balances are compared at the start and after each monthly deposit. Equality at one of these times requires \(\frac{x}{b-a}\) to be a nonnegative integer. 4. If \(x>0\), increasing \(b\) increases the denominator and decreases the calculated \(n\). If \(x=0\), the balances already match at \(n=0\). 5. An exact match after a whole number of months is not guaranteed when \(b\) changes. For example, \(x=12\), \(a=2\), and \(b=5\) give \(n=4\); increasing \(b\) to \(7\) gives \(n=\frac{12}{5}\), not a whole number.

a) \(x+an=bn\), so \(n=\frac{x}{b-a}\). The answer represents a whole number of months only if this is a nonnegative integer. b) The calculated \(n\) decreases if \(x>0\), and stays at \(0\) if \(x=0\). An exact match after a whole number of months is not guaranteed: with \(x=12\) and \(a=2\), raising \(b\) from \(5\) to \(7\) changes \(n\) from \(4\) to \(\frac{12}{5}\).

- 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 \(a = 3\).

- How many bottles can all the fillers produce together in one minute? - How do total bottles, production rate, and elapsed time relate? - Compare the units of the bottle count with those of the machine rates.

1. Divide \(F=t(4x+5y)\) by the whole factor \(4x+5y\): \(t=\frac{F}{4x+5y}\). 2. Substitute: \(t=\frac{4000}{4\cdot25+5\cdot20}=\frac{4000}{200}=20\) minutes. 3. The quantity \(4x+5y\) is the number of bottles all machines together fill each minute.

a) \(t=\frac{F}{4x+5y}\) b) \(20\) minutes c) \(4x+5y\) is the combined rate in bottles per minute.

- How are the area, crew's hourly rate, and mowing time related? - How many experienced workers and assistants are in the new crew? - For the same lawn, what must change for the crew to finish sooner?

1. Divide by the total crew rate: \(d=\frac{A}{2f+3h}\). 2. The new crew has one experienced worker and five assistants: \(A=d_{\text{new}}(f+5h)\), so \(d_{\text{new}}=\frac{A}{f+5h}\). 3. For the same positive area, less time requires the larger rate: \(f+5h>2f+3h\), or \(2h>f\). 4. The original time is \(d=\frac{19{,}200}{2\cdot1200+3\cdot800}=\frac{19{,}200}{4800}=4\) hours.

a) \(d=\frac{A}{2f+3h}\) b) \(A=d_{\text{new}}(f+5h)\), so \(d_{\text{new}}=\frac{A}{f+5h}\) c) When \(2h>f\) d) \(4\) hours

- Treat the combined rate as one factor when solving for time. - When solving for \(x\), which factors and terms stand between you and the target letter? - Check each rearrangement by putting it back into the original tank formula.

1. Divide by \(2x+5\): \(t=\frac{V}{2x+5}\). 2. Substitute \(V=750\) and \(x=35\): \(t=\frac{750}{2\cdot35+5}=10\) minutes. 3. To solve for \(x\), divide by \(t\): \(\frac{V}{t}=2x+5\). 4. Subtract \(5\), then divide by \(2\): \(x=\frac{\frac{V}{t}-5}{2}\).

a) \(t=\frac{V}{2x+5}\), for \(x>0\) b) \(10\) minutes c) \(x=\frac{\frac{V}{t}-5}{2}\), for \(t>0\)

- Which unit is used for the loaded pallet, and which unit is used for a box? - How is the total box weight related to the number of boxes? - What would make the rearranged formula give a weight per box?

1. Multiply by \(2000\): \(2000M=rkm+50\). 2. Subtract \(50\): \(2000M-50=rkm\). 3. Divide by \(rk\): \(m=\frac{2000M-50}{rk}\). 4. Substitute: \(m=\frac{2000(0.905)-50}{5\cdot8}=\frac{1760}{40}=44\) pounds. 5. Dividing pounds by \(2000\) converts them to US tons.

a) \(m=\frac{2000M-50}{rk}\) b) \(m=44\,\text{lb}\) c) It converts pounds to US tons.

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

1. Subtract \(\frac{1}{R_2}\): \(\frac{1}{R_1}=\frac{1}{R}-\frac{1}{R_2}\). 2. Combine the right side: \(\frac{1}{R_1}=\frac{R_2-R}{RR_2}\). 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 numerical fraction before isolating a variable. - Check any expression used as a divisor in the solved forms.

1) Replacing \(h\) with \(2h\) gives \(A_{\text{new}}=\frac{a+c}{2}(2h)=2A\). The area doubles. 2) Replacing \(a+c\) with \(\frac{1}{2}(a+c)\) gives \(A_{\text{new}}=\frac{1}{2}A\). 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) \(h=\frac{2A}{a+c}\), where \(a+c\ne0\) 4) \(a=\frac{2A}{h}-c\), where \(h\ne0\)

- Clear the fixed denominator before isolating the requested variable. - For each rearrangement, identify the complete product you would divide by. - Convert the decimal rate to a percentage only after solving the symbolic formula.

a) Multiply by 360: \(360I=Prd\). Divide by \(rd\): \(P=\frac{360I}{rd}\). A unique solution for \(P\) requires \(rd\ne0\). b) From \(360I=Prd\), divide by \(Pr\): \(d=\frac{360I}{Pr}\). A unique solution for \(d\) requires \(Pr\ne0\). c) Solve for the rate: \(r=\frac{360I}{Pd}\), which requires \(Pd\ne0\) for a unique solution. Substitute: \(r=\frac{360\cdot100}{5000\cdot180}=0.04\). As a percentage, the annual rate is \(4\%\).

a) \(P=\frac{360I}{rd}\), for \(rd\ne0\) b) \(d=\frac{360I}{Pr}\), for \(Pr\ne0\) c) \(r=0.04=4\%\); the solved form \(r=\frac{360I}{Pd}\) requires \(Pd\ne0\)

- Which amount is shared among the members? - How could you rewrite the formula to show the total amount they pay together? - Does your formula for the grant make the members' payment larger or smaller when the grant grows?

1. Multiply by \(s\): \(Ps=K-Z\). 2. Isolate the grant: \(Z=K-Ps\). 3. Substitute: \(Z=540-28\cdot15=120\) dollars. 4. A larger grant leaves less of the fixed total cost for the members to split.

a) \(Z=K-Ps\) b) \(Z=\$120\) c) A larger grant lowers the remaining shared cost.

- Expand the right side once, then decide which symbol is the target in each rearrangement. - When solving for \(k\), pay attention to its negative coefficient. - Use each symbolic solved form before substituting the numerical value requested next.

1. Expand the right side: \(5x-k=3x+12\). 2. Solve for \(k\): subtract \(5x\) to get \(-k=-2x+12\), then multiply by \(-1\): \(k=2x-12\). 3. For \(x=2\), \(k=2(2)-12=-8\). 4. Solve the original equation for \(x\): from \(5x-k=3x+12\), subtract \(3x\) and add \(k\) to get \(2x=k+12\). Divide by \(2\): \(x=\frac{k+12}{2}\). 5. For \(k=2\), \(x=\frac{14}{2}=7\).

a) \(k=2x-12\) b) \(k=-8\) c) \(x=\frac{k+12}{2}\) d) \(x=7\)

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

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

- Try tracing the sugar, not just the liquid: which ingredient contains it? - How do grams per liter and liters combine to give total grams? - Check that the final units of \(s\) are grams per liter.

1. Multiply by \(x+y\): \(q(x+y)=xs\). 2. Since \(x>0\), divide by \(x\): \(s=\frac{q(x+y)}{x}\). 3. Substitute: \(s=\frac{30(0.4+1.6)}{0.4}=150\) grams per liter. 4. The mixture has \(q(x+y)\) grams of sugar because \(q\) is measured per liter of the whole mixture. All that sugar comes from the \(x\) liters of concentrate, so dividing by \(x\) gives its sugar content per liter.

a) \(s=\frac{q(x+y)}{x}\), for \(x>0\) b) \(150\) grams per liter c) \(q\) describes sugar per liter of the whole \(x+y\) liters. The total sugar is \(q(x+y)\), and that sugar came from the concentrate alone.

- Compare how long each runner has been moving when Blake has run for \(t\) seconds. - At the catch-up point, the two distances must match. - After collecting the terms containing \(t\), factor \(t\) before dividing.

a) When Blake has run for \(t\) seconds, Alex has run for \(t+90\) seconds. Therefore, \(s_A=v_A(t+90)\). b) Blake's distance is \(s_B=v_Bt\). c) At the catch-up point, the distances are equal: \(v_A(t+90)=v_Bt\). Distribute: \(v_At+90v_A=v_Bt\). Then \(90v_A=t(v_B-v_A)\), so \(t=\frac{90v_A}{v_B-v_A}\). Because \(v_B>v_A\), the denominator is positive.

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

- Distribute and combine like terms before isolating \(y\). - Once \(y\) is written in terms of \(x\), choose convenient values of \(x\) to generate ordered pairs. - Relate the number of algebraic solutions to the number of points on the resulting line.

a) Distribute: \(3x+6y-5=4x+7y-2\). Rearrange: \(6y-7y=4x-3x-2+5\). Thus \(-y=x+3\), so \(y=-x-3\). b) When \(x=0\), \(y=-3\), giving \((0, -3)\). When \(x=1\), \(y=-4\), giving \((1, -4)\). c) 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.

- Choose a common denominator that eliminates both fractions at once. - Be careful with the subtraction when distributing through the second expression. - For an integer solution, choose an \(x\)-value that makes the numerator in part b) divisible by 11.

a) Multiply the entire equation by the least common denominator, 12: \(4(x+2y)-3(2x-y)=12\). Distribute and combine like terms: \(4x+8y-6x+3y=12\), so \(-2x+11y=12\). b) From \(-2x+11y=12\), add \(2x\): \(11y=2x+12\). Therefore, \(y=\frac{2x+12}{11}\). c) Choose \(x=5\). Then \(y=\frac{2\cdot5+12}{11}=2\), 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 \(x=\frac{a^2-b^2+ab}{a-b}\).

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

- How does the remaining amount depend on the hours that have passed? - How can the equation be rewritten with the unknown time on its own? - Which amount is left in the tank when it becomes empty?

1. Rearrange \(A=W-kh\) to \(W-A=kh\). 2. Divide by \(k\): \(h=\frac{W-A}{k}\). 3. Substitute: \(A=500-12\cdot8=404\) gallons. 4. The tank is empty when \(A=0\), giving \(h=\frac{W}{k}\). 5. Between the start and the empty-tank time, \(0\le h\le\frac{W}{k}\).

a) \(h=\frac{W-A}{k}\) b) \(404\) gallons c) \(h=\frac{W}{k}\) d) \(0\le h\le\frac{W}{k}\)

- What is the amount already pumped when the tank still needs \(R\) gallons? - How does the combined pumping rate relate to the time? - Can you check both rearrangements in the given equation?

1. Subtract \(R\): \(V-R=(x+y)t\). 2. Divide by the combined rate: \(t=\frac{V-R}{x+y}\). 3. Substitute: \(t=\frac{630-210}{15+20}=12\) minutes. 4. Rearrange the original equation for \(R\): \(R=V-(x+y)t\).

a) \(t=\frac{V-R}{x+y}\) b) \(12\) minutes c) \(R=V-(x+y)t\)

- Clear both denominators before collecting terms. - Distribute the subtraction carefully through the second numerator. - Use the symbolic result from part a) when applying the specified value of \(x\).

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

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

- Which lengths in the drawing are the parallel sides? Which is the height? - How might you find the area using shapes whose areas you already know? - After writing the area formula, which letters are fixed when you solve for one of them?

1. The parallel sides measure \(3x\) and \(x\); the perpendicular height is \(h\). 2. The trapezoid area formula gives \(A=\frac{3x+x}{2}h=2xh\). 3. Divide by \(2x\): \(h=\frac{A}{2x}\). For \(x=5\,\text{in}\) and \(A=70\,\text{in}^2\), \(h=7\,\text{in}\). 4. Dividing \(A=2xh\) by \(2h\) instead gives \(x=\frac{A}{2h}\).

a) \(A=2xh\) b) \(h=\frac{A}{2x}\); \(h=7\,\text{in}\) c) \(x=\frac{A}{2h}\)

- Compare the structure of \(ax+bx\) with the structure of a single product. - What common factor appears in both terms on the right? - After factoring, identify the entire coefficient multiplying \(x\).

1. The right side is a sum of two products, not the single product \(abx\). Dividing successively by \(a\) and \(b\) therefore does not undo the right side. 2. Factor the common target variable: \(Q=x(a+b)\). 3. Since \(a+b\ne0\), divide by \(a+b\): \(x=\frac{Q}{a+b}\).

a) \(ax+bx\) is \(x(a+b)\), not \(abx\), so division by \(ab\) is not an equivalent step. b) \(x=\frac{Q}{a+b}\)

- 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.

a) Multiply by \(2\): \(2A=(a+c)h\). Divide by \(h\): \(\frac{2A}{h}=a+c\). Subtract \(a\): \(c=\frac{2A}{h}-a\). b) Substitute: \(c=\frac{2\cdot40}{5}-10=16-10=6\,\text{cm}\). c) Replacing \(A\) by \(2A\) gives \(c_{\text{new}}=\frac{4A}{h}-a\). 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) \(c_{\text{new}}=2c+a\)

- 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.

a) Subtract \(\frac{1}{b}\): \(\frac{1}{g}=\frac{1}{f}-\frac{1}{b}=\frac{b-f}{fb}\). Taking reciprocals gives \(g=\frac{fb}{b-f}\). The restrictions are \(f\ne0\), \(b\ne0\), and \(b-f\ne0\). b) Multiply by \(R_i+R_a\): \(I(R_i+R_a)=U\). Divide by \(I\), then subtract \(R_a\): \(R_i=\frac{U}{I}-R_a\). The restrictions are \(I\ne0\) and \(U\ne0\), which ensure \(R_i+R_a=\frac{U}{I}\ne0\) 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 \(U \ne 0\)

- 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 an original denominator or a later 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 \(\frac{u}{v}=u+2\). Because the quotient requires \(v\ne0\), multiply by \(v\): \(u=uv+2v\). Rearrange and factor: \(u(1-v)=2v\). For \(v\ne1\), \(u=\frac{2v}{1-v}\). Thus the solved form is valid for \(v\ne0,1\). 3) The quotient excludes \(v=0\). When \(v=1\), the original equation becomes \(u=u+2\), so it has no solution. 4) If \(v=-1\), then \(u=\frac{-2}{2}=-1\), giving \((-1, -1)\).

1) Yes. 2) \(u=\frac{2v}{1-v}\), for \(v\ne0,1\) 3) \(v=0\) is excluded; \(v=1\) gives no solution. 4) For example, \((-1, -1)\)

- Divide by every factor multiplying the target variable. - Solve the same formula for \(P\) before analyzing the change. - Compare the denominator before and after the time is halved.

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

a) \(t=\frac{I}{Pr}\), where \(P\ne0\) and \(r\ne0\) b) For \(r\ne0\) and \(t\ne0\), the principal must double: \(P_{\text{new}}=2P\).

- Treat \(k\) as a fixed value while solving each equation for \(x\). - Compare the two expressions you obtain for \(x\). - The same solution means those two expressions must be equal.

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

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

- What do the units tell you about the amount subtracted from \(M\)? - How does the mass left change as more days pass? - Test whether the formula predicts more or fewer days when only \(S\) changes.

1. Subtract \(M\), then change signs: \(M-R=\frac{kMt}{S}\). 2. Multiply by \(S\): \(S(M-R)=kMt\). 3. Divide by \(kM\): \(t=\frac{S(M-R)}{kM}\). 4. Substitute: \(t=\frac{1200(3000-2250)}{10\cdot3000}=30\) days. 5. At the same dollar spending per day, a higher price for a full bin means a smaller mass used daily, so reaching the same remaining mass takes longer.

a) \(t=\frac{S(M-R)}{kM}\) b) \(30\) days c) The time increases. The same daily spending buys fewer pounds when the price per pound is higher.

- Which fraction contains the amount you need to find? - How can you use the change in each person's payment to connect the two fractions? - Compare the per-person costs to check your result.

1. Move the second fraction: \(\frac{B+s}{n+m}=\frac{B}{n}-C\). 2. Multiply by \(n+m\): \(B+s=(n+m)\left(\frac{B}{n}-C\right)\). 3. Subtract \(B\): \(s=(n+m)\left(\frac{B}{n}-C\right)-B\). 4. Substitute: \(s=25(20-2)-400=50\). 5. The new bus costs \(\$450\). Before, each person pays \(\$400\div20=\$20\); afterward, each pays \(\$450\div25=\$18\).

a) \(s=(n+m)\left(\frac{B}{n}-C\right)-B\) b) \(s=50\) c) The new rental is \(\$450\). Each person pays \(\$20\) before and \(\$18\) afterward.

- Isolate the reciprocal containing \(g\) and combine the other fractions. - Check every denominator in both the original equation and the solved form. - For the trend, track how the reciprocal involving \(b\) changes as \(b\) grows.

1. 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. For this solved form to be valid in the original equation, \(f\ne0\), \(b\ne0\), and \(b\ne f\). 3. The correct formula gives \(g=\frac{6\cdot10}{10-6}=15\). Mariam's claim gives \(g=6-10=-4\), so the claim is false. 4. In \(\frac{1}{g}=\frac{1}{f}-\frac{1}{b}\), the value \(\frac{1}{b}\) decreases as \(b\) increases. Therefore, \(\frac{1}{g}\) increases, so \(g\) decreases. 5. As \(b\) becomes very large, \(\frac{1}{b}\) approaches \(0\), so \(g\) approaches \(f\) from above.

a) \(g=\frac{fb}{b-f}\), where \(f\ne0\), \(b\ne0\), and \(b\ne f\) 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 \(c\ne a\)

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

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

- Which part of the formula prevents you from isolating the second speed? - Which distance term belongs to the first part of the ride? - After rearranging, check that the two distances divided by total time give the stated average.

1. Multiply by the total time: \(v_{\text{avg}}(t_1+t_2)=v_1t_1+v_2t_2\). 2. Subtract \(v_1t_1\): \(v_{\text{avg}}(t_1+t_2)-v_1t_1=v_2t_2\). 3. Divide by \(t_2\): \(v_2=\frac{v_{\text{avg}}(t_1+t_2)-v_1t_1}{t_2}\). 4. Substitute: \(v_2=\frac{14(2+3)-10\cdot2}{3}=\frac{50}{3}\) miles per hour. 5. Check: total distance is \(10\cdot2+\frac{50}{3}\cdot3=70\) miles, so average speed is \(\frac{70}{5}=14\) miles per hour.

a) \(v_2=\frac{v_{\text{avg}}(t_1+t_2)-v_1t_1}{t_2}\) b) \(v_2=\frac{50}{3}\) mph \(=16\frac{2}{3}\) mph c) \(\frac{10\cdot2+\frac{50}{3}\cdot3}{2+3}=\frac{70}{5}=14\,\text{mph}\).

- Clear the original denominator before isolating either target variable. - Collect all terms containing the target variable before factoring. - Use the stated nonzero conditions to check the original denominator after solving.

a) Multiply by \(x+m\): \(y(x+m)=kx\). Divide by \(x\): \(k=\frac{y(x+m)}{x}\). b) Expand \(y(x+m)=kx\): \(yx+ym=kx\). Collect and factor the \(x\)-terms: \(ym=x(k-y)\). Divide by \(k-y\): \(x=\frac{ym}{k-y}\). 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 \(k\ne y\)