Algebra 2 world problem help?!?
A painter can be paid in one of two ways.
Plan A: $500.00 plus $9.00 per hour
Plan B: $250.00 plus $14.00 per hour
Suppose the job takes n hours. For what value of n is Plan A the same as Plan B?
I don't understand how to solve this problem or what its asking. I set up like this:
500+9(n)
and
250+14(n)
I dont know what to do or if I even set it up right?
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Since you know your equations and you said you want the value the same, why don't you set them as equal to one another and solve.
permit x be the form of black marbles. permit y be the form of yellow marbles. Now convert the information into equations: 4x = 3y + 17 6x = 10y - 2 Now it particularly is an basic situation of fixing a device of equations. i could start up via tripling the 1st equation and doubling the 2d: 12x = 9y + fifty one 12x = 20y - 4 From this, you get 9y + fifty one = 20y - 4 resolve that to get y = 5 Use that y-fee in any of the above equations, then grind via some basic algebra to choose that x = 8. maximum suitable answer: there have been 8 black marbles and 5 yellow marbles. desire that facilitates!
Make up an amount of hours until you find the amount of hours where plan A and plan B cost the same.