I am setting the week hourly rate to x, and the weekend to y. Here is how the equation is set up:
13x + 14y = $250.90
15x + 8y = $204.70
This is a system of equations, and we can solve it by multiplying the top equation by 4, and the bottom equation by -7. Now it equals:
52x + 56y = $1003.60
-105x - 56y = -$1432.90
Now we add these two equations together to get:
-53x = -$429.30 --> 53x = $429.30 --> (divide both sides by 53) x = 8.10. This is how much she makes per hour on a week day.
Now we can plug in our answer for x to find y. I am going to use the first equation, but you could use either.
$105.30 + 14y = $250.90. Subtract $105.30 from both sides --> 14y = $145.60 divide by 14 --> y = $10.40
Now we know that she makes $8.10 per hour on the week days, and $10.40 per hour on the weekends. Subtracting 8.1 from 10.4, we figure out that she makes $2.30 more per hour on the weekends than week days.
Mr. Jackson invested $800 at 6% per year and $ 2400 at 4 % per year
<h3><u>Solution:</u></h3>
Mr. Jackson invested a sum of money at 6% per year, and 3 times as much at 4% per year.
Let the sum invested be ‘a’ and ‘3a’ at 6% per year and 4 % per year respectively
Also, his annual return totaled $144
We can form following equation on the basis of question:-
a = $800
The amount of money invested at 6% = a = 800
The amount of money invested at 4 % = 3a = 3(800) = 2400
So, the amount of money invested at 6% is $800 and the amount of money invested at 4% is $ 2400
You have the following expressions given in the problem above:
(y^2/y-3)(y^2-y-6/y^2+y)
By applying the exponents properties, you can simplify it, as it shown below:
(y^2/y-3)(y^2-y-6/y^2+y)
(y^4-y^3-6y^2)/(y^3+y^2-3y2-3y)
(y^4-y^3-6y^2)/(y^3-2y2-3y)
Then, you have:
y^2(y^2-y-6)/y(y^2-2y-3)
(y^2-y-6)/(y^2-2y-3)
The answer is: (y^2-y-6)/(y^2-2y-3)
The correct answer is x=3
