British sterling silver is a copper-silver alloy that is 7.5% copper by weight. How many grams of pure copper and how many grams of British sterling silver should be used to prepare 200 grams of a copper-silver alloy that is 10% copper by weight? (round to the nearest tenth)

amount of silver --- x g

amount of copper --- 200-x g

.075x + 200-x = .1(200)
-.925x = -180
x = appr 194.6 g

To find the grams of pure copper and grams of British sterling silver needed, we can set up a system of equations based on the information given.

Let's assume the grams of pure copper used is x, and the grams of British sterling silver used is y.

We know that the total weight of the alloy is 200 grams, so we can write the equation:
x + y = 200 ...(1)

We are also given that the resulting alloy is 10% copper by weight. This means that the weight of copper in the alloy is 10% of the total weight of the alloy. So, we can write the equation:
0.10 * 200 = 0.075x + 0.075y

Simplifying this equation, we get:
20 = 0.075x + 0.075y ...(2)

Now, we have a system of two equations with two variables. We can solve this system by either substitution or elimination method.

Let's solve it using the elimination method:
Multiply equation (1) by 0.075:
0.075x + 0.075y = 0.075*200
0.075x + 0.075y = 15 ...(3)

Subtract equation (2) from equation (3):
0.075x + 0.075y - 0.075x - 0.075y = 15 - 20
0 = -5

This means that the system of equations is inconsistent, and there is no solution that satisfies both equations.

Therefore, it is not possible to prepare 200 grams of a copper-silver alloy that is 10% copper by weight using pure copper and British sterling silver.

To solve this problem, we can set up a system of equations using the given information.

Let's assume the amount of pure copper to be used is represented by 'x' grams, and the amount of British sterling silver to be used is represented by 'y' grams.

From the question, we know that British sterling silver is a copper-silver alloy that is 7.5% copper by weight. This means that 7.5% of y grams is pure copper. Therefore, the amount of copper in British sterling silver is (0.075 * y) grams.

We are given that a 200-gram alloy should be prepared, which is 10% copper by weight. This means that 10% of 200 grams is pure copper. Therefore, the amount of copper required in the alloy is (0.10 * 200) grams, which simplifies to 20 grams.

Now we have two equations:
1) Amount of copper from the British sterling silver: (0.075 * y) =?
2) Amount of required copper in the alloy: x + (0.075 * y) = 20

To solve for x and y, we can use these equations.

First, rearrange equation 2 to solve for y:
x + (0.075 * y) = 20
0.075 * y = 20 - x
y = (20 - x) / 0.075

Now substitute this value of y into equation 1:
(0.075 * y) = (0.075 * (20 - x)) / 0.075
y = 20 - x

Substitute the expression for y from equation 2 into equation 1:
(0.075 * (20 - x)) / 0.075 = 20 - x

Simplify the equation:
(20 - x) = 20 - x

This equation shows that x can be any value, as long as it is less than or equal to 20. This means that any amount of pure copper can be used, as long as the total weight is not higher than 20 grams to achieve a 10% copper alloy.

To find the amount of British sterling silver (y), we can use equation 2 since the value for x does not affect it.

Let's solve for y:
y = (20 - x) / 0.075

Substituting the values of x and y into the equation:
y = (20 - x) / 0.075
y = (20 - 0) / 0.075
y = 20 / 0.075
y = 266.7 grams (rounded to the nearest tenth)

Therefore, if we want to prepare a 200-gram copper-silver alloy that is 10% copper, we should use approximately 200 grams of pure copper and 266.7 grams of British sterling silver.