I’ll do an example problem, and I challenge you to do this on your own!
4x+6y=23
7y-8x=5
Solving for y in 4x+6y=23, we can separate the y by subtracting both sides by 4x (addition property of equality), resulting in 6y=23-4x. To make the y separate from everything else, we divide by 6, resulting in (23-4x)/6=y. To solve for x, we can do something similar - subtract 6y from both sides to get 23-6y=4x. Next, divide both sides by 4 to get (23-6y)/4=x.
Since we know that (23-4x)/6=y, we can plug that into 7y-8x=5, resulting in
7*(23-4x)/6-8x=5
= (161-28x)/6-8x
Multiplying both sides by 6, we get 161-28x-48x=30
= 161-76x
Subtracting 161 from both sides, we get -131=-76x. Next, we can divide both sides by -76 to separate the x and get x=131/76. Plugging that into 4x+6y=23, we get 4(131/76)+6y=23. Subtracting 4(131/76) from both sides, we get
6y=23-524/76. Lastly, we can divide both sides by 6 to get y=(23-524/76)/6
Good luck, and feel free to ask any questions!
Answer: 4
step by step explanation: 3+1=4
Answer:
4x - 8 + 4y
Step-by-step explanation:
4(x - 2 + y)
4x - 8 + 4y
You multiply each term in the parenthesis by the number in front of it.
Sorry if I'm wrong - This is how I learned to subtract fractions in the 7th grade
First, I would line up the fractions side-by-side:
5/6 1/8
Then, my teacher would tell me to look at the numerators and subtract those - The answer would be 4
Next, she would say to do the same to the denominators - The answer is 2
4/2 is your answer I believe
Hope this helped! :)
The formula you can use for the withdrawals is that of an annuity. You have interest adding to the balance at the same time withdrawals are reducing the balance.
The formula I remember for annuities is
.. A = Pi/(1 -(1 +i)^-n) . . . . . i is the interest for each of the n intervals; A is the withdrawal, P is the initial balance.
This formula works when the withdrawal is at the end of the interval. To find the principal amount required at the time of the first withdrawal, we will compute for 3 withdrawals and then add the 7500 amount of the first withdrawal.
.. 7500 = P*.036/(1 -1.036^-3)
.. 7500 = P*0.357616
.. 7500/0.0347616 = P = 20,972.20
so the college fund balance in 4 years needs to be
.. 20,972.20 +7,500 = 28,472.20
Since the last payment P into the college fund earns interest, its value at the time of the first withdrawal is P*1.036. Each deposit before that earns a year's interest, so the balance in the fund after 4 deposits is
.. B = P*1.036*(1.036^4 -1)/(1.036 -1)
We want this balance to be the above amount, so the deposit (P) is
.. 28,472.20*0.036/(1.036*(1.036^4 -1)) = 6510.62
You must make 4 annual deposits of $6,510.62 starting now.
