Algebra II: Solving Systems of Equations in Three Variables by Substitution
 This is a free lesson from our course in Algebra II In this section, you'll learn the method of solving system of equations in three variables by substitution. This method for solving this system is analogous to the methods used to solve the two variable systems.To solve a system in three variables, it must have three different equations which relate the unknowns. Explaination is provided with the help of some examples presented with video, in instructor's voice and own handwriting to how to solve the system of equations which will clear the concept in detail. (More text below video...)
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(Continued from above)
Procedure:
Let, we have three equations equation 1, equation 2, and equation 3 having variables x, y, and z. The substitution method has these basic steps:
� solve equation1 for z.
� substitute the whole right-hand side of the equation (the one you just wrote) in place of "z" in equation 2. You should use parentheses around the substituted expression to keep everything orderly.
� solve the equation you wrote in step 2 for y.
� substitute the whole right-hand side of the equation you wrote in step 1 in place of "z" in equation3. Then substitute the whole right-hand side of the equation you wrote in step 3 in place of "y" in equation 3.
� solve the equation you wrote in step 4 for x. You should get a numeric value of x.
� use this value of x in the equation you wrote in step 3 to solve for y. You should get a numeric value of y.
� use these values of x and y in the equation you wrote in step 1 to solve for z. You should get a numeric value of z.
For example, the system of equations x + y + z = 6, 3x + 2y + z = 10, and 2x + 4y + 3z = 19, simplifies and yields to x = 1, y = 2 and z = 3.
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