Solving Simultaneous Equations

When we have two equations in two variables and are trying to find a solution which fits both, we often solve one equation for one of the variables and substitute into the second equation.

Note that if we have two unknowns (two variables) which we are trying to find, we must have two equations--that is, two relationships between the unknowns. Similarly, three unknowns requires three equations, and so forth. The group of equations is known as a system of equations, and we solve by finding the simultaneous solutions to all of the equations in the system.

Examples
  1. Solve for x and y:
    y + (x/2) = 3, 2(x+y) = 1-y
    Solving the first equation for y gives y = 3 - (x/2), so, substituting for y in the second equation we get
    2(x+(3-(x/2))) = 1 - (3-(x/2)), or 2x+6-x=-2+x/2, so x+6 = -2+x/2, which is (x/2) = -8, so x = -16.
    Then
    y = 3 - (-16/2) = 3 + 8 = 11.
    So the solution that solves both equations simultaneously is x=-16, y=11.
  2. Solve for x and y:
    y = x-1, x^2 + y^2 = 5
    Plugging in the expression x-1 for y in the second equation, we get
    x^2 + (x-1)^2 = 5, or x^2 + x^2 - 2x + 1 = 5, so 2x^2 - 2x - 4 = 0, or x^2 - x - 2 = 0, which is (x-2)(x+1) = 0
    Thus x=2 or x=-1. If x=2, then y = 2-1 = 1, and if x=-1, then y = 2-(-1) = 3. Thus the solutions to the system are x=2 and y=1, or x=-1 and y=3.
  3. Solve for Q and a:
    90.7 = Q a^(10), 91 = Q a^(13)
    Solve the first equation for Q:
    Q = (90.7/a^(10))
    Then, plugging this in to the second equation,
    91 = (90.7/a^(10)) a^(13) = 90.7 a^3, so a^3 = 91/90.7, and a = (91/90.7)^(1/3) approx 1.0011.
    Then
    Q approx 90.7/(1.0011)^(10) = 89.7083
    Note that in this case we have found an approximate solution to the system of equations:
    a approx 1.0011, Q approx 89.7083.
  4. Find the points of intersection for the graphs in each of the figures below
    [graph of y=x+1 and y=x^2-1] [graph of y=x+1 and y=3^(x-1)]
    In both cases we solve the equations simultaneously by setting the y-values equal to one another. For the left graph, we are solving the system of equations
    y=x+1, y=x^2-1
    So
    x^2-1 = x+1, or x^2 - x - 2 = 0, so (x-2)(x+1) = 0
    and thus x=2 or x=-1. To get the corresponding y values we can use either equation. For x=2, we get y=3, and for x=-1, y=0. Therefore the points of intersection are (2,3) and (-1,0).
    For the second graph, we are solving
    y=x+1, y=3^(x-1)
    And so want
    x + 1 = 3^(x-1)
    However, we cannot use algebraic techniques to solve this because the variable x appears both in the exponential and on the left-hand side. We can guess one solution, however, by noting that
    2+1 = 3^(2-1) = 3
    so that x=2, y=3 is a solution to the system. There is a second solution, as shown in the graph, which we can estimate by tracing the graphs on a calculator. This is
    x approx -0.8721, y approx 0.1279.
    Therefore the solutions are
    x=2 and y=3, or x approx -0.8721 and y approx 0.1279.
Practice

Be sure you've gone through each step in the examples above before doing these. Once you've worked them until you're sure that you understand them, go on to the next section. There are more problems of this type in the section test at the end of the section.

Note that you can get new practice problems by clicking the "Refresh" button at the bottom of the practice set.

  1. Solve the following equations to find simultaneous solutions for x and y:
    a x   +   b y   =   c
    x   +   d y   =   g
    Answer:   x =   and   x =
    (Enter answers, then click: . Answer message: )
  2. Solve the following equations to find simultaneous solutions for x and y:
    n x   +   j y   =   - p
    y   -   x ^2   =   q x
    Answer:   x =   and   x =   , or
        x =   and   x =    
    (Enter answers, then click: . Answer message: )
For more practice, click: .
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precal: 7.6 - solving simultaneous eqns
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