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A+b+c=90 and 12a+10b+6c=870 solve for a,b, and c

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5 votes

Answer:

Explanation:

You can solve for a, b, and c by using a system of linear equations.

Starting with the first equation:

a + b + c = 90

You can isolate a by subtracting b and c:

a = 90 - b - c

Next, substitute the expression for a into the second equation:

12(90 - b - c) + 10b + 6c = 870

Expanding the equation:

1080 - 12b - 12c + 10b + 6c = 870

Combining like terms:

1070 - 2b + -6c = 870

Adding 2b and 6c to both sides:

1070 = 870 + 2b + 6c

Subtracting 870 from both sides:

200 = 2b + 6c

Dividing both sides by 2:

100 = b + 3c

Finally, subtracting 3c from both sides:

100 - 3c = b

To find c, substitute the expressions for a and b back into the first equation:

a + b + c = 90

90 - b - c + b + c = 90

90 - c = 90

-c = 0

Therefore, c = 0.

To find b, substitute c = 0 into the expression for b:

100 - 3c = b

100 = b

Therefore, b = 100.

Finally, to find a, substitute b = 100 and c = 0 into the expression for a:

a = 90 - b - c

a = 90 - 100 - 0

a = -10

Note that negative values for a, b, and c are possible, but not necessarily meaningful for this specific problem or context.

The solution is:

a = -10

b = 100

c = 0

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