Problem 7.
Solving systems of equations in three variables is very similar to a system of equations in 2 variables.
a. Choose 1 variable to eliminate from equation 1, 2, 3. What is it?
b. Choose 2 equations and eliminate the variable you chose in part a using the elimination method.
c. Choose 2 different equations and eliminate the variable you chose in part a using the elimination method.
d. Using your answer from part b and part c, you have a system of equations in 2 variables. Use any method to solve this system of equation in 2 variables.
e. Plug your answers to one of the original equations 1, 2, or 3 and solve for the third variable.
+9673 (a) I would choose to eliminate x from the equation. (Actually it is your choice to eliminate which variable)
(b) I chose equation 2 and equation 3 because this makes life easier :P.
\(x + 3y + 2z = 5 -----(2)\\ 4x + 4y + 3z = 13----\;(3)\\ \text{From equation 2: } 4x + 12y + 8z = 20-----(4)\\ \text{Subtract equation 3 from equation 4: } 8y + 5z = 7\)
(c) This time I would choose equation 1 and equation 2.
\(-3x+2y+z=-6 -----(1)\\ x+3y+2z =5 \quad\;\;-----(2)\\ \text{From equation 2: }3x + 9y + 6z = 15 -----(5)\\ \text{Add equation 1 and equation 5: }11y + 7z = 9\)
(d)
\(\begin{cases}8y+5z = 7\\11y+7z=9\end{cases}\\ 88y + 55z = 77\\ 88y + 56z = 72\\ z = 72 - 77 = -5\\ 88y + 55(-5) = 77\\ y =4\)
(e) Because (y, z) = (4, -5),
\(x + 3y + 2z = 5\\ x + 12 - 10 = 5\\ x = 3\)
Therefore (x, y, z) = (3, 4, -5).
:D
+9673 (a) I would choose to eliminate x from the equation. (Actually it is your choice to eliminate which variable)
(b) I chose equation 2 and equation 3 because this makes life easier :P.
\(x + 3y + 2z = 5 -----(2)\\ 4x + 4y + 3z = 13----\;(3)\\ \text{From equation 2: } 4x + 12y + 8z = 20-----(4)\\ \text{Subtract equation 3 from equation 4: } 8y + 5z = 7\)
(c) This time I would choose equation 1 and equation 2.
\(-3x+2y+z=-6 -----(1)\\ x+3y+2z =5 \quad\;\;-----(2)\\ \text{From equation 2: }3x + 9y + 6z = 15 -----(5)\\ \text{Add equation 1 and equation 5: }11y + 7z = 9\)
(d)
\(\begin{cases}8y+5z = 7\\11y+7z=9\end{cases}\\ 88y + 55z = 77\\ 88y + 56z = 72\\ z = 72 - 77 = -5\\ 88y + 55(-5) = 77\\ y =4\)
(e) Because (y, z) = (4, -5),
\(x + 3y + 2z = 5\\ x + 12 - 10 = 5\\ x = 3\)
Therefore (x, y, z) = (3, 4, -5).
:D
