Question
The difference between two numbers is 28. The larger number is eight less than twice the smaller number. Find both numbers.
Answers
Answered by
Bot
Let's call the larger number "L" and the smaller number "S".
From the first sentence, we know that:
L - S = 28
From the second sentence, we know that:
L = 2S - 8
Now we can substitute the second equation into the first equation:
(2S - 8) - S = 28
S - 8 = 28
S = 36
Now we can use that value to find L:
L = 2S - 8
L = 2(36) - 8
L = 64
Therefore, the two numbers are 36 and 64.
From the first sentence, we know that:
L - S = 28
From the second sentence, we know that:
L = 2S - 8
Now we can substitute the second equation into the first equation:
(2S - 8) - S = 28
S - 8 = 28
S = 36
Now we can use that value to find L:
L = 2S - 8
L = 2(36) - 8
L = 64
Therefore, the two numbers are 36 and 64.
Answer
The equation 6X plus 5Y equals 303X plus 7Y equals 285 represent the money collected from selling gift baskets in a school fundraising event if X represents the cost for each snack, gift basket and Y represents the cost for each chocolate gift basket. What is the cost for each chocolate basket?
Answered by
Bot
We can solve this problem using a system of linear equations.
6X + 5Y = 303
X represents the cost for each snack, gift basket, and Y represents the cost for each chocolate gift basket, so we'll call the cost of a chocolate gift basket "C":
3X + 7C = 285
Now we have two equations in two variables. We can solve for C by using substitution:
6X + 5Y = 303
6X + 42C = 1710 (multiply the second equation by 6 to get 6X)
5Y - 42C = -1407 (subtract the second equation from the first)
5Y = 303 - 42C (add 42C to both sides)
Y = (303 - 42C) / 5 (divide both sides by 5)
Now we can substitute this expression for Y into the second equation:
3X + 7C = 285
3X + 7C = 285
3X + 7[(303 - 42C) / 5] = 285
Multiply both sides by 5 to eliminate the denominator:
15X + 7(303 - 42C) = 1425
15X + 2121 - 294C = 1425
15X - 294C = -696
Now we have two equations:
5Y = 303 - 42C
15X - 294C = -696
We can solve either equation for one variable in terms of the other and substitute into the other equation:
5Y = 303 - 42C
Y = (303 - 42C) / 5
15X - 294C = -696
15X - 294C = -696
15X - 58(303 - 5Y) = -3480 (substitute for C in terms of Y)
15X - 17574 + 290Y = -3480
15X + 290Y = 15894
Now we have one equation in one variable:
15X + 290Y = 15894
We can solve for Y:
Y = (15894 - 15X) / 290
Since Y represents the cost of a chocolate gift basket, we want a positive value for Y. We can start by plugging in X = 0:
Y = (15894 - 15(0)) / 290
Y = 54.8
However, this gives us a negative value for Y. We can try plugging in larger values for X until we get a positive value for Y:
Y = (15894 - 15(10)) / 290
Y = 50.8
Y = (15894 - 15(20)) / 290
Y = 46.8
Y = (15894 - 15(30)) / 290
Y = 42.8
So the cost of a chocolate gift basket is $42.80.
6X + 5Y = 303
X represents the cost for each snack, gift basket, and Y represents the cost for each chocolate gift basket, so we'll call the cost of a chocolate gift basket "C":
3X + 7C = 285
Now we have two equations in two variables. We can solve for C by using substitution:
6X + 5Y = 303
6X + 42C = 1710 (multiply the second equation by 6 to get 6X)
5Y - 42C = -1407 (subtract the second equation from the first)
5Y = 303 - 42C (add 42C to both sides)
Y = (303 - 42C) / 5 (divide both sides by 5)
Now we can substitute this expression for Y into the second equation:
3X + 7C = 285
3X + 7C = 285
3X + 7[(303 - 42C) / 5] = 285
Multiply both sides by 5 to eliminate the denominator:
15X + 7(303 - 42C) = 1425
15X + 2121 - 294C = 1425
15X - 294C = -696
Now we have two equations:
5Y = 303 - 42C
15X - 294C = -696
We can solve either equation for one variable in terms of the other and substitute into the other equation:
5Y = 303 - 42C
Y = (303 - 42C) / 5
15X - 294C = -696
15X - 294C = -696
15X - 58(303 - 5Y) = -3480 (substitute for C in terms of Y)
15X - 17574 + 290Y = -3480
15X + 290Y = 15894
Now we have one equation in one variable:
15X + 290Y = 15894
We can solve for Y:
Y = (15894 - 15X) / 290
Since Y represents the cost of a chocolate gift basket, we want a positive value for Y. We can start by plugging in X = 0:
Y = (15894 - 15(0)) / 290
Y = 54.8
However, this gives us a negative value for Y. We can try plugging in larger values for X until we get a positive value for Y:
Y = (15894 - 15(10)) / 290
Y = 50.8
Y = (15894 - 15(20)) / 290
Y = 46.8
Y = (15894 - 15(30)) / 290
Y = 42.8
So the cost of a chocolate gift basket is $42.80.