Here's the question you clicked on:
sassyj
What is the solution of the system? Use substitution. x = –2y + 10 –6y = 6x – 6 z = –3x + 10y
A. (6, 12) B. (–12, –6) C. (–6, –12) D. (2, 4)
Solve two at a time.
1. Label the three equations x = –2y + 10 -(1) –6y = 6x – 6 -(2) z = –3x + 10y -(3) 2. Sub equation (1) into equation (2). Then solve it, you can get the value of y... Hmm... Are you posting the right question?
A. x = 9, y = –8, z = 114 B. x = –8, y = 9, z = 114 C. x = –8, y = 114, z = 9 D. x = 114, y = 9, z = –8
yeah, now options are correct....
I posted the wrong options- now, would it be A?
No, it won't be A ..
What did you get on solving the equations? x,y.z?
@sassyj Can you solve the first two equations or do you need help?
I dont get any of this-even reading the textbook
x = –2y + 10 -(1) –6y = 6x – 6 -(2) z = –3x + 10y -(3) Sub equation (1) into equation (2), that means when you see x, replace x by –2y + 10. Do you get this part?
So, try it. Sub. (1) into (2), what do you get?
If you really don't understand how substitution works, I suggest you to watch a short video on Khan Academy first http://www.khanacademy.org/math/algebra/systems-of-eq-and-ineq/v/practice-using-substitution-for-systems