Solving Systems of Linear equations with 3 Variables<br />To solve for three variables, we need a system of three independent equations.<br />
Independent, Dependent and Inconsistent Equations<br />Systems of three equations in three variables follow the same characteristics of systems of equations in two variables<br />Independent equations have one solution <br />Dependent equations have an infinite number of solutions<br />Inconsistent equations have no solution<br />Solving a system of equations in three variables involves a few more steps, but is essentially the same process as for systems of two equations in two variables.<br />
Steps to solving a system of equations in 3 Variables<br />Ensure that the equations are in standard form: <br /> Ax + By + Cz = D<br />Remove any decimals or fractions from the equations.<br />Eliminate one of the variables using two of the three equations. Result will be a new equation with two variables.<br />Eliminate the same variable using another set of two equations. Result will be a second equation in two variables.<br />Solve the new system of two equations.<br />Using the solution for the two variables, substitute the values into one of the original equations to solve for the third variable.<br />Check the solution set in the remaining two original equations.<br />
Example system of three equations<br />Example: solve for x, y and z<br />First, we need to ensure that all equations are in standard form, i.e. all variables are on the left side of the equation. Note that equation #3 is not in standard form. We need to get the y and z terms to the left side of the equation. <br />
Put equations in Standard Form<br />Rewriting Equation 3 into standard form:<br />Add y to both sizes<br />Add 6z to both sides<br />Our 3 equations are now:<br />But, note that equations #1 contains a fraction, and #2 contains decimals. If we eliminate the fractions and decimals, the equations will be easier to work with.<br />
Remove Fractions and Decimals<br />We can eliminate the fraction in #1 by multiplying both sides of the equation by 4.<br />This is our new version of #1<br /><ul><li>We can eliminate the decimals in #2 by multiplying both sides of the equation by 100.</li></ul>Note that all terms have a common factor of 5<br /><ul><li> So, Let’s divide both sides by 5, so we have smaller numbers to work with. </li></ul>This becomes our new #2<br />
Eliminate One Variable<br />We have rewritten our three equations as follows:<br />Now, to solve, we need to first get two equations with two variables. To do so, we eliminate one of the variables from two of the equations. Then eliminate the same variable from another set of two equations. <br />In the first column, we eliminated y from equations 1 & 2, resulting in equation A.<br />In the second column, we eliminated y from equations 2 & 3, resulting in equation B.<br />
Solve the system in two variables<br />To solve our new system of equations A & B, the first step is to eliminate one of the variables <br />If we choose to eliminate x by addition, we must get the equations in a form such that the coefficients of x in the two equations are inverses of one another (for example: +1 and -1 or +5 and -5). To do so we need to multiply each of the equations by some factor.<br />Before we determine that factor, we need to determine what the resulting coefficient of x will be. It will need to be a common multiple of the current coefficients, 8 and 10. The least common multiple of 8 and 10 is 40, so we will want the coefficients of the x terms to be 40 and -40. <br />If we multiply equation A by 5 and equation B by 4, the resulting coefficient of x will be 40. However, we need one of those coefficients to be negative, so we’ll multiply the second equation by -4. <br />Equation A Equation B<br />
Solve first for z, then for x<br />Add the new set of equations A and B to eliminate x, and then solve for z:<br />Add the two equations to eliminate x<br />Divide both sides by 102 and reduce<br />Now substitute the value for of z into equation A or B to solve for x; We are using equation B:<br />Substitute 1/6 for z<br />Simplify<br />Subtract 2 from both sides<br />Divide both sides by 10 and reduce<br />
Use one of the original equations to solve for y<br />Next we need to substitute the values of x and z into one of the original equations to solve for y. We’ll use equation 2 (after decimals were removed):<br />Substitute the values for x and z<br />Simplify<br />Subtract 3 from both sizes<br />Multiply by (-1)<br />The solution for the system of equations is the ordered triple: <br />
Check your answer<br />Substitute the values for x, y and z into the other two original equations (1 and 3)<br />Equation 1:<br />Substitute the values for x, y, z<br />Simplify<br />The resulting statement is true.<br />Equation 3:<br />Substitute the values for x, y, z<br />Simplify<br />The resulting statement is true.<br />
Summary<br />In our example problem, all three variables were present in each of the equations. Your assignments may include some problems where one of the variables is missing in some of the equations as in the example below. <br />These problems will be solved in the same manner. However, some of the steps to eliminate a variable have already been done for you. Note that we always need 3 equations in order to solve for 3 variables.<br />To solve a problem like this one, It is important to keep your work organized and to write the equations so that terms with like variables are lined up with one another.<br />x + 2y = 8<br /> 2y + 3z = 1<br />3x – z = -3<br />
Summary<br />Our example problem resulted in one unique solution set. However, this may not always be the case. If the variables all drop out when adding two of the equations together, a unique solution is not possible. See the following examples.<br />-5x + 6y – 3z = 17<br />5x – 6y + 3z = -2<br /> 0 = 15<br />In this example, the variables all cancel out, and the result is a false statement. This indicates that there is no solution, and we have an inconsistent set of equations.<br />In this example, the variables all cancel out, and the result is a true statement. This indicates that the equations are dependent and we have an infinite number of solutions. <br />2x + y – z = 6<br />-2x – y + z = -6<br /> 0 = 0<br />
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