Solve the system of equations algebraically to determine where the two cost options are equivalent, showing all work.
First equation is y=$224+$6(x-35)
Second Y=$7(x)

Question

Solve the system of equations algebraically to determine where the two cost options are equivalent, showing all work.
First equation is y=$224+$6(x-35)
Second Y=$7(x)

anonymous · Answer

y is total cost and x is the number of hours

anonymous · Answer

x=14
y=98
i hope that helps?

anonymous · Answer

ty for ur help i appreciate it

anonymous · Answer

your welcome
(:

directrix · Answer

@jannine245

The problem statement dictates that all work be shown so that is what we will do. 

There are two cost options for purchasing the same item, or at least that's what I get from the problem. Those two options will be equivalent at the value of x for which both equations are true.

directrix · Answer

y = 224 + 6*(x-35)
y = 7x
============

Set the two expressions for y equal.

So,

7x = 224 + 6*(x-35)

7x = 224 + 6x -  210

7x = 14 + 6x

x = ?

directrix · Answer

Once you solve for x, get the corresponding y-value by multiplying the value of x you got by 7. And, that's where the two cost options are equivalent.