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Study PortalMD-001Systems Equations Thre…
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    MD-001

    Math Deficiency - I

    Progress0 / 38 done
    Topics38
    1Definition, Representation, and Operations
    2Graphical Transformation of Functions3Properties of Functions4Composition and Inverses of Functions5Domain and Range of Functions6Maximum and Minimum Values of Functions7Increasing and Decreasing Functions8Zeros and Intercepts of Functions9Piecewise Functions10Continuity and Discontinuity of Functions11Polynomials and Rational Functions12Polynomial Long Division and Synthetic Division13Solution of Rational Functions14Absolute Valued Functions and Their Properties
    15Horizontal, Vertical, and Oblique16Exponential Functions and Their Properties17Logarithmic Functions and Their Properties
    18Two Equations and Two Unknowns19Three Equations and Three Unknowns
    20Addition, Subtraction, and Multiplication21Row Operations and Row Echelon Forms22Augmented Matrices
    232x2 and Higher Order24Cramer's Rule25Inverse Matrices26Series and Sequences
    27Angles in Radians and Degrees28Right Triangle Trigonometry29Law of Cosines and Sines30Area of a Triangle31Graphs of Trigonometric Functions32Graphs of Inverse Trigonometric Functions33Basic Trigonometric Identities34Trigonometric Equations
    35Parabolas, Circles, Ellipses, and Hyperbolas36Degenerate Conics37Polar and Parametric Equations38Polar and Rectangular Coordinates
    12 min read
    Topic 19 of 38
    Topic Study Note Illustration
    19

    Systems of Equations: Three Equations and Three Unknowns

    A system of three equations and three unknowns involves three linear equations with three variables. The general form is:

    Systems of Equations: Three Equations and Three Unknowns

    A system of three equations and three unknowns involves three linear equations with three variables. The general form is:

    {a1x+b1y+c1z=d1a2x+b2y+c2z=d2a3x+b3y+c3z=d3\begin{cases} a_1x + b_1y + c_1z = d_1 \\ a_2x + b_2y + c_2z = d_2 \\ a_3x + b_3y + c_3z = d_3 \end{cases}⎩⎨⎧​a1​x+b1​y+c1​z=d1​a2​x+b2​y+c2​z=d2​a3​x+b3​y+c3​z=d3​​

    Where:

    • xxx, yyy, and zzz are the unknown variables.
    • a1,a2,a3a_1, a_2, a_3a1​,a2​,a3​, etc., are constants, as are b1,b2,b3b_1, b_2, b_3b1​,b2​,b3​, and c1,c2,c3c_1, c_2, c_3c1​,c2​,c3​.
    • d1,d2,d3d_1, d_2, d_3d1​,d2​,d3​ are the constants on the right-hand side of the equations.

    The goal is to find the values of xxx, yyy, and zzz that satisfy all three equations simultaneously.


    Methods for Solving Systems of Equations with Three Unknowns

    There are several methods to solve systems of three equations with three unknowns. The most common methods are:

    1. Substitution Method
    2. Elimination Method
    3. Matrix Method (Gaussian Elimination)

    Each method can be used to reduce the system to simpler equations and ultimately solve for the unknowns.


    1. Substitution Method

    The substitution method involves solving one of the equations for one variable and substituting that expression into the other two equations. This reduces the system to a system of two equations in two unknowns, which can then be solved.

    Steps:

    1. Solve one equation for one variable in terms of the others (e.g., solve for xxx, yyy, or zzz).
    2. Substitute the expression into the other two equations, eliminating the variable.
    3. Solve the resulting system of two equations in two unknowns.
    4. Once you find the value of one variable, substitute it back into one of the original equations to find the other variables.

    Example:

    Solve the system:

    {x+y+z=62x−y+3z=143x+2y−z=7\begin{cases} x + y + z = 6 \\ 2x - y + 3z = 14 \\ 3x + 2y - z = 7 \end{cases}⎩⎨⎧​x+y+z=62x−y+3z=143x+2y−z=7​
    • Step 1: Solve the first equation for xxx:

      x=6−y−zx = 6 - y - zx=6−y−z
    • Step 2: Substitute x=6−y−zx = 6 - y - zx=6−y−z into the second and third equations:

      • Second equation:

        2(6−y−z)−y+3z=14⇒12−2y−2z−y+3z=14⇒12−3y+z=142(6 - y - z) - y + 3z = 14 \quad \Rightarrow \quad 12 - 2y - 2z - y + 3z = 14 \quad \Rightarrow \quad 12 - 3y + z = 142(6−y−z)−y+3z=14⇒12−2y−2z−y+3z=14⇒12−3y+z=14

        Simplify:

        −3y+z=2(Equation 4)-3y + z = 2 \quad \text{(Equation 4)}−3y+z=2(Equation 4)
      • Third equation:

        3(6−y−z)+2y−z=7⇒18−3y−3z+2y−z=7⇒18−y−4z=73(6 - y - z) + 2y - z = 7 \quad \Rightarrow \quad 18 - 3y - 3z + 2y - z = 7 \quad \Rightarrow \quad 18 - y - 4z = 73(6−y−z)+2y−z=7⇒18−3y−3z+2y−z=7⇒18−y−4z=7

        Simplify:

        −y−4z=−11⇒y+4z=11(Equation 5)-y - 4z = -11 \quad \Rightarrow \quad y + 4z = 11 \quad \text{(Equation 5)}−y−4z=−11⇒y+4z=11(Equation 5)
    • Step 3: Solve the new system of two equations:

      {−3y+z=2y+4z=11\begin{cases} -3y + z = 2 \\ y + 4z = 11 \end{cases}{−3y+z=2y+4z=11​
      • Multiply the second equation by 3 to eliminate yyy:

        3y+12z=333y + 12z = 333y+12z=33
      • Add the two equations:

        (−3y+z)+(3y+12z)=2+33(-3y + z) + (3y + 12z) = 2 + 33(−3y+z)+(3y+12z)=2+33

        Simplify:

        13z=35⇒z=351313z = 35 \quad \Rightarrow \quad z = \frac{35}{13}13z=35⇒z=1335​
    • Step 4: Substitute z=3513z = \frac{35}{13}z=1335​ into one of the original equations (e.g., y+4z=11y + 4z = 11y+4z=11):

      y+4(3513)=11⇒y+14013=11⇒y=11−14013=14313−14013=313y + 4\left(\frac{35}{13}\right) = 11 \quad \Rightarrow \quad y + \frac{140}{13} = 11 \quad \Rightarrow \quad y = 11 - \frac{140}{13} = \frac{143}{13} - \frac{140}{13} = \frac{3}{13}y+4(1335​)=11⇒y+13140​=11⇒y=11−13140​=13143​−13140​=133​
    • Step 5: Finally, substitute y=313y = \frac{3}{13}y=133​ and z=3513z = \frac{35}{13}z=1335​ into the equation x=6−y−zx = 6 - y - zx=6−y−z:

      x=6−313−3513=6−3813=7813−3813=4013x = 6 - \frac{3}{13} - \frac{35}{13} = 6 - \frac{38}{13} = \frac{78}{13} - \frac{38}{13} = \frac{40}{13}x=6−133​−1335​=6−1338​=1378​−1338​=1340​

    Thus, the solution is:

    x=4013,y=313,z=3513x = \frac{40}{13}, \quad y = \frac{3}{13}, \quad z = \frac{35}{13}x=1340​,y=133​,z=1335​

    2. Elimination Method

    The elimination method involves adding or subtracting equations in such a way that one of the variables is eliminated. This method reduces the system to a simpler system, often starting with two equations in two unknowns.

    Steps:

    1. Multiply the equations (if necessary) so that the coefficients of one of the variables are opposites (i.e., one is the negative of the other).
    2. Add or subtract the equations to eliminate one variable.
    3. Solve the resulting system of two equations in two unknowns.
    4. Once one variable is found, substitute it back into one of the original equations to solve for the other variables.

    Example:

    Solve the system:

    {x+y+z=62x−y+3z=143x+2y−z=7\begin{cases} x + y + z = 6 \\ 2x - y + 3z = 14 \\ 3x + 2y - z = 7 \end{cases}⎩⎨⎧​x+y+z=62x−y+3z=143x+2y−z=7​
    • Step 1: Eliminate zzz by adding the first and second equations:

      (x+y+z)+(2x−y+3z)=6+14⇒3x+4z=20(Equation 4)(x + y + z) + (2x - y + 3z) = 6 + 14 \quad \Rightarrow \quad 3x + 4z = 20 \quad \text{(Equation 4)}(x+y+z)+(2x−y+3z)=6+14⇒3x+4z=20(Equation 4)
    • Step 2: Eliminate zzz by adding the first and third equations:

      (x+y+z)+(3x+2y−z)=6+7⇒4x+3y=13(Equation 5)(x + y + z) + (3x + 2y - z) = 6 + 7 \quad \Rightarrow \quad 4x + 3y = 13 \quad \text{(Equation 5)}(x+y+z)+(3x+2y−z)=6+7⇒4x+3y=13(Equation 5)
    • Step 3: Now solve the system of two equations:

      {3x+4z=204x+3y=13\begin{cases} 3x + 4z = 20 \\ 4x + 3y = 13 \end{cases}{3x+4z=204x+3y=13​

      Solve this system using methods for two equations (substitution or elimination), and then substitute back to find the values of xxx, yyy, and zzz.


    3. Matrix Method (Gaussian Elimination)

    The matrix method, particularly Gaussian elimination, is an efficient algorithm for solving systems of linear equations. It involves transforming the system of equations into an augmented matrix and performing row operations to reduce the matrix to row echelon form or reduced row echelon form.

    Steps:

    1. Write the system of equations as an augmented matrix.
    2. Use row operations (swap rows, multiply a row by a constant, add/subtract rows) to reduce the matrix to row echelon form.
    3. Back-substitute to solve for the unknowns.

    Types of Solutions

    Like systems of two equations, a system of three equations can have:

    • One solution (Unique Solution): The system is consistent and independent, and the three equations intersect at a single point.
    • No solution: The system is inconsistent (the equations represent parallel planes that never meet).
    • Infinite solutions: The system is dependent (the three equations represent the same plane or coincide in some way, giving infinitely many solutions).

    Conclusion

    Solving systems of three equations with three unknowns can be done using various methods, such as substitution, elimination, or matrix methods. While the substitution and elimination methods are often straightforward and intuitive, the matrix method (Gaussian elimination) provides a more systematic and efficient approach, especially for larger systems. Understanding the different methods and when to use each will help you handle complex systems of equations in algebra and other areas of mathematics.

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    Systems of Equations: Two Equations and Two Unknowns
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    Matrix Algebra: Addition, Subtraction, and Multiplication

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