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11.3: Elimination by Substitution

  • Page ID
    49411
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    When Substitution Works Best

    We know how to solve a linear equation in one variable. We shall now study a method for solving a system of two linear equations in two variables by transforming the two equations in two variables into one equation in one variable.

    To make this transformation, we need to eliminate one equation and one variable. We can make this elimination by substitution.

    When Substitution Works Best

    The substitution method works best when either of these conditions exists:

    1. One of the variables has a coefficient of 1,  or
    2. One of the variables can be made to have a coefficient of 1 without introducing fractions.

    The Substitution Method

    The Substitution Method

    To solve a system of two linear equations in two variables,

    1. Solve one of the equations for one of the variables.
    2. Substitute the expression for the variable chosen in step 1 into the other equation.
    3. Solve the resulting equation in one variable.
    4. Substitute the value obtained in step 3 into the equation obtained in step 1 and solve to obtain the value of the other variable.
    5. Check the solution in both equations.
    6. Write the solution as an ordered pair.

    Sample Set A

    Example \(\PageIndex{1}\)

    Solve the system

    \(\left\{\begin{array}{r}
    2x + 3y = 14 \\
    3x + y = 7
    \end{array}\right.\)

    Step 1: Since the coefficient of \(y\) in equation 2 is \(1\), we will solve equation 2 for \(y\).

    \(y = -3x + 7\)

    Step 2: Substitute the expression \(-3x + 7\) for \(y\) in equation 1.

    \(2x + 3(-3x + 7) = 14\)

    Step 3: Solve the equation obtained in step 2.

    \(\begin{aligned}
    2x + 3(-3x + 7) &= 14\\
    2x - 9x + 21 &= 14\\
    -7x + 21 &= 14\\
    -7x &= -7\\
    x &= 1
    \end{aligned}\)

    Step 4: Substitute \(x = 1\) into the equation obtained in step 1, \(y = -3x + 7\).

    \(\begin{aligned}
    y &= -3(1) + 7\\
    y &= -3 + 7\\
    y &= 4
    \end{aligned}\)

    Step 5: Substitute \(x = 1, y = 4\) into each of the original equations for a check.

    1)

    \(\begin{aligned}
    2 x+3 y &=14 \\
    2(1)+3(4) &=14 \text { Is this correct? } \\
    2+12 &=14 \text { Is this correct? } \\
    14 &=14 \text { Yes, this is correct. }
    \end{aligned}\)

    2)

    \(\begin{aligned}
    3x + y &= 7\\
    3(1) + (4) &= 7 \text{ Is this correct? }\\
    3 + 4 &= 7 \text{ Is this correct? }\\
    7 &= 7 \text{ Yes, this is correct. }
    \end{aligned}\)

    Step 6: The solution is \((1, 4)\). The point \((1, 4)\) is the point of intersection of the two lines of the system.

    Practice Set A

    Practice Problem \(\PageIndex{1}\)

    Solve the system

    \(\left\{\begin{array}{r}
    5x - 8y = 18 \\
    4x + y = 7
    \end{array}\right.\)

    Answer

    The point \((2, -1)\) is the point of intersection of the two lines.

    Substitution And Parallel Lines

    The following rule alerts us to the fact that the two lines of a system are parallel.

    Substitution and Parallel Lines

    If computations eliminate all the variables and produce a contradiction, the two lines of a system are parallel, and the system is called inconsistent.

    Sample Set B

    Example \(\PageIndex{2}\)

    Solve the system

    \(\left\{\begin{array}{r}
    2x - y = 1 \\
    4x - 2y = 4
    \end{array}\right.\)

    Step 1: Solve equation for \(y\).

    \(\begin{aligned}
    2x - y &= 1\\
    -y &= -2x + 1\\
    y &= 2x - 1
    \end{aligned}\)

    Step 2: Substitute the expression \(2x - 1\) for \(y\) into equation 2.

    \(4x - 2(2x - 1) = 4\)

    Step 3: Solve the equation obtained in step 2.

    \(\begin{aligned}
    4x - 2(2x - 1) &= 4\\
    4x - 4x + 2 &= 4\\
    2 &\not= 4
    \end{aligned}\)

    Computations have eliminated all the variables and produce a contradiction. These lines are parallel.

    A graph of two parallel lines. One line is labeled with the equation two x minus y is equal to one and passes through the points one, one, and zero, negative one. A second line is labeled with the equation four x minus two y is equal to four and passes through the points one, zero, and zero, negative two.

    This system is inconsistent.

    Practice Set B

    Practice Problem \(\PageIndex{2}\)

    Solve the system

    \(\left\{\begin{array}{r}
    7x - 3y = 2 \\
    14x - 6y = 1
    \end{array}\right.\)

    Answer

    Substitution produces \(4 \not= 1\), or \(\dfrac{1}{2} \not = 2\), a contradiction. These lines are parallel and the system is inconsistent.

    Substitution And Coincident Lines

    The following rule alerts us to the fact that the two lines of a system are coincident.

    Substituion and Coincident Lines

    If computations eliminate all the variables and produce an identity, the two lines of a system are coincident and the system is called dependent.

    Sample Set C

    Example \(\PageIndex{3}\)

    Solve the system

    \(\left\{\begin{array}{r}
    4x + 8y = 8 \\
    3x + 6y = 6
    \end{array}\right.\)

    Step 1: Divide equation 1 by \(4\) and solve for \(x\).

    \(\begin{aligned}
    4x + 8y &= 8\\
    x + 2y &= 2\\
    x &= -2y + 2
    \end{aligned}\)

    Step 2: Substitute the expression \(-2y + 2\) for \(x\) in equation 2.

    \(3(-2y + 2) + 6y = 6\)

    Step 3: Solve the equation obtained in step 2.

    \(\begin{aligned}
    3(-2y + 2) + 6y &= 6\\
    -6y + 6 + 6y &= 6\\
    6 &= 6
    \end{aligned}\)

    Computations have eliminated all the variables and produced an identity. These lines are coincident.

    A graph of two coincident lines. The line is labeled with the equation x plus two y is equal to two and a second label with the equation three x plus six y is equal to six. The lines pass through the points zero, one and two, zero. Since the lines are coincident, they have the same graph.

    This system is dependent.

    Practice Set C

    Practice Problem \(\PageIndex{3}\)

    Solve the system

    \(\left\{\begin{array}{r}
    4x + 3y = 1 \\
    -8x - 6y = -2
    \end{array}\right.\)

    Answer

    Computations produce \(-2 = -2\), an identity. These lines are coincident and the system is dependent.

    Systems in which a coefficient of one of the variables is not \(1\) or cannot be made to be \(1\) without introducing fractions are not well suited for the substitution method. The problem in Sample Set D illustrates this “messy” situation.

    Sample Set D

    Example \(\PageIndex{4}\)

    Solve the system

    \(\left\{\begin{array}{r}
    3x + 2y = 1 \\
    4x - 3y = 3
    \end{array}\right.\)

    Step 1: We will solve equation (1) for \(y\).

    \(\begin{aligned}
    3x + 2y &= 1\\
    2y &= -3x + 1\\
    y &= \dfrac{-3}{2}x + \dfrac{1}{2}
    \end{aligned}\)

    Step 2: Substitute the expression \(\dfrac{-3}{2}x + \dfrac{1}{2}\) for \(y\) in equation (2).

    \(4x - 3(\dfrac{-3}{2}x + \dfrac{1}{2}) = 3\)

    Step 3: Solve the equation obtained in step 2.

    \(\begin{aligned}
    4x - 3(\dfrac{-3}{2}x + \dfrac{1}{2}) &= 3 \text{ Multiply both sides by the LCD, } 2\\
    4x + \dfrac{9}{2}x - \dfrac{3}{2} &= 3\\
    8x + 9x - 3 &= 6\\
    17x - 3 &= 6\\
    17x &= 9\\
    x &= \dfrac{9}{17}
    \end{aligned}\)

    Step 4: Substitute \(x = \dfrac{9}{17}\) into the equation obtained in step 1, \(y = \dfrac{-3}{2}x + \dfrac{1}{2}\)

    \(y = \dfrac{-3}{2}(\dfrac{9}{17}) + \dfrac{1}{2}\\
    y = \dfrac{-27}{34} + \dfrac{17}{34} = \dfrac{-10}{34} = \dfrac{-5}{17}\).

    We now have \(x = \dfrac{9}{17}\) and \(y = \dfrac{-5}{17}\)\

    Step 5: Substitution will show that these values of \(x\) and \(y\) check.

    Step 6: The solution is \((\dfrac{9}{17}, \dfrac{-5}{17})\)

    Practice Set D

    Practice Problem \(\PageIndex{4}\)

    Solve the system

    \(\left\{\begin{array}{r}
    9x - 5y = -4 \\
    2x + 7y = -9
    \end{array}\right.\)

    Answer

    These lines intersect at the point \((−1,−1)\).

    Exercises

    For the following problems, solve the systems by substitution.

    Exercise \(\PageIndex{1}\)

    \(\left\{\begin{array}{r}
    3x + 2y = 9 \\
    y = -3x + 6
    \end{array}\right.\)

    Answer

    \((1, 3)\)

    Exercise \(\PageIndex{2}\)

    \(\left\{\begin{array}{r}
    5x - 3y = -6 \\
    y = -4x + 12
    \end{array}\right.\)

    Exercise \(\PageIndex{3}\)

    \(\left\{\begin{array}{r}
    2x + 2y = 0 \\
    x = 3y - 4
    \end{array}\right.\)

    Exercise \(\PageIndex{4}\)

    \(\left\{\begin{array}{r}
    9x - 5y = -4 \\
    2x + 7y = -9
    \end{array}\right.\)

    Answer

    \((-1, 1)\)

    Exercise \(\PageIndex{5}\)

    \(\left\{\begin{array}{r}
    3x + 5y = 9 \\
    x = 4y - 14
    \end{array}\right.\)

    Exercise \(\PageIndex{6}\)

    \(\left\{\begin{array}{r}
    -3x + y = -4\\
    2x + 3y = 10
    \end{array}\right.\)

    Answer

    \((2,2)\)

    Exercise \(\PageIndex{7}\)

    \(\left\{\begin{array}{r}
    -4x + y = -7 \\
    2x + 5y = 9
    \end{array}\right.\)

    Exercise \(\PageIndex{8}\)

    \(\left\{\begin{array}{r}
    6x - 6 = 18 \\
    x + 3y = 3
    \end{array}\right.\)

    Answer

    \((4, -\dfrac{1}{3})\)

    Exercise \(\PageIndex{9}\)

    \(\left\{\begin{array}{r}
    -x - y = 5 \\
    23x + y = 5
    \end{array}\right.\)

    Exercise \(\PageIndex{10}\)

    \(\left\{\begin{array}{r}
    -5x + y = 4 \\
    10x - 2y = -8
    \end{array}\right.\)

    Answer

    Dependent (same line)

    Exercise \(\PageIndex{11}\)

    \(\left\{\begin{array}{r}
    x + 4y = 1 \\
    -3x - 12y = -1
    \end{array}\right.\)

    Exercise \(\PageIndex{12}\)

    \(\left\{\begin{array}{r}
    4x - 2y = 8 \\
    6x + 3y = 0
    \end{array}\right.\)

    Answer

    \((1,−2)\)

    Exercise \(\PageIndex{13}\)

    \(\left\{\begin{array}{r}
    2x + 3y = 12 \\
    2x + 4y = 18
    \end{array}\right.\)

    Exercise \(\PageIndex{14}\)

    \(\left\{\begin{array}{r}
    3x - 9y = 6 \\
    6x - 18y = 5
    \end{array}\right.\)

    Answer

    inconsistent (parallel lines)

    Exercise \(\PageIndex{15}\)

    \(\left\{\begin{array}{r}
    -x + 4y = 8 \\
    3x - 12y = 10
    \end{array}\right.\)

    Exercise \(\PageIndex{16}\)

    \(\left\{\begin{array}{r}
    x + y = -6 \\
    x - y = 4
    \end{array}\right.\)

    Answer

    \((−1,−5)\)

    Exercise \(\PageIndex{17}\)

    \(\left\{\begin{array}{r}
    2x + y = 0 \\
    x - 3y = 0
    \end{array}\right.\)

    Exercise \(\PageIndex{18}\)

    \(\left\{\begin{array}{r}
    4x - 2y = 7 \\
    y = 4
    \end{array}\right.\)

    Answer

    \((\dfrac{15}{4}, 4)\)

    Exercise \(\PageIndex{19}\)

    \(\left\{\begin{array}{r}
    x + 6y = 11 \\
    x = -1
    \end{array}\right.\)

    Exercise \(\PageIndex{20}\)

    \(\left\{\begin{array}{r}
    2x - 4y = 10 \\
    3x = 5y + 12
    \end{array}\right.\)

    Answer

    \((−1,−3)\)

    Exercise \(\PageIndex{21}\)

    \(\left\{\begin{array}{r}
    y + 7x + 4 = 0 \\
    x = -7y + 28
    \end{array}\right.\)

    Exercise \(\PageIndex{22}\)

    \(\left\{\begin{array}{r}
    x + 4y = 0 \\
    x + \dfrac{2}{3}y = \dfrac{10}{3}
    \end{array}\right.\)

    Answer

    \((4,−1)\)

    Exercise \(\PageIndex{23}\)

    \(\left\{\begin{array}{r}
    x = 24 - 5y \\
    x - \dfrac{5}{4}y = \dfrac{3}{2}
    \end{array}\right.\)

    Exercise \(\PageIndex{24}\)

    \(\left\{\begin{array}{r}
    x = 11 - 6y \\
    3x + 18y = -33
    \end{array}\right.\)

    Answer

    inconsistent (parallel lines)

    Exercise \(\PageIndex{25}\)

    \(\left\{\begin{array}{r}
    2x + \dfrac{1}{3}y = 4 \\
    3x + 6y = 39
    \end{array}\right.\)

    Exercise \(\PageIndex{26}\)

    \(\left\{\begin{array}{r}
    \dfrac{4}{5}x + \dfrac{1}{2}y = \dfrac{3}{10} \\
    \dfrac{1}{3} + \dfrac{1}{2}y = \dfrac{-1}{6}
    \end{array}\right.\)

    Answer

    \((1,−1)\)

    Exercises For Review

    Exercise \(\PageIndex{27}\)

    Find the quotient: \(\dfrac{x^2 - x - 12}{x^2 - 2x - 15} \div \dfrac{x^2 - 3x - 10}{x^2 - 2x - 8}\)

    Answer

    \(\dfrac{(x-4)^2}{(x-5)^2}\)

    Exercise \(\PageIndex{28}\)

    Find the difference: \(\dfrac{x + 2}{x^2 + 5x + 6} - \dfrac{x + 1}{x^2 + 4x + 3}\)

    Exercise \(\PageIndex{29}\)

    Simplify \(-\sqrt{81x^8y^5z^4}\)

    Answer

    \(-9x^4y^2z^2 \sqrt{y}\)

    Exercise \(\PageIndex{30}\)

    Use the quadratic formula to solve \(2x^2 + 2x - 3 = 0\)

    Exercise \(\PageIndex{31}\)

    Solve by graphing:

    \(\left\{\begin{array}{r}
    x - y = 1 \\
    2x + y = 5
    \end{array}\right.\)

    An xy coordinate plane with gridlines labeled negative five and five with increments of one unit for both axes.

    Answer

    \((2,1)\)

    A graph of two lines intersecting at a point with coordinates negative two, one. One of the lines is passing through a point with coordinates zero, five and the other line is passing through a point with coordinates zero, negative one.


    This page titled 11.3: Elimination by Substitution is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Denny Burzynski & Wade Ellis, Jr. (OpenStax CNX) .

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