Back-substitute the values found fortwo variables into one of the original equations to find the value of the third variable. We can now use any one of the original equations and back-substitute the values of x and y to find the value for z. We will use Equation 2. 3x + 3y + 2z = -3 Equation 2 3(-1) + 3(2) + 2z = -3 Substitute -1 for x and 2 for y. 3 + 2z -3 Multiply and then add: 3(-1) + 3(2) = -3 + 6 = 3. 2z = -6 Subtract 3 from both sides. z = -3 Divide both sides by 2. Withx = -1, y = 2, and z = -3, the proposed solution is the ordered triple (-1, 2, -3).

Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Back-substitute the values found fortwo variables into one of the original equations
to find the value of the third variable. We can now use any one of the original equations
and back-substitute the values of x and y to find the value for z. We will use Equation 2.
3x + 3y + 2z = -3
Equation 2
3(-1) + 3(2) + 2z = -3
Substitute -1 for x and 2 for y.
3 + 2z
-3
Multiply and then add:
3(-1) + 3(2) = -3 + 6 = 3.
2z = -6
Subtract 3 from both sides.
z = -3
Divide both sides by 2.
Withx = -1, y = 2, and z =
-3, the proposed solution is the ordered triple (-1, 2, -3).
Transcribed Image Text:Back-substitute the values found fortwo variables into one of the original equations to find the value of the third variable. We can now use any one of the original equations and back-substitute the values of x and y to find the value for z. We will use Equation 2. 3x + 3y + 2z = -3 Equation 2 3(-1) + 3(2) + 2z = -3 Substitute -1 for x and 2 for y. 3 + 2z -3 Multiply and then add: 3(-1) + 3(2) = -3 + 6 = 3. 2z = -6 Subtract 3 from both sides. z = -3 Divide both sides by 2. Withx = -1, y = 2, and z = -3, the proposed solution is the ordered triple (-1, 2, -3).
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