Elementary Row Operations Interchange two equations Multiply one equation by a nonzero constant Add a multple of one equation to another equation

Equation Properties Must be of the first degree (linear equation) Must be of the form ax+by+cz=h If solving for (n) variables, must have (n) equations.

2y + 2x + 2z = 0 2x + 4z - 4y = 8 Process of Elimination We have three equations and we are solving for three unknowns (x, y, and z) -2z + 4y + 2x = 4

Setting Up For Elimination Of the equation, ax+by+cz=h, (where a, b, c, and h are known, and x, y, and z are unknown) the unknown variables must be placed under the same column

Example of Setting Up For Elimination 2y + 2x + 2z = 2 2x + 4z + 4y = 8 -2z + 4y + 2x = 4 2x + 2y + 2z = 0 2x - 4y + 4z = 8 2x + 4y - 2z = 4 Green is original equation. Blue is set up equation. X YZH

Rules for Elimination After setting up the equations in the form previously shown, decide which variable will be the first to be reduced to zero

Subtract the first equation from the last equation. (This way it is the x variable that reduces to a value of zero.) 2x + 4y – 2z = 4 -2x – 2y – 2z = 0 2y – 4z = 4 The first two equations are the same as the original, but the last equation has been modified in order to reduce the number of unknowns. 2x + 2y + 2z = 0 2x – 4y + 4z = 8 2y – 4z = 4

Rules for Elimination We have now two equations with three unknown variables and one with two unknowns. Next step is to reduce another two equations to produce another equation with two unknowns.

We’re going to get rid of another term by subtracting the first from the second. 2x – 4y + 4z = 8 - 2x – 2y – 2z = 0 -6y + 2z = 8 The first equation is the same, but the second and third equations now have two unknowns. 2x + 2y +2z = 0 - 6y + 2z = 8 2y – 4z = 4

Rules for Elimination With two equations and two unknowns, we can figure a way to reduce them to have one equation, one unknown. We do this in the following way:

To get rid of one of the unknowns on either the second or the third equation we multiply the third equation by 3, this way they both have the same coefficient in front of the variable(y) we are reducing 3 (2y – 4z = 4) -6y+ 2z = 8 The result after multiplying by three is : 6y – 12z = 12 -6y+ 2z = 8 Now we add the second equation to the third to approach an equation with one unknown. 6y – 12z =12 + (-6y + 2z = 8) -10z = 20 Note: The bottom equation can be multiplied by two if the (z) variable was to be reduce.

Rules for Elimination We now are set to solve for the variables.

The equations that we are left with are: 2x + 2y + 2z = 0 - 6y + 2z = z = 20 We can solve for the third term because its one equation one unknown.( A simple use of algebra results in the following result.) z = -2 We can use the value for z and use it to in the second equation to make it into one equation one unknown.( With the use of substitution leads to the following equation.) -6y – 4 = 8 Solving for the variable that is unknown leads to: y = -2

The last step is to use the values that have previously been solved for (y and z) to make the first equation into one equation one unknown. 2x – 4 – 4 = 0 Solving with algebra leads to: x = 4 The results then are: x = 4z = -2y = -2

To Conclude Equations of the form ax+by+cz=h, who have x, y, and z unknown can be solved in the following way: With (n) unknowns and (n) equations, decide on a variable to exclude from an upcoming equation that will have (n-1) unknowns. Now choose two equations that will be added to eliminate the chosen variable

To Conclude Multiply one variable’s coefficient, of one equation, with the coefficient of the same variable in the other equation, and do the same for the other, except multiplying by the opposite of the coefficient. (ex. 5x and 2x, multiply 5x by 2 and 2x by -5) Now add them and you will have eliminated a variable

To Conclude Continue this process until you have one equation one unknown. Known solve for the variable and use the result as a substitution for the equations with more than one unknowns.