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2.2: Use a General Strategy to Solve Linear Equations

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    32403
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    Summary

    By the end of this section, you will be able to:

    • Use the commutative and associative properties
    • Use the properties of identity, inverse, and zero
    • Simplify expressions using the Distributive Property

    Before you get started, take this readiness quiz.

    1. Simplify: \(\frac{3}{2}(12x+20)\).
    2. Simplify: \(5−2(n+1)\).
    3. Find the LCD of \(\frac{5}{6}\) and \(\frac{1}{4}\).

    Solve Linear Equations Using a General Strategy

    Solving an equation is like discovering the answer to a puzzle. The purpose in solving an equation is to find the value or values of the variable that makes it a true statement. Any value of the variable that makes the equation true is called a solution to the equation. It is the answer to the puzzle!

    Definition: SOLUTION OF AN EQUATION

    A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.

    To determine whether a number is a solution to an equation, we substitute the value for the variable in the equation. If the resulting equation is a true statement, then the number is a solution of the equation.

    how-to.png DETERMINE WHETHER A NUMBER IS A SOLUTION TO AN EQUATION

    1. Substitute the number for the variable in the equation.
    2. Simplify the expressions on both sides of the equation.
    3. Determine whether the resulting equation is true.
      • If it is true, the number is a solution.
      • If it is not true, the number is not a solution.

    Example \(\PageIndex{1}\)

    Determine whether the values are solutions to the equation: \(5y+3=10y−4\).

    1. \(y=\frac{3}{5}\)
    2. \(y=\frac{7}{5}\)
    Solution

    Since a solution to an equation is a value of the variable that makes the equation true, begin by substituting the value of the solution for the variable.

    a.

      \(5 y+3=10 y-4\)
    Substitute \(\color{rec}\frac{3}{5}\) for \(y\) \(5\left( \color{red} \frac{3}{5} \color{black}\right)+3 \stackrel{?}{=} 10\left( \color{red}\frac{3}{5} \color{black}\right)-4\)
    Multiply. \(3+3\stackrel{?}{=} 6-4\)
    Simplify. \(6 \neq 2\)

    Since \(y=\frac{3}{5}\) does not result in a true equation, \(y=\frac{3}{5}\) is not a solution to the equation \(5y+3=10y−4.\)

    b.

      \(5 y+3=10 y-4\)
    Substitute \(\color{red} \frac{7}{5}\) for \(y\) \(5\left(\color{red} \frac{7}{5} \color{black}\right)+3 \stackrel{?}{=} 10\left(\color{red}\frac{7}{5}\color{back}\right)-4\)
    Multiply. \(7+3 \stackrel{?}{=} 14-4\)
    Simplify. \(10=10 \checkmark\)

    Since \(y=\frac{7}{5}\) results in a true equation, \(y=\frac{7}{5}\) is a solution to the equation \(5y+3=10y−4.\)

    try-it.png \(\PageIndex{1A}\)

    Determine whether the values are solutions to the equation: \(9y+2=6y+3.\)

    1. \(y=\frac{4}{3}\)
    2. \(y=\frac{1}{3}\)
    Answer a

     no 

    Answer b

    yes

    try-it.png \(\PageIndex{1B}\)

    Determine whether the values are solutions to the equation: \(4x−2=2x+1\).

    1. \(x=\frac{3}{2}\)
    2. \(x=−\frac{1}{2}\)
    Answer a

    yes

    Answer b

    no

    There are many types of equations that we will learn to solve. In this section we will focus on a linear equation.

    Definition: LINEAR EQUATION

    A linear equation is an equation in one variable that can be written, where \(a\) and \(b\) are real numbers and \(a≠0\), as:

    \[ax+b=0 \nonumber\]

    To solve a linear equation it is a good idea to have an overall strategy that can be used to solve any linear equation. In the next example, we will give the steps of a general strategy for solving any linear equation. Simplifying each side of the equation as much as possible first makes the rest of the steps easier.

    EXAMPLE \(\PageIndex{2}\)

    Solve: \(7(n−3)−8=−15\)

    Solution
    CNX_IntAlg_Figure_02_01_003a_img.jpg
    Figure_02_01_003b_img.jpg" />
    CNX_IntAlg_Figure_02_01_003c_img.jpg
    Figure_02_01_003d_img.jpg" />

    CNX_IntAlg_Figure_02_01_003e_img.jpg

    try-it.png \(\PageIndex{2A}\)

    Solve: \(2(m−4)+3=−1.\)

    Answer

    \(m=2\)

    try-it.png \(\PageIndex{2B}\)

    Solve: \(5(a−3)+5=−10.\)

    Answer

    \(a=0\)

    These steps are summarized in the General Strategy for Solving Linear Equations below.

    how-to.png SOLVE LINEAR EQUATIONS USING A GENERAL STRATEGY

    1. Simplify each side of the equation as much as possible.
      Use the Distributive Property to remove any parentheses.
      Combine like terms.
    2. Collect all the variable terms on one side of the equation.

      Use the Addition or Subtraction Property of Equality.

    3. Collect all the constant terms on the other side of the equation.

      Use the Addition or Subtraction Property of Equality.

    4. Make the coefficient of the variable term equal to 1.

      Use the Multiplication or Division Property of Equality.

      State the solution to the equation.

    5. Check the solution.

      Substitute the solution into the original equation to make sure the result is a true statement.

    EXAMPLE \(\PageIndex{3}\)

    Solve: \(\frac{2}{3}(3m−6)=5−m\).

    Answer
      \(\frac{2}{3}(3 m-6)=5-m\)
    Distribute. \(2 m-4=5-m\)
    Add m to both sides to get the variables only on the left. CNX_IntAlg_Figure_02_01_004h_img.jpg
    Simplify. \(3 m-4=5\)
    Add 4 to both sides to get constants only on the right. CNX_IntAlg_Figure_02_01_004j_img.jpg
    Simplify. \(3 m=9\)
    Divide both sides by three. CNX_IntAlg_Figure_02_01_004l_img.jpg
    Simplify. \(m=3\)

     

    Check: CNX_IntAlg_Figure_02_01_004f_img.jpg
    Let m=3 CNX_IntAlg_Figure_02_01_004b_img.jpg
      CNX_IntAlg_Figure_02_01_004c_img.jpg
      CNX_IntAlg_Figure_02_01_004d_img.jpg
      CNX_IntAlg_Figure_02_01_004e_img.jpg

    try-it.png \(\PageIndex{3A}\)

    Solve: \(\frac{1}{3}(6u+3)=7−u\).

    Answer

    \(u=2\)

    try-it.png \(\PageIndex{3B}\)

    Solve: \(\frac{2}{3}(9x−12)=8+2x\).

    Answer

    \(x=4\)

    We can solve equations by getting all the variable terms to either side of the equal sign. By collecting the variable terms on the side where the coefficient of the variable is larger, we avoid working with some negatives. This will be a good strategy when we solve inequalities later in this chapter. It also helps us prevent errors with negatives.

    EXAMPLE \(\PageIndex{4}\)

    Solve: \(4(x−1)−2=5(2x+3)+6\).

    Answer
          \(4(x-1)-2=5(2 x+3)+6\)
    Distribute.    

    \(4 x-4-2=10 x+15+6\)

    Combine like terms.     \(4 x-6=10 x+21\)
    Subtract \(4x\) from each side to get the variables only on the right since \(10>4\).     CNX_IntAlg_Figure_02_01_005j_img.jpg
    Simplify.     \(-6=6 x+21\)
    Subtract 21 from each side to get the constants on left.     CNX_IntAlg_Figure_02_01_005l_img.jpg
    Simplify.     \(-27=6 x\)
    Divide both sides by 6.     CNX_IntAlg_Figure_02_01_005n_img.jpg
    Simplify.     \(-\frac{9}{2}=x\)
    Check: CNX_IntAlg_Figure_02_01_005g_img.jpg      
    Let \(x=−92.\) CNX_IntAlg_Figure_02_01_005b_img.jpg      
      CNX_IntAlg_Figure_02_01_005c_img.jpg    
      CNX_IntAlg_Figure_02_01_005d_img.jpg      
      CNX_IntAlg_Figure_02_01_005e_img.jpg      
      CNX_IntAlg_Figure_02_01_005f_img.jpg      

    try-it.png \(\PageIndex{4A}\)

    Solve: \(6(p−3)−7=5(4p+3)−12.\)

    Answer

    \(p=−2\)

    try-it.png \(\PageIndex{4B}\)

    Solve: \(8(q+1)−5=3(2q−4)−1.\)

    Answer

    \(q=−8\)

    EXAMPLE \(\PageIndex{5}\)

    Solve: \(10[3−8(2s−5)]=15(40−5s)\).

    Answer
      \(10[3-8(2 s-5)]=15(40-5 s)\)
    Simplify from the innermost parentheses first. \(10[3-16 s+40]=15(40-5 s)\)
    Combine like terms in the brackets. \(10[43-16 s]=15(40-5 s)\)
    Distribute. \(430-160 s=600-75 s\)
    Add \(160s\) to both sides to get the variable to the right. CNX_IntAlg_Figure_02_01_006l_img.jpg
    Simplify. \(430=600+85 s\)
    Subtract 600 from both sides to get the constant to the left. CNX_IntAlg_Figure_02_01_006n_img.jpg
    Simplify. \(-170=85 s\)
    Divide both sides by 85. CNX_IntAlg_Figure_02_01_006p_img.jpg
    Simplify. \(\-2=s)
    Check: \(10[3-8(2 s-5)]=15(40-5 s)\)  
    Let \(s=−2\). CNX_IntAlg_Figure_02_01_006b_img.jpg  
      CNX_IntAlg_Figure_02_01_006c_img.jpg  
      CNX_IntAlg_Figure_02_01_006d_img.jpg  
      CNX_IntAlg_Figure_02_01_006e_img.jpg  
      CNX_IntAlg_Figure_02_01_006f_img.jpg  
      CNX_IntAlg_Figure_02_01_006g_img.jpg  

    try-it.png \(\PageIndex{5A}\)

    Solve: \(6[4−2(7y−1)]=8(13−8y)\).

    Answer

    \(y=−\frac{17}{5}\)

    try-it.png \(\PageIndex{5B}\)

    Solve: \(12[1−5(4z−1)]=3(24+11z).\)

    Answer

    \(z=0\)

    Classify Equations

    Whether or not an equation is true depends on the value of the variable. The equation \(7x+8=−13\) is true when we replace the variable, \(x\), with the value \(−3\), but not true when we replace \(x\) with any other value. An equation like this is called a conditional equation. All the equations we have solved so far are conditional equations.

    Definition: CONDITIONAL EQUATION

    An equation that is true for one or more values of the variable and false for all other values of the variable is a conditional equation.

    Now let’s consider the equation \(7y+14=7(y+2)\). Do you recognize that the left side and the right side are equivalent? Let’s see what happens when we solve for \(y\).

    Solve:

      \(7 y+14=7(y+2)\)
    Distribute. \(7 y+14=7 y+14\)
    Subtract \(7y\) to each side to get the \(y\)'s to one side. \(7 y \color{red}-7 y \color{black} +14=7 y \color{red} -7 y \color{black}+14\)
    Simplify—the \(y\)’s are eliminated. \(14=14\)
      But \(14=14\) is true.

    This means that the equation \(7y+14=7(y+2)\) is true for any value of y. We say the solution to the equation is all of the real numbers. An equation that is true for any value of the variable is called an identity.

    Definition: IDENTITY

    An equation that is true for any value of the variable is called an identity.

    The solution of an identity is valid for all real numbers.

    What happens when we solve the equation \(−8z=−8z+9\)?

    Solve:

      \(-8 z=-8 z+9\)
    Add 8z to both sides to leave the constant alone on the right. \(-8 z \color{red} +8 z \color{black}=-8 z \color{red}+8 z \color{black} +9\)
    Simplify—the \(z\)'s are eliminated. \(0 \neq 9\)
      But \(0≠9\).

    Solving the equation \(−8z=−8z+9\) led to the false statement \(0=9\). The equation \(−8z=−8z+9\) will not be true for any value of z. It has no solution. An equation that has no solution, or that is false for all values of the variable, is called a contradiction.

    definition: CONTRADICTION

    An equation that is false for all values of the variable is called a contradiction.

    A contradiction has no solution.

    The next few examples will ask us to classify an equation as conditional, an identity, or as a contradiction.

    EXAMPLE \(\PageIndex{6}\)

    Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: \(6(2n−1)+3=2n−8+5(2n+1)\).

    Answer
      \(6(2 n-1)+3=2 n-8+5(2 n+1)\)
    Distribute. \(12 n-6+3=2 n-8+10 n+5\)
    Combine like terms. \(12 n-3=12 n-3\)
    Subtract 12n from each side to get the \(n\)’s to one side. CNX_IntAlg_Figure_02_01_009d_img.jpg
    Simplify. \(-3=-3\)
    This is a true statement. The equation is an identity.
      The solution is all real numbers.

    try-it.png \(\PageIndex{6A}\)

    Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: \(4+9(3x−7)=−42x−13+23(3x−2).\)

    Answer

    identity; all real numbers

    Exercise \(\PageIndex{6B}\)

    Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: \(8(1−3x)+15(2x+7)=2(x+50)+4(x+3)+1.\)

    Answer

    identity; all real numbers

    EXAMPLE \(\PageIndex{7}\)

    Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: \(8+3(a−4)=0\).

    Answer
      CNX_IntAlg_Figure_02_01_010a_img.jpg
    Distribute. CNX_IntAlg_Figure_02_01_010b_img.jpg
    Combine like terms. CNX_IntAlg_Figure_02_01_010c_img.jpg
    Add 4 to both sides. CNX_IntAlg_Figure_02_01_010d_img.jpg
    Simplify. CNX_IntAlg_Figure_02_01_010e_img.jpg
    Divide. CNX_IntAlg_Figure_02_01_010f_img.jpg
    Simplify. CNX_IntAlg_Figure_02_01_010g_img.jpg
    The equation is true when \(a=\frac{4}{3}\). This is a conditional equation.
     

    The solution is \(a=\frac{4}{3}\).

    Exercise \(\PageIndex{7A}\)

    Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: \(11(q+3)−5=19\).

    Answer

    conditional equation; \(q=−\frac{9}{11}\)

    Exercise \(\PageIndex{7B}\)

    Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: \(6+14(k−8)=95\).

    Answer

    conditional equation; \(k=\frac{201}{14}\)

    EXAMPLE \(\PageIndex{8}\)

    Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: \(5m+3(9+3m)=2(7m−11)\).

    Answer
      CNX_IntAlg_Figure_02_01_011a_img.jpg
    Distribute. CNX_IntAlg_Figure_02_01_011b_img.jpg
    Combine like terms. CNX_IntAlg_Figure_02_01_011c_img.jpg
    Subtract \(14m\) from both sides. CNX_IntAlg_Figure_02_01_011d_img.jpg
    Simplify. CNX_IntAlg_Figure_02_01_011e_img.jpg
    But \(27≠−22\). The equation is a contradiction.
      It has no solution.

    Exercise \(\PageIndex{8A}\)

    Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: \(12c+5(5+3c)=3(9c−4)\).

    Answer

    contradiction; no solution

    Exercise \(\PageIndex{8B}\)

    Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution:\(4(7d+18)=13(3d−2)−11d\).

    Answer

    contradiction; no solution

    We summarize the methods for classifying equations in the table.

    Type of equation What happens when you solve it? Solution
    Conditional Equation True for one or more values of the variables and false for all other values One or more values
    Identity True for any value of the variable All real numbers
    Contradiction False for all values of the variable No solution
     

    Solve Equations with Fraction or Decimal Coefficients

    We could use the General Strategy to solve the next example. This method would work fine, but many students do not feel very confident when they see all those fractions. So, we are going to show an alternate method to solve equations with fractions. This alternate method eliminates the fractions.

    We will apply the Multiplication Property of Equality and multiply both sides of an equation by the least common denominator (LCD) of all the fractions in the equation. The result of this operation will be a new equation, equivalent to the first, but without fractions. This process is called clearing the equation of fractions.

    To clear an equation of decimals, we think of all the decimals in their fraction form and then find the LCD of those denominators.

    EXAMPLE \(\PageIndex{9}\): How to Solve Equations with Fraction or Decimal Coefficients

    Solve: \(\frac{1}{12}x+\frac{5}{6}=\frac{3}{4}\).

    Answer
    CNX_IntAlg_Figure_02_01_012a_img.jpg

    CNX_IntAlg_Figure_02_01_012c_img.jpg

    Exercise \(\PageIndex{9A}\)

    Solve: \(\frac{1}{4}x+\frac{1}{2}=\frac{5}{8}\).

    Answer

    \(x=\frac{1}{2}\)

    Exercise \(\PageIndex{9B}\)

    Solve: \(\frac{1}{8}x+\frac{1}{2}=\frac{1}{4}\).

    Answer

    \(x=−2\)

    Notice in the previous example, once we cleared the equation of fractions, the equation was like those we solved earlier in this chapter. We changed the problem to one we already knew how to solve. We then used the General Strategy for Solving Linear Equations.

    how-to.png SOLVE EQUATIONS WITH FRACTION OR DECIMAL COEFFICIENTS.

    1. Find the least common denominator (LCD) of all the fractions and decimals (in fraction form) in the equation.
    2. Multiply both sides of the equation by that LCD. This clears the fractions and decimals.
    3. Solve using the General Strategy for Solving Linear Equations.

    EXAMPLE \(\PageIndex{10}\)

    Solve: \(5=\frac{1}{2}y+\frac{2}{3}y−\frac{3}{4}y\).

    Answer

    We want to clear the fractions by multiplying both sides of the equation by the LCD of all the fractions in the equation.

    Find the LCD of all fractions in the equation. CNX_IntAlg_Figure_02_01_013e_img.jpg
    The LCD is 12.  
    Multiply both sides of the equation by 12. CNX_IntAlg_Figure_02_01_013f_img.jpg
    Distribute. CNX_IntAlg_Figure_02_01_013g_img.jpg
    Simplify—notice, no more fractions. CNX_IntAlg_Figure_02_01_013h_img.jpg
    Combine like terms. CNX_IntAlg_Figure_02_01_013i_img (1).jpg
    Divide by five. CNX_IntAlg_Figure_02_01_013j_img.jpg
    Simplify. CNX_IntAlg_Figure_02_01_013k_img.jpg
    Check: CNX_IntAlg_Figure_02_01_013a_img.jpg  
    Let \(y=12\). CNX_IntAlg_Figure_02_01_013b_img.jpg  
      CNX_IntAlg_Figure_02_01_013c_img.jpg  
      CNX_IntAlg_Figure_02_01_013d_img.jpg  

    Exercise \(\PageIndex{10A}\)

    Solve: \(7=\frac{1}{2}x+\frac{3}{4}x−\frac{2}{3}x\).

    Answer

    \(x=12\)

    Exercise \(\PageIndex{10B}\)

    Solve: \(−1=\frac{1}{2}u+\frac{1}{4}u−\frac{2}{3}u\).

    Answer

    \(u=−12\)

    In the next example, we’ll distribute before we clear the fractions.

    EXAMPLE \(\PageIndex{11}\)

    Solve: \(\frac{1}{2}(y−5)=\frac{1}{4}(y−1)\).

    Answer
      CNX_IntAlg_Figure_02_01_014c_img.jpg
    Distribute. CNX_IntAlg_Figure_02_01_014d_img.jpg
    Simplify. CNX_IntAlg_Figure_02_01_014e_img.jpg
    Multiply by the LCD, four. CNX_IntAlg_Figure_02_01_014f_img.jpg
    Distribute. CNX_IntAlg_Figure_02_01_014g_img.jpg
    Simplify. CNX_IntAlg_Figure_02_01_014h_img.jpg
    Collect the variables to the left. CNX_IntAlg_Figure_02_01_014i_img.jpg
    Simplify. CNX_IntAlg_Figure_02_01_014j_img.jpg
    Collect the constants to the right. CNX_IntAlg_Figure_02_01_014k_img.jpg
    Simplify. CNX_IntAlg_Figure_02_01_014l_img.jpg
    An alternate way to solve this equation is to clear the fractions without distributing first. If you multiply the factors correctly, this method will be easier.
      CNX_IntAlg_Figure_02_01_014m_img.jpg
    Multiply by the LCD, 4. CNX_IntAlg_Figure_02_01_014n_img.jpg
    Multiply four times the fractions. CNX_IntAlg_Figure_02_01_014o_img.jpg
    Distribute. CNX_IntAlg_Figure_02_01_014p_img.jpg
    Collect the variables to the left. CNX_IntAlg_Figure_02_01_014q_img (1).jpg
    Simplify. CNX_IntAlg_Figure_02_01_014r_img.jpg
    Collect the constants to the right. CNX_IntAlg_Figure_02_01_014s_img.jpg
    Simplify. CNX_IntAlg_Figure_02_01_014t_img.jpg
    Check: CNX_IntAlg_Figure_02_01_014a_img.jpg  
    Let \(y=9\). CNX_IntAlg_Figure_02_01_014b_img.jpg  
    Finish the check on your own.

    Exercise \(\PageIndex{11A}\)

    Solve: \(\frac{1}{5}(n+3)=\frac{1}{4}(n+2)\).

    Answer

    \(n=2\)

    Exercise \(\PageIndex{11B}\)

    Solve: \(\frac{1}{2}(m−3)=\frac{1}{4}(m−7)\).

    Answer

    \(m=−1\)

    When you multiply both sides of an equation by the LCD of the fractions, make sure you multiply each term by the LCD—even if it does not contain a fraction.

    EXAMPLE \(\PageIndex{12}\)

    Solve: \(\frac{4q+3}{2}+6=\frac{3q+5}{4}\)

    Answer
      \(\frac{4 q+3}{2}+6=\frac{3 q+5}{4}\)
    Multiply both sides by the LCD, 4. CNX_IntAlg_Figure_02_01_015d_img.jpg
    Distribute. CNX_IntAlg_Figure_02_01_015e_img.jpg
    Simplify. \(\2(4 q+3)+24=3 q+5)
      \(8 q+6+24=3 q+5\)
      \(8 q+30=3 q+5\)
    Collect the variables to the left. CNX_IntAlg_Figure_02_01_015i_img.jpg
    Simplify. \(5 q+30=5\)
    Collect the constants to the right. CNX_IntAlg_Figure_02_01_015k_img.jpg
    Simplify. \(5 q=-25\)
    Divide both sides by five. CNX_IntAlg_Figure_02_01_015m_img.jpg
    Simplify. \(q=-5\)
    Check: \(\frac{4 q+3}{2}+6=\frac{3 q+5}{4}\)  
    Let \(q=−5.\) CNX_IntAlg_Figure_02_01_015b_img.jpg  
    Finish the check on your own.  

    Exercise \(\PageIndex{12A}\)

    Solve: \(\frac{3r+5}{6}+1=\frac{4r+3}{3}\).

    Answer

    \(r=3\)

    Exercise \(\PageIndex{12B}\)

    Solve: \(\frac{2s+3}{2}+1=\frac{3s+2}{4}\).

    Answer

    \(s=−8\)

    Some equations have decimals in them. This kind of equation may occur when we solve problems dealing with money or percentages. But decimals can also be expressed as fractions. For example, \(0.7=\frac{7}{10}\) and \(0.29=\frac{29}{100}\). So, with an equation with decimals, we can use the same method we used to clear fractions—multiply both sides of the equation by the least common denominator.

    The next example uses an equation that is typical of the ones we will see in the money applications in a later section. Notice that we will clear all decimals by multiplying by the LCD of their fraction form.

    EXAMPLE \(\PageIndex{13}\)

    Solve: \(0.25x+0.05(x+3)=2.85\).

    Answer

    Look at the decimals and think of the equivalent fractions:

    \[0.25=\frac{25}{100}, \; \; \; \;\;\;\;\; 0.05=\frac{5}{100}, \;\;\;\;\;\;\;\; 2.85=2\frac{85}{100}.\]

    Notice, the LCD is 100. By multiplying by the LCD we will clear the decimals from the equation.

      CNX_IntAlg_Figure_02_01_016a_img.jpg
    Distribute first. CNX_IntAlg_Figure_02_01_016b_img.jpg
    Combine like terms. CNX_IntAlg_Figure_02_01_016c_img.jpg
    To clear decimals, multiply by 100. CNX_IntAlg_Figure_02_01_016d_img.jpg
    Distribute. CNX_IntAlg_Figure_02_01_016e_img.jpg
    Subtract 15 from both sides. CNX_IntAlg_Figure_02_01_016f_img.jpg
    Simplify. CNX_IntAlg_Figure_02_01_016g_img.jpg
    Divide by 30. CNX_IntAlg_Figure_02_01_016h_img.jpg
    Simplify. CNX_IntAlg_Figure_02_01_016i_img.jpg
    Check it yourself by substituting \(x=9\) into the original equation.

    Exercise \(\PageIndex{13A}\)

    Solve: \(0.25n+0.05(n+5)=2.95.\)

    Answer

    \(n=9\)

    Exercise \(\PageIndex{13B}\)

    Solve: \(0.10d+0.05(d−5)=2.15.\)

    Answer

    \(d=16\)

    Key Concepts

    • How to determine whether a number is a solution to an equation
      1. Substitute the number in for the variable in the equation.
      2. Simplify the expressions on both sides of the equation.
      3. Determine whether the resulting equation is true.

        If it is true, the number is a solution.

        If it is not true, the number is not a solution.

    • How to Solve Linear Equations Using a General Strategy
      1. Simplify each side of the equation as much as possible.

        Use the Distributive Property to remove any parentheses.

        Combine like terms.

      2. Collect all the variable terms on one side of the equation.

        Use the Addition or Subtraction Property of Equality.

      3. Collect all the constant terms on the other side of the equation.

        Use the Addition or Subtraction Property of Equality.

      4. Make the coefficient of the variable term equal to 1.

        Use the Multiplication or Division Property of Equality.

        State the solution to the equation.

      5. Check the solution.

        Substitute the solution into the original equation to make sure the result is a true statement.

    • How to Solve Equations with Fraction or Decimal Coefficients
      1. Find the least common denominator (LCD) of all the fractions and decimals (in fraction form) in the equation.
      2. Multiply both sides of the equation by that LCD. This clears the fractions and decimals.
      3. Solve using the General Strategy for Solving Linear Equations.

    Glossary

    conditional equation
    An equation that is true for one or more values of the variable and false for all other values of the variable is a conditional equation.
    contradiction
    An equation that is false for all values of the variable is called a contradiction. A contradiction has no solution.
    identity
    An equation that is true for any value of the variable is called an Identity. The solution of an identity is all real numbers.
    linear equation
    A linear equation is an equation in one variable that can be written, where a and b are real numbers and \(a≠0\), as \(ax+b=0\).
    solution of an equation
    A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.

    This page titled 2.2: Use a General Strategy to Solve Linear Equations is shared under a CC BY license and was authored, remixed, and/or curated by OpenStax.