SOLUTION: Please help solve the system of equations y=50+0.15(x-500) and y=105+0.10(x-650) Solve the system of linear equations algebraically using either the substitution method or the

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: Please help solve the system of equations y=50+0.15(x-500) and y=105+0.10(x-650) Solve the system of linear equations algebraically using either the substitution method or the       Log On


   

Question 479610: Please help solve the system of equations
y=50+0.15(x-500) and y=105+0.10(x-650)
Solve the system of linear equations algebraically using either the substitution method or the elimination method to determine where the two cost options are equivalent, showing all work.
a. Explain each step used to solve the system of linear equations. Include the following in your explanation:
• The algebraic method used to solve the system of equations
• All mathematical operations used to solve the system of equations
• The solution of the system of equations
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
y=50+0.15(x-500)
y= 50+0.15x-75
y-0.15x=-25 ........1

y=105+0.10(x-650)
y= 105 +0.10x-65
y-0.10x=40............2
Use the elimination method to find x & y
-0.15 x + 1 y = -25 .............1
-0.1 x + 1 y = 40 .............2
Eliminate y
multiply (1)by -1
Multiply (2) by 1
0.15 x -1 y = 25
-0.1 x + 1 y = 40
Add the two equations
0.05 x = 65
/ 0.05
x = 1300
plug value of x in (1)
-0.15 x + 1 y = -25
-195 + 1 y = -25
1 y = -25 + 195
1 y = 170
y = 170
The solution of system of equations (1300,170)
m.ananth@hotmail.ca