At a concession stand, seven hot dogs and two hamburgers cost $11.00, two hot dogs and seven hamburgers cost $16.00. Find the cost of one hot dog and the cost of one hamburger.
7d+2b = 11
2d+7b = 16
5d + 2(d+b) = 11
5b + 2(d+b) = 16
add the equations and you get
5(d+b) + 4(d+b) = 27
9(d+b) = 27
d+b = 3
similarly, b-d = 1
b=2
d=1
dog costs $1
burger costs $2
To find the cost of one hot dog and one hamburger, we can set up a system of equations based on the given information.
Let's assume the cost of one hot dog is 'x' dollars, and the cost of one hamburger is 'y' dollars.
According to the first statement, seven hot dogs and two hamburgers cost $11.00:
7x + 2y = 11 --------- (Equation 1)
According to the second statement, two hot dogs and seven hamburgers cost $16.00:
2x + 7y = 16 --------- (Equation 2)
We now have a system of equations to solve. Let's solve them using the substitution method:
From Equation 1, we can rearrange it to express 'x' in terms of 'y':
7x = 11 - 2y
x = (11 - 2y) / 7 ------- (Equation 3)
Substitute the expression for 'x' from Equation 3 into Equation 2:
2((11 - 2y) / 7) + 7y = 16
Multiply both sides of the equation by 7 to eliminate the denominator:
2(11 - 2y) + 49y = 112
22 - 4y + 49y = 112
45y = 90
y = 90 / 45
y = 2
Now substitute the value of 'y' into Equation 3 to find the value of 'x':
x = (11 - 2(2)) / 7
x = (11 - 4) / 7
x = 7 / 7
x = 1
Therefore, the cost of one hot dog is $1.00, and the cost of one hamburger is $2.00.
To find the cost of one hot dog and one hamburger, we can set up a system of equations based on the given information.
Let's assume the cost of one hot dog is 'h', and the cost of one hamburger is 'b'.
According to the first statement, "seven hot dogs and two hamburgers cost $11.00", we can write the equation:
7h + 2b = 11.00 ----(Equation 1)
According to the second statement, "two hot dogs and seven hamburgers cost $16.00", we can write the equation:
2h + 7b = 16.00 ----(Equation 2)
Now, we have a system of two equations with two variables. We can solve these equations using any method of our choice, such as substitution or elimination.
Let's use the method of elimination to solve the system.
Multiply Equation 1 by 2, so that the coefficients of 'h' will cancel each other when we subtract the equations.
2(7h + 2b) = 2(11.00)
14h + 4b = 22.00 ----(Equation 3)
Now, subtract Equation 3 from Equation 2 to eliminate 'h':
(2h + 7b) - (14h + 4b) = 16.00 - 22.00
2h + 7b - 14h - 4b = -6.00
-12h +3b = -6.00 ----(Equation 4)
Now, we have a new equation with only the variable 'h'. Let's solve this equation for 'h'.
Multiply Equation 4 by 4, so that the coefficients of 'b' will cancel each other when we subtract the equations.
4(-12h + 3b) = 4(-6.00)
-48h + 12b = -24.00 ----(Equation 5)
Now, subtract Equation 5 from Equation 3 to eliminate 'b':
(14h + 4b) - (-48h + 12b) = 22.00 - (-24.00)
14h + 4b + 48h - 12b = 22.00 + 24.00
62h - 8b = 46.00 ----(Equation 6)
Now, we have a new equation with only the variable 'h'. Let's solve this equation for 'h'.
Add Equation 6 and Equation 4 to eliminate 'b':
(-12h + 3b) + (62h - 8b) = -6.00 + 46.00
-12h + 3b + 62h - 8b = 40.00
50h - 5b = 40.00 ----(Equation 7)
Now, we have a new equation with only the variable 'h'. Let's solve this equation for 'h'.
Divide Equation 7 by 5:
(50h - 5b) / 5 = 40.00/5
10h - b = 8.00 ----(Equation 8)
Now, we have a new equation with only the variable 'h'. Let's solve this equation for 'h'.
Add Equation 8 and Equation 4 to eliminate 'h':
(-12h + 3b) + (10h - b) = -6.00 + 8.00
-12h + 10h + 3b - b = 2.00
-2h + 2b = 2.00 ----(Equation 9)
Now, we have a new equation with only the variable 'b'. Let's solve this equation for 'b'.
Multiply Equation 9 by 3, so that the coefficients of 'h' will cancel each other when we subtract the equations.
3(-2h + 2b) = 3(2.00)
-6h + 6b = 6.00 ----(Equation 10)
Now, subtract Equation 10 from Equation 8 to eliminate 'h':
(10h - b) - (-6h + 6b) = 8.00 - 6.00
10h - b + 6h - 6b = 2.00
16h - 7b = 2.00 ----(Equation 11)
Now, we have a new equation with only the variable 'b'. Let's solve this equation for 'b'.
Add Equation 11 and Equation 9 to eliminate 'b':
(-2h + 2b) + (16h - 7b) = 2.00 + 2.00
-2h + 16h + 2b - 7b = 4.00
14h - 5b = 4.00 ----(Equation 12)
Now, we have a new equation with only the variable 'b'. Let's solve this equation for 'b'.
Now, we have two equations with only the variable 'h' and 'b':
Equation 8: 10h - b = 8.00 ----(Equation 8)
Equation 12: 14h - 5b = 4.00 ----(Equation 12)
We can solve this system of equations by either substitution or elimination.
Let's use the method of elimination.
Multiply Equation 8 by 5 and Equation 12 by -1 to eliminate 'b':
5(10h - b) = 5(8.00)
14h - 5b = 4.00
-1(14h - 5b) = -1(4.00)
-14h + 5b = -4.00
Add the two equations to eliminate 'b':
(14h - 5b) + (-14h + 5b) = 4.00 + (-4.00)
0h + 0b = 0
The resulting equation, 0 = 0, indicates that the system of equations is dependent.
This means that there are infinite solutions to the problem, and we cannot determine the exact values for the cost of one hot dog and one hamburger.
Therefore, with the given information, we cannot find the exact cost of one hot dog and one hamburger.