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Difficulty: 1 – 5

5548789

The force formula is \(F=ma\). Solve the formula for \(a\), assuming \(m\ne0\).

Hints

- 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.

Solution

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

Answer

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

5125409

Solve the formula \(0.8z + 4k = 12k - 1.6\) for \(z\). The result may contain \(k\).

Hints

- 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 \(z\).

Solution

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

Answer

\(z = 10k - 2\)

5134869

In physics, power \(P\) is work \(W\) divided by time \(t\): \(P=\frac{W}{t}\). a) Before substituting any numerical values, solve the formula for \(t\) in terms of \(W\) and \(P\). b) A motor produces a constant power of \(500\,\text{W}\). Use your solved formula to find the time needed to perform \(2500\,\text{J}\), \(5000\,\text{J}\), and \(10{,}000\,\text{J}\) of work. c) Explain why the solved formula is useful when the work value changes but the power stays fixed.

Hints

- 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.

Solution

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.

Answer

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 \(W\).

5135049

In each equation, \(M\) stands for a missing expression. Find \(M\) and state the condition needed for the result to be uniquely determined. a) \(A=Mh\) b) \(\frac{s}{t}=M\) c) \(Mr=C\)

Hints

- 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.

Solution

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 \(r\ne0\).

Answer

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

5140599

Solve each formula for the indicated variable. a) \(P=UI\), for \(U\) b) \(s=vt\), for \(t\) c) \(F=ma\), for \(a\)

Hints

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

Solution

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 \(m\ne0\).

Answer

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

5140609

Solve each equation for the indicated variable. a) \(y=3x-12\), for \(x\) b) \(V=\frac{1}{3}Gh\), for \(G\) c) \(u=a+b+c\), for \(b\)

Hints

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

Solution

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 \(c\): \(b=u-a-c\).

Answer

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

5140619

Solve each formula for the indicated variable. a) \(A=\frac{a+c}{2}h\), for \(a\) b) \(I=Prt\), for \(r\) c) \(S=2B+L\), for \(B\)

Hints

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

Solution

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: \(B=\frac{S-L}{2}\).

Answer

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

5154019

Solve each equation for the indicated variable. a) \(y=mx+n\), for \(x\) b) \(A=\frac{1}{2}(a+c)h\), for \(c\) c) \(A_f=A_0(1+r)\), for \(r\)

Hints

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

Solution

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 \(A_0\ne0\).

Answer

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 \(A_0\ne0\)

5234019

Solve the equation \(8(x-a)=16b+24a\) for \(x\).

Hints

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

Solution

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

Answer

\(x=4a+2b\)

5234029

Solve the formula \(\frac{4y-12a}{3}=8b+4a\) for \(y\).

Hints

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

Solution

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

Answer

\(y=6a+6b\)

5239239

Density is modeled by \(\rho=\frac{m}{V}\), where \(m\) is mass and \(V\) is volume. 1) How does \(\rho\) change if \(V\) is multiplied by 4 while \(m\) stays fixed? 2) How does \(\rho\) change if \(m\) is divided by 2 while \(V\) stays fixed? 3) Solve the formula for \(m\). 4) Solve the formula for \(V\).

Hints

- 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.

Solution

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 \(m\ne0\).

Answer

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 \(m\ne0\)

5241139

Given \(P=\frac{a+b}{c}d\), assume \(c\ne0\), \(d\ne0\), \(a+b\ne0\), and \(P\ne0\). Solve the formula: a) for \(a\) b) for \(d\) c) for \(c\)

Hints

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

Solution

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 \(P\): \(c=\frac{d(a+b)}{P}\).

Answer

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

5279229

A class rents a bus. The total cost \(K\) consists of a fixed fee \(G\) plus \(p\) dollars for each of \(n\) participants: \( K=G+np. \) a) Rearrange the formula to solve for \(n\). b) Find \(n\) when \(K=468\), \(G=120\), and \(p=14.50\). c) Rearrange the original formula to solve for \(p\).

Hints

- Remove the fixed fee before dividing by a per-person quantity. - Keep the product \(np\) intact until the additive term is removed. - Use the same starting equation for both rearrangements. - Check the numerical result in the original cost formula.

Solution

1. Subtract \(G\): \( K-G=np. \) 2. Divide by \(p\): \( n=\frac{K-G}{p}. \) 3. Substitute: \( n=\frac{468-120}{14.50}=\frac{348}{14.50}=24. \) 4. To solve for \(p\), again subtract \(G\) and then divide by \(n\): \( p=\frac{K-G}{n}. \)

Answer

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

5279709

Solve \(5(x-2)+k=2(x+1)\) for \(k\) in terms of \(x\). Then use your formula to find the value of \(k\) when \(x=6\).

Hints

- 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.

Solution

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 \(x=6\): \(k=12-18=-6\).

Answer

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

5548799

Solve \(A=xy+zy\) for \(y\). Assume \(x+z\ne0\).

Hints

- 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.

Solution

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

Answer

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

5134879

The formula \(\frac{a}{x-b}=c\) uses positive values \(a\), \(b\), and \(c\), with \(x>b\). Solve the formula for \(x\). Show the algebraic steps.

Hints

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

Solution

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 \(x>b\).

Answer

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

5134989

Solve each formula for the variable in parentheses. a) \(A = \frac{(a + c)h}{2}\quad (c)\) b) \(v = \frac{s_2 - s_1}{t}\quad (s_1)\) c) \(p = \frac{F}{A}\quad (A)\)

Hints

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

Solution

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.

Answer

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 \(F \ne 0\)

5135019

For nonzero real numbers \(x\) and \(y\), suppose \(\frac{1}{x}+\frac{1}{y}=1\). a) Find \(y\) when \(x=2\) and when \(x=5\). b) Solve the equation for \(y\). c) Explain mathematically why no value of \(x\) works when \(y=1\).

Hints

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

Solution

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.

Answer

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.

5135029

The product of two rational numbers \(a\) and \(b\) is twice their sum, so \(ab=2(a+b)\). a) Solve the equation for \(b\) in terms of \(a\), and state the value of \(a\) for which this solved form is not defined. Check what the original equation says when \(a\) has that value. b) Use your solved formula to find two different ordered pairs \((a,b)\) that satisfy the relationship. c) Use the solved formula to find \(b\) when \(a=4\).

Hints

- 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.

Solution

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 \(a=4\), \(b=\frac{2\cdot4}{4-2}=4\).

Answer

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) \(b=4\)

5135059

In each formula, \(M\) stands for a missing expression. Find \(M\) and state the conditions needed for it to be uniquely determined and for every original fraction to be defined. a) \(V=\frac{1}{3}GM\) b) \(\rho=\frac{m}{M}\) c) \(\frac{a+c}{2}M=A\) d) \(\frac{pV}{M}=T\)

Hints

- 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 \(M\).

Solution

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 \(V\ne0\).

Answer

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 \(V\ne0\)

5135069

In each equation, \(M\) stands for a missing expression. Find \(M\) and state the conditions under which it is uniquely determined and every original fraction is defined. a) \(\frac{a}{b}=\frac{c}{M}\) b) \(\frac{1}{x}M=\frac{y}{z}\) c) \(\frac{x}{y}+1=\frac{M}{y}\) d) \(\frac{k}{M}=\frac{m}{n}\)

Hints

- 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 \(M\).

Solution

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 \(n\ne0\).

Answer

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 \(n\ne0\)

5135079

Find three ordered pairs of rational numbers \((x, y)\) whose sum equals their product. 1) Write an equation and solve it for \(y\). 2) State the restriction on \(x\). 3) Give three ordered pairs that satisfy the condition.

Hints

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

Solution

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 \(\left(-1, \frac{1}{2}\right)\).

Answer

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

5135089

Two rational numbers \(a\) and \(b\) satisfy this condition: twice their sum equals the quotient \(\frac{a}{b}\). 1) Write the equation. 2) Solve the equation for \(a\). 3) Which values are not allowed for \(b\)? 4) Give three ordered pairs \((a, b)\) that satisfy the condition.

Hints

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

Solution

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 \(\left(-\frac{8}{3}, 2\right)\).

Answer

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 \(\left(-\frac{8}{3}, 2\right)\)

5135279

For a particular delivery-drone route, the return speed \(v_2\), outbound speed \(v_1\), and whole-trip average speed \(v_g\) satisfy \(v_2=1.5v_1\) and \(v_g=1.2v_1\). a) Solve \(v_g=1.2v_1\) for \(v_1\) in terms of \(v_g\). b) Substitute your result into \(v_2=1.5v_1\) to write \(v_2\) in terms of \(v_g\). c) If \(v_g=54\,\text{mph}\), find \(v_1\) and \(v_2\).

Hints

- 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.

Solution

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 \(v_2=1.25\cdot54=67.5\,\text{mph}\).

Answer

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

5135539

Solve the equation \(\frac{1}{x}=\frac{1}{y}-\frac{1}{z}\) for \(z\). State the necessary restrictions.

Hints

- 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.

Solution

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.

Answer

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

5135549

The volume formula for a pyramid is \(V=\frac{1}{3}Ah\). a) Solve the formula for \(A\). b) The volume remains constant. How does \(A\) change if the height \(h\) is doubled? Justify your answer.

Hints

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

Solution

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.

Answer

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

5135589

Density is defined by \(\rho=\frac{m}{V}\), where \(m\) is mass and \(V\) is volume. a) Solve the formula for \(m\). b) Solve the original formula for \(V\). c) A liquid has density \(0.84\,\text{g}/\text{cm}^3\). Use your formula from part a) to find the mass of \(1.2\,\text{L}\) of the liquid. Give the answer in kilograms.

Hints

- 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 \(\text{g}/\text{cm}^3\).

Solution

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 \(m=\rho V\): \(m=0.84\cdot1200=1008\,\text{g}=1.008\,\text{kg}\).

Answer

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

5135599

Two metal cylinders each have mass \(540\,\text{g}\). Cylinder 1 has density \(\rho_1=9.0\,\text{g}/\text{cm}^3\), and cylinder 2 is aluminum with density \(\rho_2=2.7\,\text{g}/\text{cm}^3\). a) Solve \(\rho=\frac{m}{V}\) for \(V\). b) Find the volumes \(V_1\) and \(V_2\). c) By what factor is the aluminum cylinder’s volume greater than the first cylinder’s volume?

Hints

- 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.

Solution

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 \(\frac{V_2}{V_1}=\frac{200}{60}=\frac{10}{3}\approx3.33\).

Answer

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 \(3.33\)

51356010

A concrete foundation slab for a backyard shed is \(15\,\text{ft}\) long, \(12\,\text{ft}\) wide, and \(0.5\,\text{ft}\) thick. Concrete has density about \(150\,\text{lb}/\text{ft}^3\). Use \(\rho=\frac{m}{V}\). a) Find the slab's volume in cubic feet. b) Solve the formula for \(m\) and find the concrete's total mass in short tons. c) A small truck can carry at most \(1.5\) tons per trip. What is the minimum number of trips needed?

Hints

- Start with the rectangular-prism volume of the slab. - Rearrange the density formula for mass before substituting. - After converting pounds to short tons, decide how a non-whole number of truckloads affects the minimum trip count.

Solution

1. The slab volume is \(V=15\cdot12\cdot0.5=90\,\text{ft}^3\). 2. Rearrange the density formula to \(m=\rho V\). Then \(m=150\cdot90=13{,}500\,\text{lb}\). 3. Since \(2000\,\text{lb}=1\) short ton, the mass is \(6.75\) short tons. 4. The number of truckloads is \(6.75\div1.5=4.5\). Since a partial trip cannot carry the remaining concrete, at least \(5\) trips are needed.

Answer

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

5138969

For a positive length \(x>0\), a circle has circumference \(C=10\pi x\). a) Express the radius \(r\) and diameter \(d\) in terms of \(x\). b) Write and simplify a formula for the circle's area \(A\) in terms of \(x\).

Hints

- 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.

Solution

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: \(A=\pi(5x)^2=25\pi x^2\).

Answer

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

5140569

Ohm’s law relates voltage \(U\), resistance \(R\), and current \(I\) by \(R=\frac{U}{I}\). a) Solve the formula for \(I\), and state the conditions for a unique solution that is valid in the original formula. b) Find the current when \(U=12\,\text{V}\) and \(R=5\,\Omega\). c) If \(I\) remains constant and \(R\) doubles, how does \(U\) change? Justify your answer algebraically.

Hints

- 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 \(R\).

Solution

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.

Answer

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

5140579

Pressure \(p\) on an area \(A\) is \(p=\frac{F}{A}\), where \(F\) is force. a) Solve the formula for \(A\), and state the conditions for a unique solution that is valid in the original formula. b) An object exerts \(450\,\text{N}\) of force and produces pressure \(1500\,\text{Pa}\). Find the contact area in square meters. c) If the force triples while pressure remains constant, how must the area change?

Hints

- 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.

Solution

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.

Answer

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

5140589

Average speed is related to distance and time by \(v=\frac{s}{t}\). a) Solve the formula for \(t\), and state the conditions for a unique solution that is valid in the original formula. b) A vehicle travels \(90\) miles at a constant \(60\,\text{mph}\). Find the travel time. c) A second vehicle travels twice the distance, \(2s\), in four times the time, \(4t\). Express its speed \(v_{\text{new}}\) in terms of \(v\).

Hints

- 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.

Solution

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 \(v=\frac{s}{t}\), \(v_{\text{new}}=\frac{1}{2}v\).

Answer

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

5225449

A water tank initially contains \(B\) gallons. Pipe A adds \(a\) gallons per minute, Pipe B adds \(b\) gallons per minute, and a drain removes \(c\) gallons per minute. After \(m\) minutes, the tank contains \(W\) gallons: \( W=B+m(a+b-c). \) a) Rearrange the formula to solve for \(m\). b) The tank's capacity is \(V\). Write \(V-W\) in terms of \(B,m,a,b,c\), and explain what that expression represents. c) State the physical conditions on \(m\) under which the model is meaningful if the tank cannot contain less than \(0\) or more than \(V\) gallons.

Hints

- Isolate the term containing \(m\) before dividing. - Treat \(a+b-c\) as one grouped net-rate factor. - Capacity minus current contents is unused capacity. - Separate the algebraic rearrangement from the physical-domain restrictions.

Solution

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 formula for \(W\): \( V-W=V-[B+m(a+b-c)]. \) This is the unused tank capacity after \(m\) minutes. 4. Physically, \(m\ge0\) and \( 0\le B+m(a+b-c)\le V. \)

Answer

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

5225649

Two savings accounts grow each month. Account A starts with \(\$x\) and receives \(\$a\) each month. Account B starts at \(\$0\) and receives \(\$b\) each month, where \(b>a\). a) Derive a formula for the number of months \(n\) until the accounts have equal balances. State the condition for this to happen after a whole number of months. b) Based on the formula, explain how \(n\) changes when \(b\) increases while \(x\) and \(a\) stay fixed. Include the case \(x=0\).

Hints

- Compare the two balance expressions after the same number of months. - The monthly catch-up amount is determined by the difference between the two monthly deposits. - In part b), consider separately whether the fixed starting amount \(x\) is positive or zero.

Solution

a) After \(n\) months, Account A has \(x+an\), and Account B has \(bn\). Set the balances equal: \(x+an=bn\). Then \(x=n(b-a)\), so \(n=\frac{x}{b-a}\). Equality occurs after a whole number of months only when \(\frac{x}{b-a}\) is a nonnegative integer. b) Because \(b>a\), the denominator \(b-a\) is positive. If \(x>0\), increasing \(b\) increases the denominator while the numerator stays fixed, so \(n\) decreases. If \(x=0\), then \(n=0\) for every allowed value of \(b\), so increasing \(b\) does not change \(n\).

Answer

a) \(n=\frac{x}{b-a}\). For equality after a whole number of months, \(\frac{x}{b-a}\) must be a nonnegative integer. b) If \(x>0\), increasing \(b\) decreases \(n\), so Account B catches up sooner. If \(x=0\), \(n=0\) and does not change.

5229319

Solve \(3(x - 2a) + 5a = 14a\) for \(x\). What value of the parameter \(a\) makes the solution \(x = 15\)?

Hints

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

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 \(5\): \(a = 3\).

Answer

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

5238279

A beverage plant uses \(4\) Type A machines and \(5\) Type B machines at the same time. Each Type A machine fills \(x\) bottles per minute, each Type B machine fills \(y\) bottles per minute, the order contains \(F\) bottles, and the filling time is \(t\) minutes. The quantities satisfy \( F=t(4x+5y). \) a) Rearrange the formula to solve explicitly for \(t\). b) Find \(t\) when \(x=25\), \(y=20\), and \(F=4000\). c) Explain why the denominator in your rearranged formula represents a rate.

Hints

- Treat \(4x+5y\) as one complete factor multiplying \(t\). - Use the same division on both sides to isolate \(t\). - After rearranging, substitute values only into the solved formula. - Track the units of \(F\) and \(4x+5y\).

Solution

1. Start with \(F=t(4x+5y)\). 2. Divide both sides by the entire factor \(4x+5y\): \( t=\frac{F}{4x+5y}. \) 3. Substitute the values: \( t=\frac{4000}{4\cdot25+5\cdot20} =\frac{4000}{200}=20. \) 4. The quantity \(4x+5y\) is the combined number of bottles filled per minute by all nine machines, so it is a rate.

Answer

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

5238289

A landscaping crew has \(2\) experienced workers and \(3\) assistants. Each experienced worker mows \(f\) square feet per hour, each assistant mows \(h\) square feet per hour, the lawn area is \(A\) square feet, and the work takes \(d\) hours. For the original crew, \( A=d(2f+3h). \) a) Rearrange the formula to solve for \(d\). b) One experienced worker is replaced by two assistants. Write the corresponding area equation using a new time \(d_{\text{new}}\), then rearrange it to solve for \(d_{\text{new}}\). c) Use the two solved formulas to determine when the new crew is faster than the original crew. d) Find the original time when \(f=1200\), \(h=800\), and \(A=19{,}200\).

Hints

- Isolate a time variable by dividing by the entire crew-rate factor. - Recount each worker type after the replacement before writing the new equation. - For the same positive area, a larger rate gives a smaller time. - Compare the two denominators before substituting numerical values.

Solution

1. Divide \(A=d(2f+3h)\) by \(2f+3h\): \( d=\frac{A}{2f+3h}. \) 2. The new crew has \(1\) experienced worker and \(5\) assistants, so \( A=d_{\text{new}}(f+5h). \) Dividing by \(f+5h\) gives \( d_{\text{new}}=\frac{A}{f+5h}. \) 3. For the same positive area \(A\), the new crew is faster when its denominator is larger: \( f+5h>2f+3h. \) Thus \(2h>f\). 4. For the given original crew, \( d=\frac{19{,}200}{2\cdot1200+3\cdot800} =\frac{19{,}200}{4800}=4. \)

Answer

a) \(d=\frac{A}{2f+3h}\) b) \(A=d_{\text{new}}(f+5h)\), so \(d_{\text{new}}=\frac{A}{f+5h}\) c) The new crew is faster when \(2h>f\). d) \(4\) hours

5238329

An empty water tank holds \(V\) gallons. Pipe A adds \(x\) gallons per minute, and Pipe B adds \(5\) gallons per minute more than Pipe A. If both pipes run for \(t\) minutes, then \( V=t(2x+5). \) a) Rearrange the formula to solve for \(t\). b) Find \(t\) when \(V=750\) and \(x=35\). c) Rearrange the same formula to solve for \(x\) in terms of \(V\) and \(t\).

Hints

- Treat \(2x+5\) as one factor when isolating \(t\). - For part c), undo multiplication by \(t\) first. - Then undo the added \(5\) before dividing by \(2\). - Keep the symbolic quantities intact until the requested variable is isolated.

Solution

1. Divide both sides of \(V=t(2x+5)\) by \(2x+5\): \( t=\frac{V}{2x+5}. \) 2. Substitute \(V=750\) and \(x=35\): \( t=\frac{750}{70+5}=\frac{750}{75}=10. \) 3. To solve for \(x\), divide by \(t\): \(\frac{V}{t}=2x+5\). 4. Subtract \(5\): \(\frac{V}{t}-5=2x\). 5. Divide by \(2\): \( x=\frac{\frac{V}{t}-5}{2}. \)

Answer

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

5238449

Boxes are stacked on a pallet in \(r\) rows with \(k\) boxes per row. Each box weighs \(m\) pounds, and the empty pallet weighs \(50\) pounds. If the loaded pallet weighs \(M\) tons, then \( M=\frac{rkm+50}{2000}. \) a) Rearrange the formula to solve for the box weight \(m\). b) Find \(m\) when \(M=0.905\), \(r=5\), and \(k=8\). c) Explain the purpose of the factor \(2000\) in the original formula.

Hints

- Clear the denominator before moving the pallet's fixed weight. - Subtract the \(50\)-pound pallet weight before dividing by the number of boxes. - The product \(rk\) is the total number of boxes. - Check the units after rearranging.

Solution

1. Multiply both sides by \(2000\): \( 2000M=rkm+50. \) 2. Subtract \(50\): \( 2000M-50=rkm. \) 3. Divide by \(rk\): \( m=\frac{2000M-50}{rk}. \) 4. Substitute: \( m=\frac{2000(0.905)-50}{5\cdot8} =\frac{1810-50}{40} =\frac{1760}{40}=44. \) 5. The denominator \(2000\) converts the total weight from pounds to tons because \(1\) ton is \(2000\) pounds.

Answer

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

5239229

For two resistors in parallel, the equivalent resistance \(R\) satisfies \(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}\). Solve the equation for \(R_1\) and state the necessary restrictions.

Hints

- 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.

Solution

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.

Answer

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

5239249

The area of a trapezoid is \(A=\frac{a+c}{2}h\), where \(a\) and \(c\) are the parallel side lengths and \(h\) is the height. 1) How does \(A\) change if \(h\) doubles while \(a\) and \(c\) stay fixed? 2) How does \(A\) change if \(a+c\) is halved while \(h\) stays fixed? 3) Solve the formula for \(h\). 4) Solve the formula for \(a\).

Hints

- 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.

Solution

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 \(h\ne0\).

Answer

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 \(h\ne0\)

5239319

For simple interest calculated with a 360-day year, use \(I=\frac{Prd}{360}\), where \(I\) is interest, \(P\) is principal, \(r\) is the annual rate as a decimal, and \(d\) is the number of days. a) Solve the formula for \(P\), stating the condition for a unique solution. b) Solve the formula for \(d\), stating the condition for a unique solution. c) What annual interest rate is needed for a principal of \(\$5000\) to earn \(\$100\) in 180 days? First solve the formula for \(r\), stating the condition for a unique solution.

Hints

- 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.

Solution

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 \(4\%\).

Answer

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 \(Pd\ne0\)

5239349

A sports club buys equipment costing \(\$K\). A city grant contributes \(\$Z\), and the remaining cost is divided equally among \(s\) active members. If each member pays \(P\) dollars, then \( P=\frac{K-Z}{s}. \) a) Rearrange the formula to solve for \(Z\). b) Find the grant \(Z\) when \(K=540\), \(s=15\), and \(P=28\). c) Explain why increasing \(Z\) lowers \(P\) when \(K\) and \(s\) stay fixed.

Hints

- Clear the denominator first. - Keep track of the subtraction sign attached to \(Z\). - Isolate \(Z\) only after multiplying by \(s\). - Interpret the formula after rearranging, not only numerically.

Solution

1. Multiply both sides by \(s\): \(Ps=K-Z\). 2. Add \(Z\) to both sides and subtract \(Ps\): \(Z=K-Ps\). 3. Substitute the values: \(Z=540-28\cdot15=540-420=120\). 4. A larger grant removes more of the fixed total cost before the remainder is divided among members, so each member pays less.

Answer

a) \(Z=K-Ps\) b) \(Z=\$120\) c) A larger grant reduces the amount left for the members to share.

5239709

Consider \(5x-k=3(x+4)\), where \(k\) is a parameter. a) Solve the equation for \(k\) in terms of \(x\). b) Use your formula from part a) to find \(k\) when \(x=2\). c) Solve the original equation for \(x\) in terms of \(k\). d) Use your formula from part c) to find \(x\) when \(k=2\).

Hints

- 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.

Solution

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 \(k=2\), \(x=\frac{14}{2}=7\).

Answer

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

5239739

Solve \(k(x-m)=5(x+n)\) for \(x\). State the condition on \(k\) that guarantees a unique solution.

Hints

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

Solution

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 \(k\ne5\).

Answer

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

5239829

Solve \(4(x-2a)+3b=2(b-x)-6a\) for \(x\). Simplify the result.

Hints

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

Solution

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

Answer

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

5240659

A fruit-juice mixture contains \(x\) liters of concentrate and \(y\) liters of water. The concentrate contains \(s\) grams of sugar per liter. If the final mixture has sugar concentration \(q\) grams per liter, then \( q=\frac{xs}{x+y}. \) a) Rearrange the formula to solve for \(s\). b) Find \(s\) when \(q=30\), \(x=0.4\), and \(y=1.6\). c) Explain why the factor \(x+y\) appears in your rearranged formula.

Hints

- Clear the denominator before trying to isolate \(s\). - Treat \(x+y\) as one grouped quantity. - Divide by the coefficient multiplying \(s\) only after clearing the fraction. - Use the units to interpret \(x+y\).

Solution

1. Multiply both sides by \(x+y\): \( q(x+y)=xs. \) 2. Divide by \(x\): \( s=\frac{q(x+y)}{x}. \) 3. Substitute the values: \( s=\frac{30(0.4+1.6)}{0.4}=\frac{60}{0.4}=150. \) 4. The factor \(x+y\) is the total volume of the final mixture.

Answer

a) \(s=\frac{q(x+y)}{x}\) b) \(150\) grams per liter c) \(x+y\) is the total mixture volume.

5240739

Two runners, Alex and Blake, train on a long, straight path. Alex runs at a constant speed of \(v_A\,\text{ft/s}\). Blake starts from the same point \(90\) seconds later and runs at \(v_B\,\text{ft/s}\), where \(v_B > v_A\). a) Write an expression for the distance \(s_A\) that Alex has run when Blake has been running for \(t\) seconds. b) Write an expression for Blake's distance \(s_B\) after \(t\) seconds. c) Solve for the time \(t\) Blake needs to catch Alex in terms of \(v_A\) and \(v_B\).

Hints

- 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.

Solution

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.

Answer

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

5242059

Consider the linear equation in two variables \(3(x+2y)-5=4x+7y-2\). a) Rewrite the equation so that \(y\) is expressed in terms of \(x\) in the form \(y=mx+b\). b) Give two different ordered pairs \((x, y)\) that satisfy the equation. c) Explain how many solutions the equation has and why.

Hints

- 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.

Solution

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 \(y=-x-3\).

Answer

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.

5242069

Consider the equation \(\frac{x+2y}{3}-\frac{2x-y}{4}=1\). a) Rewrite the equation in the form \(ax+by=c\), where \(a\), \(b\), and \(c\) are integers. b) Solve the equation for \(y\). c) Find one solution \((x, y)\) in which both coordinates are integers.

Hints

- 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.

Solution

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.

Answer

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

5244049

Solve \(\frac{x-a}{b}=\frac{x-b}{a}+1\) for \(x\). Write the result as a single fraction. Assume \(a\ne b\) and \(a,b\ne0\).

Hints

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

Solution

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

Answer

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

52453211

A free-fall experiment is performed from heights \(h_1=45\,\text{m}\) and \(h_2=180\,\text{m}\). a) Find the fall times \(t_1\) and \(t_2\) in seconds. Use \(h=\frac{1}{2}gt^2\) with \(g=9.81\,\frac{\text{m}}{\text{s}^2}\), and round to the nearest hundredth. b) By what factor does the fall time change when the height is multiplied by \(4\)?

Hints

- Isolate \(t^2\) before taking the physically meaningful positive square root. - Keep the two calculations unrounded until the final requested hundredths. - For the factor comparison, compare the heights inside the square root rather than relying only on rounded times.

Solution

1. Solve the model for positive time: \(t=\sqrt{\frac{2h}{g}}\). 2. For \(h_1=45\,\text{m}\), \(t_1=\sqrt{\frac{2\cdot45}{9.81}}\,\text{s}\approx3.03\,\text{s}\). 3. For \(h_2=180\,\text{m}\), \(t_2=\sqrt{\frac{2\cdot180}{9.81}}\,\text{s}\approx6.06\,\text{s}\). 4. Exactly, \(\frac{t_2}{t_1}=\sqrt{\frac{180}{45}}=\sqrt{4}=2\). Multiplying the height by \(4\) doubles the fall time.

Answer

a) \(t_1\approx3.03\,\text{s}\); \(t_2\approx6.06\,\text{s}\) b) The fall time is multiplied by \(2\).

5279249

A water tank initially contains \(W\) gallons. A leak drains water at a constant rate of \(k\) gallons per hour. If \(A\) is the amount of water remaining after \(h\) hours, then \( A=W-kh. \) a) Rearrange the formula to solve for \(h\). b) Find \(A\) when \(W=500\), \(k=12\), and \(h=8\). c) Use your rearranged formula to give the empty-tank time in terms of \(W\) and \(k\). d) State the physically meaningful interval of \(h\) while the tank is draining.

Hints

- Treat \(A=W-kh\) as a literal equation and isolate the term containing \(h\). - After moving the constant term, account carefully for the negative sign on \(kh\). - For an empty tank, substitute \(A=0\) into the formula you already solved for \(h\). - The model stops being physical when the remaining amount would become negative.

Solution

1. Start with \(A=W-kh\). Subtract \(W\): \(A-W=-kh\). 2. Multiply by \(-1\): \(W-A=kh\). Divide by \(k\): \( h=\frac{W-A}{k}. \) 3. For the given values, \(A=500-12\cdot8=404\) gallons. 4. The tank is empty when \(A=0\), so the rearranged formula gives \(h=\frac{W}{k}\). 5. During the physical draining process, \(0\le h\le\frac{W}{k}\).

Answer

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

5280329

Two pumps fill a tank. Pump A moves \(x\) gallons per minute and Pump B moves \(y\) gallons per minute. After both pumps have run for \(t\) minutes, the tank still needs \(R\) gallons to reach its capacity \(V\). The quantities satisfy \( V=(x+y)t+R. \) a) Rearrange the formula to solve for \(t\). b) Find \(t\) when \(V=630\), \(x=15\), \(y=20\), and \(R=210\). c) Rearrange the original formula to solve for \(R\).

Hints

- Remove the additive term \(R\) before dividing. - Treat \(x+y\) as a single factor multiplying \(t\). - For part c), isolate \(R\) with one subtraction. - Check that the units of your solved formula for \(t\) are minutes.

Solution

1. Subtract \(R\): \( V-R=(x+y)t. \) 2. Divide by \(x+y\): \( t=\frac{V-R}{x+y}. \) 3. Substitute: \( t=\frac{630-210}{15+20}=\frac{420}{35}=12. \) 4. To solve for \(R\), subtract \((x+y)t\) from both sides: \( R=V-(x+y)t. \)

Answer

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

5280439

Consider the equation \(\frac{2x-a}{3}-\frac{x+a}{2}=a\). a) Solve for \(x\) in terms of \(a\). b) What value of \(a\) makes the solution \(x=22\)? Justify your answer.

Hints

- 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 \(x\).

Solution

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 \(a=2\).

Answer

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

5365809

The diagram shows a trapezoid. The two horizontal sides are parallel; use the dimension labels shown in the diagram. a) Write and simplify a formula for the area \(A\) in terms of \(x\) and \(h\). b) Solve the formula for \(h\), and then find \(h\) when \(x=5\,\text{in.}\) and \(A=70\,\text{in.}^2\). c) Solve the formula for \(x\).

Figure for problem 536580

Hints

- Read the two parallel-side labels and the perpendicular-height label from the diagram. - Substitute the two displayed base expressions into the trapezoid area formula before simplifying. - Isolate the requested variable symbolically before using the numerical values in part b).

Solution

1. From the diagram, the parallel sides have lengths \(3x\) and \(x\), and the perpendicular height is \(h\). 2. Use the trapezoid area formula: \(A=\frac{a+b}{2}h\). Substitute the displayed base lengths: \(A=\frac{3x+x}{2}h=2xh\). 3. Divide by \(2x\): \(h=\frac{A}{2x}\). For \(x=5\,\text{in.}\) and \(A=70\,\text{in.}^2\), \(h=\frac{70}{2\cdot5}\,\text{in.}=7\,\text{in.}\). 4. Solve \(A=2xh\) for \(x\) by dividing by \(2h\): \(x=\frac{A}{2h}\).

Answer

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

5441229

A regression model is \(\hat y=6x+4\), where \(x\) is elapsed time in hours. A new variable \(m\) measures the same elapsed time in minutes. Rewrite the model using \(m\) as the input variable.

Hints

- Express hours in terms of minutes. - Replace the original input variable rather than changing the response. - Check that sixty minutes produces the same predicted change as one hour.

Solution

1. Because \(m=60x\), the hour value is \(x=\frac{m}{60}\). 2. Substitute into the original model: \(\hat y=6\left(\frac{m}{60}\right)+4\). 3. Simplifying gives \(\hat y=0.1m+4\).

Answer

\(\hat y=0.1m+4\)

5548809

Mia is solving \(Q=ax+bx\) for \(x\). She says, “Since \(x\) is multiplied by both \(a\) and \(b\), divide by \(a\) and then by \(b\), so \(x=\frac{Q}{ab}\).” a) Explain why Mia's reasoning is incorrect. b) Solve the formula correctly for \(x\), assuming \(a+b\ne0\).

Hints

- 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 \(x\).

Solution

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 \(a+b\): \(x=\frac{Q}{a+b}\).

Answer

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

5134889

The area of a trapezoid is \(A=\frac{a+c}{2}h\), where \(a\) and \(c\) are the parallel side lengths and \(h\) is the height. a) Solve the formula for \(c\). b) A trapezoid has \(A=40\,\text{cm}^2\), \(h=5\,\text{cm}\), and \(a=10\,\text{cm}\). Find \(c\). c) If \(h\) and \(a\) stay fixed while \(A\) doubles, how does \(c\) change? Justify your answer using the formula from part a).

Hints

- 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.

Solution

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.

Answer

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

5134999

Solve each formula for the variable in parentheses. State all necessary restrictions. a) \(\frac{1}{f} = \frac{1}{b} + \frac{1}{g}\quad (g)\) b) \(I = \frac{U}{R_i + R_a}\quad (R_i)\)

Hints

- 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.

Solution

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.

Answer

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 \(U \ne 0\)

5135009

Solve the equation \(y=\frac{kx}{x+1}\) for \(x\). State the parameter conditions for a unique solution and describe the exceptional cases.

Hints

- 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 \(x=-1\).

Solution

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.

Answer

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.

5135099

Consider ordered pairs \((u, v)\) for which the quotient of \(u\) and \(v\) is 2 greater than \(u\). 1) Verify whether \((2, 0.5)\) satisfies the condition. 2) Write the general equation and solve it for \(u\). 3) Which value of \(v\) is excluded by the quotient, and for which additional value does the equation have no solution for \(u\)? Explain. 4) Find another ordered pair in which both numbers are negative.

Hints

- 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.

Solution

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 \((-1, -1)\).

Answer

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, \((-1, -1)\)

5154029

Simple interest is modeled by \(I=Prt\), where \(I\) is interest, \(P\) is principal, \(r\) is the annual interest rate as a decimal, and \(t\) is time in years. a) Solve the formula for \(t\). b) Suppose \(I\) and \(r\) remain fixed. How must \(P\) change if \(t\) is cut in half? Justify your answer using the formula solved for \(P\).

Hints

- 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.

Solution

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.

Answer

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

5229329

Consider these equations with variable \(x\) and parameter \(k\): (1) \(4(x + 2k) = 20k\) (2) \(2x - 5k = 11k\) a) Solve each equation for \(x\) in terms of \(k\). b) Find the value of \(k\) for which the two equations have the same solution for \(x\).

Hints

- 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.

Solution

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 \(k=0\).

Answer

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

5238359

A wood-pellet bin holds \(M\) pounds when full. Filling it completely costs \(\$S\), and the pellets have a constant price per pound. A heating system uses pellets worth \(\$k\) each day. After \(t\) days, \(R\) pounds remain. These quantities satisfy \( R=M-\frac{kMt}{S}. \) a) Rearrange the formula to solve for \(t\). b) Find \(t\) when \(M=3000\), \(S=1200\), \(k=10\), and \(R=2250\). c) Explain why \(S\) appears in the numerator of your solved formula for \(t\).

Hints

- First isolate the term containing \(t\). - Remove the denominator \(S\) before dividing by the other factors. - Keep \(kM\) together as the factor multiplying \(t\). - Check the direction of the relationship between \(S\) and \(t\) from the solved formula.

Solution

1. Subtract \(M\): \( R-M=-\frac{kMt}{S}. \) Equivalently, \( M-R=\frac{kMt}{S}. \) 2. Multiply by \(S\): \( S(M-R)=kMt. \) 3. Divide by \(kM\): \( t=\frac{S(M-R)}{kM}. \) 4. Substitute: \( t=\frac{1200(3000-2250)}{10\cdot3000} =\frac{1200\cdot750}{30000}=30. \) 5. A larger full-bin cost \(S\), with the same dollar use \(k\) per day, means fewer pounds are used per day, so reaching the same remaining amount takes more time.

Answer

a) \(t=\frac{S(M-R)}{kM}\) b) \(30\) days c) A larger \(S\) means each daily dollar amount buys fewer pounds, so the same mass decrease takes more days.

5239269

A youth group with \(n\) members rents a bus costing \(\$B\). The cost is shared equally. Then \(m\) more people join, the bus cost rises by \(\$s\), and each person's savings relative to the original per-person cost is \(C\) dollars. These quantities satisfy \( C=\frac{B}{n}-\frac{B+s}{n+m}. \) a) Rearrange the formula to solve for \(s\). b) Use your rearranged formula to find \(s\) when \(B=400\), \(n=20\), \(m=5\), and \(C=2\). c) Explain what your value of \(s\) means in the situation.

Hints

- First isolate the fraction that contains \(s\). - Clear its denominator only after the fraction is isolated. - Subtract \(B\) last. - Check the result by comparing the old and new per-person costs.

Solution

1. Start with \( C=\frac{B}{n}-\frac{B+s}{n+m}. \) 2. Move the second fraction to the other side: \( \frac{B+s}{n+m}=\frac{B}{n}-C. \) 3. Multiply by \(n+m\): \( B+s=(n+m)\left(\frac{B}{n}-C\right). \) 4. Subtract \(B\): \( s=(n+m)\left(\frac{B}{n}-C\right)-B. \) 5. Substitute the values: \( s=25(20-2)-400=25\cdot18-400=50. \) 6. The larger bus can cost \(\$50\) more while still giving each person a \(\$2\) savings.

Answer

a) \(s=(n+m)\left(\frac{B}{n}-C\right)-B\) b) \(s=50\) c) The new bus may cost \(\$50\) more and each person will still save \(\$2\).

5239329

The thin-lens equation relates focal length \(f\), object distance \(g\), and image distance \(b\): \(\frac{1}{f}=\frac{1}{g}+\frac{1}{b}\). a) Solve for \(g\) as a single fraction and state the restrictions for the solved form to be valid in the original equation. b) A student claims the formula can be rearranged as \(g=f-b\). Test the claim using \(f=6\) and \(b=10\). c) Assuming \(b>f>0\), describe how \(g\) changes as \(b\) increases while \(f\) remains fixed. Justify your answer.

Hints

- Isolate the reciprocal containing \(g\) and combine the other fractions. - Check every original denominator and the denominator in the solved form. - For the trend in part c), track how \(\frac{1}{b}\) changes as \(b\) grows.

Solution

a) Subtract \(\frac{1}{b}\): \(\frac{1}{g}=\frac{1}{f}-\frac{1}{b}=\frac{b-f}{fb}\). Taking reciprocals gives \(g=\frac{fb}{b-f}\). For this solved form to be valid in the original equation, \(f\ne0\), \(b\ne0\), and \(b\ne f\). b) The correct formula gives \(g=\frac{6\cdot10}{10-6}=15\). The claim gives \(g=6-10=-4\), so the claim is false. c) 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. As \(b\) becomes very large, \(\frac{1}{b}\) approaches 0, so \(g\) approaches \(f\) from above.

Answer

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.

5239749

Solve \(\frac{x}{a}-b=\frac{x}{c}\) for \(x\) in terms of \(a\), \(b\), and \(c\). State the parameter conditions required for a unique solution.

Hints

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

Solution

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 \(c\ne a\).

Answer

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

5239789

Two numbers are in the ratio \(1:n\), where the second number is larger and \(n > 1\). Their difference is \(d\). Write an expression for each number in terms of \(n\) and \(d\).

Hints

- 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.

Solution

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 \(y = nx\): \(y = \frac{nd}{n - 1}\).

Answer

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

5239819

Solve \((x+a)(x+b)-x^2=c(x+1)\) for \(x\). State the condition on \(a\), \(b\), and \(c\) that guarantees exactly one solution.

Hints

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

Solution

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 \(a+b-c\ne0\).

Answer

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

5240389

Two wheels roll side by side for a distance \(s\). Wheel A has circumference \(u\). Wheel B is larger and has circumference \(1.25u\). Write an equation and derive a formula for the distance \(s\) at which Wheel A makes exactly \(n\) more revolutions than Wheel B. Your formula for \(s\) should depend only on \(n\) and \(u\).

Hints

- 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.

Solution

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 \(5u\): \(s = 5nu\).

Answer

\(s = 5nu\)

5240639

A fruit-juice concentrate is \(p\%\) fruit juice. There are \(V\) liters of concentrate, where \(V > 0\). Water is added to make a drink that is \(q\%\) fruit juice, where \(0 < q < p \le 100\). Derive a formula for the amount of water \(w\), in liters, that must be added.

Hints

- 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 \(w\).

Solution

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

Answer

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

5240649

A laboratory has \(m\) grams of a salt solution that is \(s\%\) salt. Pure salt is added to raise the concentration to \(k\%\), where \(m > 0\) and \(0 \le s < k < 100\). Derive a formula for the number of grams of pure salt \(z\) that must be added.

Hints

- 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 \(z\).

Solution

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

Answer

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

5240689

A cyclist travels for \(t_1\) hours at \(v_1\) miles per hour and then for \(t_2\) hours at \(v_2\) miles per hour. The average speed over the whole trip is \( v_{\text{avg}}=\frac{v_1t_1+v_2t_2}{t_1+t_2}. \) a) Rearrange the formula to solve for \(v_2\). b) Find \(v_2\) when \(v_{\text{avg}}=14\), \(v_1=10\), \(t_1=2\), and \(t_2=3\). c) Check your result in the original average-speed formula.

Hints

- Clear the fraction by multiplying by the entire denominator. - Move the known distance term \(v_1t_1\) before dividing. - The last factor attached to \(v_2\) is \(t_2\). - Check using total distance divided by total time.

Solution

1. Multiply both sides by \(t_1+t_2\): \( v_{\text{avg}}(t_1+t_2)=v_1t_1+v_2t_2. \) 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: \( v_2=\frac{14(2+3)-10\cdot2}{3} =\frac{70-20}{3} =\frac{50}{3}=16\frac23. \) 5. Check: total distance \(=10\cdot2+\frac{50}{3}\cdot3=70\) miles and total time \(=5\) hours, so the average is \(70/5=14\) mph.

Answer

a) \(v_2=\frac{v_{\text{avg}}(t_1+t_2)-v_1t_1}{t_2}\) b) \(v_2=16\frac23\) mph c) Substitution gives an average speed of \(14\) mph.

5241149

Given \(y=\frac{kx}{x+m}\), where \(x+m\ne0\): a) Solve for \(k\), assuming \(x\ne0\). b) Solve for \(x\), assuming \(m\ne0\), \(k\ne0\), and \(k\ne y\).

Hints

- 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.

Solution

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 \(x+m=0\).

Answer

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

52411611

Solve \(\frac{x}{a-b}-\frac{x}{a+b}=\frac{2b}{a}\) for \(x\). Simplify the result. Assume \(a\ne0\), \(b\ne0\), \(a-b\ne0\), and \(a+b\ne0\).

Hints

- Factor the target variable from the left side before combining the rational expressions. - Use a common denominator for the two reciprocal terms. - Check which stated nonzero condition permits the final cancellation.

Solution

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

Answer

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

52411811

Solve \(\frac{a(x+b)}{a+b}+\frac{b(x-a)}{a-b}=a\) for \(x\). Assume \(a\ne b\), \(a\ne-b\), and \(a\ne0\).

Hints

- Clear both symbolic denominators with one common factor. - Expand carefully and collect all terms containing \(x\). - After collecting terms, look for a common factor that appears on both sides.

Solution

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 \(x=a\).

Answer

\(x=a\)

53606910

A closed rectangular prism has a square base with side length \(a\), height \(h\), and total surface area \(1000\,\text{cm}^2\). a) Write a function \(h(a)\) that gives the height in terms of the base side length. State the physically meaningful domain. b) Substitute \(h(a)\) into the volume formula and show that \(V(a)=250a-0.5a^3\). c) Find the volume when \(a=5\,\text{cm}\) and when \(a=15\,\text{cm}\). Which prism has the greater volume?

Hints

- Express total surface area as the two square bases plus four rectangular side faces. - After isolating \(h\), use positive dimensions to determine the physically meaningful domain. - Substitute the height expression into \(V=a^2h\) before evaluating either requested base length.

Solution

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

Answer

a) \(h(a)=\frac{250}{a}-0.5a\), for \(0<a<10\sqrt{5}\) b) \(V(a)=250a-0.5a^3\) 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.

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