Algebra I

Linear Systems: Solve by Elimination

Big Picture

Systems of linear equations with two or more variables (and equations) can be solved in various ways, includingadding, subtracting, and multiplying the equations together in order to eliminate a variable. These methods are oftenfaster than graphing the equations or using the substitution method.

Key Terms

System of Linear Equations: A set of equations that must be solved together to find the solution that fits them both.

Solve by Adding or Subtracting

It is always possible to solve by substitution, but sometimes there are faster methods of solving a linear system. Another method of solving a system of linear equations (linear system) is by combining the equations in a way so that we only have one equation with one variable.

  • Add or subtract the equations to eliminate one of the variables.
  • Solve the new equation for the other variable.
  • Substitute the value from step 2 into one of the original equations and solve for the eliminated variable.

Tips:

  • Look for x- or y-coefficients that have the same value. If the signs are the same, we can subtract the equations. If the signs are different, we can add the equations.
  • Add the equations vertically (one above the other). Line the x’s and y’s in their own columns.

Example of solving by addition: \(2x+y = 9\) and \(5x−y = 5\)

  • Add the two equations together to eliminate \(y\).
    \(\begin{array}{r}
    &2x + y = 9\\
    +\!\!\!\!\!\!&5x - y = 5\\
    \hline
    &7x + 0y =14
    \end{array}\)
  • Solve the new equation for \(x\).
    \(7x = 14\)
    \(x = 2\)
  • Substitute \(x = 2\) into one of the equations and solve for \(y\).
    \(2(2) + y = 9\)
    \(y = 9 − 4 = 5\)

Check the solution by plugging it into both equations.
\(2(2)+5 = 9\)
\(5(2)−5 = 5\)

Both equations are true, so \((2, 5)\) is the solution.

Example of solving by subtraction: \(x+4y = 11\) and \(x−y = 1\)

  • Subtract the two equations to eliminate \(x\).
    \(\begin{array}{r}
    &x + 4y = 11\\
    -\!\!\!\!\!\!&(x − y = 1)\\
    \hline
    &0x+5y = 10
    \end{array}\)

When subtracting equations, don’t forget to subtract the whole equation, not just one side of the equal sign.Be careful when distributing the negative sign!

  1. Solve the new equation for \(y\).
    \(5y = 10\)
    \(y = 2\)
  2. Substitute \(y = 2\) into one of the equations and solve for \(x\).
    \(x−2 = 1\)
    \(x = 1+2 = 3\)

Check the solution by plugging it into both equations.
\(3+4(2) = 11\)
\(3-2 = 1\)

Both equations are true, so \((3, 2)\) is the solution.

    Algebra I

    Linear Systems: Solve by Elimination cont.

    Solve by Multiplying

    If there are no coefficients that have the same value, we can still solve by addition or subtraction by multiplying one or both of the equations by a constant. This could be more work than solving by substitution, but it avoids working with fraction coefficients.

    Example of multiplying one equation: \(3x+4y = 17\) and \(7y−6x = -4\)

    1. We can see that if we multiply the first equation by \(2\), the coefficients of \(x\) will be \(+6\) and \(-6\).
      \(2(3x+4y = 17) \rightarrow 6x+8y = 34\)
    2. Add the two equations together to eliminate \(x\).
      \(\begin{array}{r}
      &6x + 8y = 34\\
      +\!\!\!\!\!\!&(-6x+7y = -4)\\
      \hline
      &0x+15y = 30
      \end{array}\)
    3. Solve the new equation for \(y\)
      \(15y = 30\)
      \(y = 2\)
    4. Substitute y into one of the equations and solve for \(x\).
      \(3x+4(2) = 17\)
      \(3x = 17-8 = 9\)
      \(x = 3\)

    Check the solution by plugging it into both equations.
    \(3(3)+4(2) = 17\)
    \(7(2)−6(3) = -4\)
    Both equations are true, so \((3, 2)\) is the solution.

    Example of multiplying both equations: \(3x+4y = -6\) and \(4x−3y = 17\)

    1. The coefficients do not match and are not simple multiples of each other. We can multiply the first equation by \(4\) and the second equation by \(3\) so that the coefficients of x are both \(+12\).
      \(4(3x+4y = -6) \rightarrow 12x+16y = -24\)
      \(3(4x-3y = 17) \rightarrow 12x-9y = 51\)
    2. Subtract the two equations to eliminate \(x\).
      \(\begin{array}{r}
      &12x+16y = -24\\
      -\!\!\!\!\!\!&(12x - 9y = 51)\\
      \hline
      &0x+25y = -75
      \end{array}\)
    3. Solve the new equation for \(y\).
      \(25y = -75\)
      \(y = -3\)
    4. Substitute \(y = -3\) into one of the equations and solve for \(x\).
      \(3x + 4(-3) = -6\)
      \(3x = -6+12 = 6\)
      \(x = 2\)
    • When solved by substitution, there is only one solution.

    Check the solution by plugging it into both equations.
    \(3(2)+4(-3) = -6\)
    \(4(2)-3(3) = 17\)
    Both equations are true, so \((2, -3)\) is the solution.

    Systems of Linear Equations in Three Variables

    Solving linear systems in three variables is not that different from solving linear systems in two variables.

    1. Put all three equations in the form \(Ax + By + Cz = D\) if needed.
    2. Eliminate one of the variables from a pair of equations to form a new equation with two variables.
    3. Eliminate the same variable from another pair of equations to form a new equation with two variables.
    4. Solve the new 2-variable system for two of the variables.
    5. Plug the variables into one of the original equations to find the third variable.
    6. Check work by plugging the solutions into the equations.