Can somebody help me with my algebra homeworks?

Question

1 - At most how much water can be added to a 7 liters of a solution that is 60% acid to obtain a solution that is at least 50% acid?

2 - The square of the sum of two numbers is 20 more than the square of their difference. One of the number is 1. What is the other number?



please show me the steps if you can. Thanks! :D

Answer 1

(7+x)*0.5=7*0.6
3.5+0.5x=4.2
0.5x=0.7
x=1.4 litres

2.1+x^2=20+1-x^2
2x^2=20
x^2=10
x=rt10

Answer 2

1. Let w = liters of water added. Then 7+w is how many liters in the final amount. 7 liters of a 60% solution has 4.2 liters of acid (7X0.60). If this is 50% of your final solution, the final amount is 8.4 L. So 7 + w = 8.4

2. If the unknown number is x, then (x+1)(x+1) = (x-1)(x-1) + 20
[20 more than means add 20; square means times itself]

Do this out by FOIL if that is what the call it at your school:

x^2 + 2x + 1 = x^2 - 2x + 1 + 20

Move the x^2 and x terms to the left and solve the resulting equation. Hope this explanation makes sense.

Answer 3

1.
If the 7 liters is 60% acid, then 4.2 liters are acid (7 x 60%).
That means 2.8 liters are water (7 - 4.2)
The most water you could add is enough to bring it up to the 4.2 liters of acid. If you add more than that, it will be less than 50% acid.
To get 4.2 liters of water, you need to add 1.4 liters (4.2 - 2.8).
The answer is 1.4 liters of water is the max.

2.
(x + y)^2 = 20 + (x - y)^2
x^2 +2xy + y^2 = 20 + x^2 - 2xy + y^2
Subtract x^2 and y^2 from both sides
2xy = 20 - 2xy
Add 2xy to both sides
4xy = 20
Divide by 4
xy = 5
One of the numbers = 1
5/1 = 1
The other number = 5
Check
(5+1)^2 - (5-1)^2 = 20
6^2 - 4^2 = 20
36 - 16 = 20
20 = 20

Answer 4

1.

(7 + X) x 0.5= 7 x 0.6
3.5 + 0.5X = 4.2
0.5X = 0.7
X = 1.4 liters



2.

(X + 1)(X + 1) = (X - 1)(X - 1) + 20
X^2 + 2X + 1 = X^2 - 2X + 1 + 20
2X + 1 = X^2 - X^2 - 2X + 1 + 20
2X + 1 = -2X + 21
2X + 2X + 1 = 21
4X = 21 - 1
4X = 20
X = 5