Skip to main content
Mathematics LibreTexts

5.4: Further Techniques in Problem Solving

  • Page ID
    49369
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    Combining Techniques in Equation Solving

    In Sections 5.2 and 5.3 we worked with techniques that involved the use of addition, subtraction, multiplication, and division to solve equations. We can combine these techniques to solve more complicated equations. To do so, it is helpful to recall that an equation is solved for a particular variable when all other numbers and/or letters have been disassociated from it and it is alone on one side of the equal sign. We will also note that

    To associate numbers and letters we use the order of operations.

    1. Multiply/divide
    2. Add/subtract

    To undo an association between numbers and letters we use the order of operations in reverse.

    1. Add/subtract
    2. Multiply/divide

    Sample Set A

    Example \(\PageIndex{1}\)

    Solve \(4x - 7 = 9\) for \(x\).

    \(\begin{array}{flushleft}
    4x-7&=&9&\text{First, undo the association between } x \text{ and } 7.\\
    &&&\text{The } 7 \text{ is associated with } x \text{ by subtraction.}\\
    &&&\text{Undo the association by adding } 7 \text{ to both sides.}\\
    4x-7+7&=&9+7\\
    4x&=&16&\text{Now, undo the association between } x { and } 4.\\
    &&&\text{The } 4 \text{ is associated with } x \text{ by multiplication.}\\
    &&&\text{Undo the association by dividing both sides by } 4.\\
    \dfrac{4x}{4}&=&\dfrac{16}{4}\\
    16-7&=&9&\text{Is this correct? }\\
    x&=&4
    \end{array}\)

    Check:
    \(\begin{array}{flushleft}
    4(4)-7&=&9&\text{Is this correct?}\\
    9&=&9&\text{Yes, this is correct}.
    \end{array}\)

    Example \(\PageIndex{2}\)

    Solve \(\dfrac{3y}{4}-5=-11\).

    \(\begin{array}{flushleft}
    \dfrac{3y}{4}-5&=&-11&-5\text{ is associated with } y \text{ by subtraction.}\\
    &&&\text{Undo the association by adding } 5 \text{ to both sides}\\
    \dfrac{3y}{4}-5+5&=&-11+5\\
    \dfrac{3y}{4}&=&-6&4\text{ is associated with } y \text{ by division}.\\
    &&&\text{Undo the association by multiplying both sides by } 4.\\
    4 \cdot \dfrac{3y}{4}&=&4(-6)\\
    3y&=&-24&3\text{ is associated with } y \text{ by multiplication.}\\
    &&&\text{ Undo the association by dividing both sides by } 3.\\
    \dfrac{3y}{3}&=&\dfrac{-24}{3}\\
    \dfrac{3y}{3}&=&-8\\
    y&=&-8
    \end{array}\)

    Check:
    \(\begin{array}{flushleft}
    \dfrac{3(-8)}{4}-5&=&-11&\text{Is this correct?}\\
    \dfrac{-24}{4}-5&=&-11&\text{Is this correct?}\\
    -6-5&=&-11&\text{Is this correct?}\\
    -11&=&-11&\text{Yes, this is correct}
    \end{array}\)

    Example \(\PageIndex{3}\)

    Solve \(\dfrac{8a}{3b} + 2m = 6m - 5\) for \(a\).

    \(\begin{array}{flushleft}
    \dfrac{8a}{3b}+2m&=&6m-5&2m\text{ is associated with } a \text{ by addition. Undo the association}\\
    &&&\text{by subtracting } 2m \text{ from both sides.}\\
    \dfrac{8a}{3b}+2m-2m&=&6m-5-2m\\
    \dfrac{8a}{3b}&=&4m-5&3b\text{associated with } a \text{ by division. Undo the association}\\
    &&&\text{by multiplying both sides by } 3b.\\
    (3b)(\dfrac{8a}{3b})&=&3b(4m-5)\\
    8a&=&12bm-15b&8\text{ is associated with } a \text{ by multiplication. Undo the}\\
    &&&\text{multiplication by dividing both sides by } 8.\\
    \dfrac{8a}{8}&=&\dfrac{12bm-15b}{8}\\
    a&=&\dfrac{12bm-15b}{8}
    \end{array}\)

    Practice Set A

    Practice Problem \(\PageIndex{1}\)

    Solve \(3y−1=11\) for \(y\).

    Answer

    \(y = 4\)

    Practice Problem \(\PageIndex{2}\)

    Solve \(\dfrac{5m}{2} + 6 = 1\) for \(m\).

    Answer

    \(m=−2\)

    Practice Problem \(\PageIndex{3}\)

    Solve \(2n+3m=4\) for \(n\).

    Answer

    \(n = \dfrac{4-3m}{2}\)

    Practice Problem \(\PageIndex{4}\)

    Solve \(\dfrac{9k}{2h}+5=p-2\) for \(k\)

    Answer

    \(k = \dfrac{2hp-14h}{9}\)

    Sometimes when solving an equation it is necessary to simplify the expressions composing it.

    Sample Set B

    Example \(\PageIndex{4}\)

    Solve \(4x+1-3x = (-2)(4)\) for \(x\).

    \(\begin{array}{flushleft}
    4x+1-3x&=&(-2)(4)\\
    x+1&=&-8\\
    x&=&-9\\
    \end{array}\)

    Check:
    \(\begin{array}{flushleft}
    4(-9)+1-3(-9)&=&-8&\text{Is this correct?}\\
    -36+1+27&=&-8&\text{Is this correct?}\\
    -8&=&-8&\text{Yes, this is correct.}
    \end{array}\)

    Example \(\PageIndex{5}\)

    Solve \(3(m-6)-2m=-4+1\) for \(m\).

    \(\begin{array}{flushleft}
    3(m-6)-2m&=&-4+1\\
    3m-18-2m&=&-3\\
    m-18&=&-3\\
    m&=&15
    \end{array}\)

    Check:
    \(\begin{array}{flushleft}
    3(15-6)-2(15)&=&-4+1&\text{Is this correct?}\\
    3(9)-30&=&-3&\text{Is this correct?}\\
    27-30&=&-3&\text{Is this correct?}\\
    -3&=&-3&\text{Yes, this is correct}
    \end{array}\)

    Practice Set B

    Solve and check each equation.

    Practice Problem \(\PageIndex{5}\)

    \(16x−3−15x=8\) for \(x\).

    Answer

    \(x=11\)

    Practice Problem \(\PageIndex{6}\)

    \(4(y−5)−3y=−1\) for \(y\).

    Answer

    \(y=19\)

    Practice Problem \(\PageIndex{7}\)

    \(-2(a^2+3a-1)+2a^2+7a=0\) for \(a\).

    Answer

    \(a=−2\)

    Practice Problem \(\PageIndex{8}\)

    \(5m(m-2a-1)-5m^2+2a(5m+3)=10\) for \(a\).

    Answer

    \(a = \dfrac{10+5m}{6}\)

    Often the variable we wish to solve for will appear on both sides of the equal sign. We can isolate the variable on either the left or right side of the equation by using the techniques of Sections 5.2 and 5.3.

    Sample Set C

    Example \(\PageIndex{6}\)

    Solve \(6x-4=2x+8\) for \(x\).

    \(\begin{array}{flushleft}
    6x-4&=&2x+8&\text{To isolate } x \text{ on the left side, subtract } 2x \text{ from both sides.}\\
    6x-4-2x&=&2x+8-2x\\
    4x-4&=&8&\text{Add } 4 \text{ to both sides.}\\
    4x-4+4&=&8+4\\
    4x&=&12&\text{Divide both sides by } 4\\
    \dfrac{4x}{4}&=&\dfrac{12}{4}\\
    x&=&3
    \end{array}\)

    Check:

    \(\begin{array}{flushleft}
    6(3)-4&=&2(3)+8&\text{Is this correct?}\\
    18-4&=&6+8&\text{Is this correct?}\\
    14&=&14&\text{Yes, this is correct.}
    \end{array}\)

    Example \(\PageIndex{7}\)

    Solve \(6(1-3x)+1=2x-[3(x-7)-20]\) for \(x\).

    On left side of an equation, arrows show that six is multiplied with each term inside the parentheses, and on right side, arrows show that three is multiplied with each term inside the parentheses.

    \(\begin{array}{flushleft}
    6-18x+1&=&2x-[3x-21-20]\\
    -18x+7&=&2x-[3x-41]\\
    -18x+7&=&2x-3x+41\\
    -18x+7&=&-x+41&\text{To isolate } x \text{ on the right side, add } 18x \text{ to both sides.}\\
    -18x+7+18x&=&-x+41+18x\\
    7&=&17x+41&\text{Subtract } 41 \text{ from both sides}.\\
    7-41&=&17x+41-41\\
    -34&=&17x&\text{Divide both sides by } 17.\\
    \dfrac{-34}{17}&=&\dfrac{17x}{17}\\
    -2&=&x&\text{Since the equation} -2=x \text{ is equivalent to the equation }\\
    &&&x=-2, \text{ we can write the answer as } x=-2.\\
    x&=&-2
    \end{array}\)

    Check:
    \(\begin{array}{flushleft}
    6(1-(3(-2))+1&=&2(-2)-[3(-2-7)-20]&\text{Is this correct?}\\
    6(1+6)+1&=&−4−[3(−9)−20]&\text{Is this correct?}\\
    6(7)+1&=&-4-[-27-20]&\text{Is this correct?}\\
    42+1&=&-4-[-47]&\text{Is this correct?}\\
    43&=&-4+47&\text{Is this correct?}\\
    43&=&43&\text{Yes, this is correct}\\
    \end{array}\)

    Practice Set C

    Practice Problem \(\PageIndex{9}\)

    Solve \(8a+5=3a−5\) for \(a\).

    Answer

    \(a=−2\)

    Practice Problem \(\PageIndex{10}\)

    Solve \(9y+3(y+6)=15y+21\) for \(y\).

    Answer

    \(y=−1\)

    Practice Problem \(\PageIndex{11}\)

    Solve \(3k+2[4(k−1)+3]=63−2k\) for \(k\).

    Answer

    \(k=5\)

    Recognizing Identities and Contradictions

    As we noted in Section 5.2, some equations are identities and some are contradictions. As the problems of Sample Set D will suggest,

    Recognizing an Identity

    1. If, when solving an equation, all the variables are eliminated and a true statement results, the equation is an identity.

    Recognizing a Contradiction

    1. If, when solving an equation, all the variables are eliminated and a false statement results, the equation is a contradiction.

    Sample Set D

    Example \(\PageIndex{8}\)

    Solve \(9x + 3(4-3x) = 12\) for \(x\)

    On left side of an equation, arrows show that three is multiplied with each term inside the parentheses.

    \(\begin{array}{flushleft}
    9x+12-9x&=&12\\
    12&=&12
    \end{array}\)

    The variable has been eliminated and the result is a true statement. The original equation is an identity.

    Example \(\PageIndex{8}\)

    Solve \(-2(10-2y)-4y+1=-18\) for \(y\)

    On left side of an equation, arrows show that negative two is multiplied with each term inside the parentheses.

    \(\begin{array}{flushleft}
    -20+4y-4y+1&=&-18\\
    -19&=&-18
    \end{array}\)

    The variable has been eliminated and the result is a false statement. The original equation is a contradiction.

    Practice Set D

    Classify each equation as an identity or a contradiction.

    Practice Problem \(\PageIndex{12}\)

    \(6x+3(1−2x)=3\)

    Answer

    identity, \(3=3\)

    Practice Problem \(\PageIndex{13}\)

    \(−8m+4(2m−7)=28\)

    Answer

    contradiction, \(−28=28\)

    Practice Problem \(\PageIndex{14}\)

    \(3(2x−4)−2(3x+1)+14=0\)

    Answer

    identity, \(0=0\)

    Practice Problem \(\PageIndex{15}\)

    \(−5(x+6)+8=3[4−(x+2)]−2x\)

    Answer

    contradiction, \(−22=6\)

    Exercises

    For the following problems, solve each conditional equation. If the equation is not conditional, identify it as an identity or a contradiction.

    Exercise \(\PageIndex{1}\)

    \(3x+1=16\)

    Answer

    \(x=5\)

    Exercise \(\PageIndex{2}\)

    \(6y−4=20\)

    Exercise \(\PageIndex{3}\)

    \(4a−1=27\)

    Answer

    \(a=7\)

    Exercise \(\PageIndex{4}\)

    \(3x+4=40\)

    Exercise \(\PageIndex{5}\)

    \(2y+7=−3\)

    Answer

    \(y=−5\)

    Exercise \(\PageIndex{6}\)

    \(8k−7=−23\)

    Exercise \(\PageIndex{7}\)

    \(5x+6=−9\)

    Answer

    \(x=−3\)

    Exercise \(\PageIndex{8}\)

    \(7a+2=−26\)

    Exercise \(\PageIndex{9}\)

    \(10y−3=−23\)

    Answer

    \(y=−2\)

    Exercise \(\PageIndex{10}\)

    \(14x+1=−55\)

    Exercise \(\PageIndex{11}\)

    \(\dfrac{x}{9} + 2 = 6\)

    Answer

    \(x=36\)

    Exercise \(\PageIndex{12}\)

    \(\dfrac{m}{7} - 8 = -11\)

    Exercise \(\PageIndex{13}\)

    \(\dfrac{y}{4}+6=12\)

    Answer

    \(y=24\)

    Exercise \(\PageIndex{14}\)

    \(\dfrac{x}{8}-2=5\)

    Exercise \(\PageIndex{15}\)

    \(\dfrac{m}{11}-15=-19\)

    Answer

    \(m=−44\)

    Exercise \(\PageIndex{16}\)

    \(\dfrac{k}{15}+20=10\)

    Exercise \(\PageIndex{17}\)

    \(6+\dfrac{k}{5}=5\)

    Answer

    \(k=−5\)

    Exercise \(\PageIndex{18}\)

    \(1-\dfrac{n}{2}=6\)

    Exercise \(\PageIndex{19}\)

    \(\dfrac{7x}{4}+6=-8\)

    Answer

    \(x=−8\)

    Exercise \(\PageIndex{20}\)

    \(\dfrac{-6m}{5}+11=-13\)

    Exercise \(\PageIndex{21}\)

    \(\dfrac{3k}{14} + 25 = 22\)

    Answer

    \(k=-14\)

    Exercise \(\PageIndex{22}\)

    \(3(x−6)+5=−25\)

    Exercise \(\PageIndex{23}\)

    \(16(y−1)+11=−85\)

    Answer

    \(y=−5\)

    Exercise \(\PageIndex{24}\)

    \(6x+14=5x−12\)

    Exercise \(\PageIndex{25}\)

    \(23y−19=22y+1\)

    Answer

    \(y=20\)

    Exercise \(\PageIndex{26}\)

    \(−3m+1=3m−5\)

    Exercise \(\PageIndex{27}\)

    \(8k+7=2k+1\)

    Answer

    \(k=−1\)

    Exercise \(\PageIndex{28}\)

    \(12n+5=5n−16\)

    Exercise \(\PageIndex{29}\)

    \(2(x−7)=2x+5\)

    Answer

    contradiction

    Exercise \(\PageIndex{30}\)

    \(−4(5y+3)+5(1+4y)=0\)

    Exercise \(\PageIndex{31}\)

    \(3x+7=−3−(x+2)\)

    Answer

    \(x=−3\)

    Exercise \(\PageIndex{32}\)

    \(4(4y+2)=3y+2[1−3(1−2y)]\)

    Exercise \(\PageIndex{33}\)

    \(5(3x−8)+11=2−2x+3(x−4)\)

    Answer

    \(x = \dfrac{19}{14}\)

    Exercise \(\PageIndex{34}\)

    \(12−(m−2)=2m+3m−2m+3(5−3m)\)

    Exercise \(\PageIndex{35}\)

    \(−4 \cdot k−(−4−3k)=−3k−2k−(3−6k)+1\)

    Answer

    \(k = 3\)

    Exercise \(\PageIndex{36}\)

    \(3[4−2(y+2)]=2y−4[1+2(1+y)]\)

    Exercise \(\PageIndex{37}\)

    \(−5[2m−(3m−1)]=4m−3m+2(5−2m)+1\)

    Answer

    \(m=2\)

    For the following problems, solve the literal equations for the indicated variable. When directed, find the value of that variable for the given values of the other variables.

    Exercise \(\PageIndex{38}\)

    Solve \(I = \dfrac{E}{R}\) for \(R\). Find the value of \(R\) when \(I = 0.005\) and \(E = 0.0035\).

    Exercise \(\PageIndex{39}\)

    Solve \(P = R - C\) for \(R\). Find the value of \(R\) when \(P = 27\) and \(C = 85\).

    Answer

    \(R=112\)

    Exercise \(\PageIndex{40}\)

    Solve \(z = \dfrac{x-\hat{x}}{s}\) for \(x\). Find the value of \(x\) when \(z = 1.96, s = 2.5,\) and \(\hat{x} = 15\).

    Exercise \(\PageIndex{41}\)

    Solve \(F = \dfrac{S_{x}^{2}}{S_{y}^{2}}\) for \(S_{x}^{2} \cdot S_{x}^{2}\) represents a single quantity. Find the value of \(S_{x}^{2}\) when \(F = 2.21\) and \(S_{y}^{2} = 3.24\)

    Answer

    \(S_{x}^{2} = F \cdot S_{y}^{2}\); \(S_{x}^{2} = 7.1604\)

    Exercise \(\PageIndex{42}\)

    Solve \(p = \dfrac{nRT}{V}\) for \(R\)

    Exercise \(\PageIndex{43}\)

    Solve \(x=4y+7\) for \(y\).

    Answer

    \(y = \dfrac{x-7}{4}\)

    Exercise \(\PageIndex{44}\)

    Solve \(y=10x+16\) for \(x\).

    Exercise \(\PageIndex{45}\)

    Solve \(2x+5y=12\) for \(y\).

    Answer

    \(y = \dfrac{-2x+12}{5}\)

    Exercise \(\PageIndex{46}\)

    Solve \(−9x+3y+15=0\) for \(y\).

    Exercise \(\PageIndex{47}\)

    Solve \(m = \dfrac{2n-h}{5}\) for \(n\).

    Answer

    \(\dfrac{5m+h}{2}\)

    Exercise \(\PageIndex{48}\)

    Solve \(t = \dfrac{Q+6P}{8}\) for \(P\).

    Exercises for Review

    Exercise \(\PageIndex{49}\)

    Simplify \((x+3)^2(x-2)^3(x-2)^4(x+3)\)

    Answer

    \((x+3)^3(x-2)^7\)

    Exercise \(\PageIndex{50}\)

    Find the product. \((x-7)(x+7)\)

    Exercise \(\PageIndex{51}\)

    Find the product. \((2x-1)^2\).

    Answer

    \(4x^2-4x+1\)

    Exercise \(\PageIndex{52}\)

    Solve the equation \(y−2=−2\).

    Exercise \(\PageIndex{53}\)

    Solve the equation \(\dfrac{4x}{5} = -3\)

    Answer

    \(x = \dfrac{-15}{4}\)


    This page titled 5.4: Further Techniques in Problem Solving is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Denny Burzynski & Wade Ellis, Jr. (OpenStax CNX) .