A math teacher drove by a playground that was full of boys and dogs.The teacher notice there were a total of 40 heads and 100 feet. Hom many boys and how many dogs were there?
10 dogs
30 kids
2-4=2-=+y=y765
120
To solve this problem, we can use a system of equations to represent the number of boys and dogs at the playground. Let's assume that there are b boys and d dogs.
Given that there are a total of 40 heads, we can write the equation:
b + d = 40
Given that there are a total of 100 feet, we can write the equation:
2b + 4d = 100
Now we have a system of equations:
b + d = 40 (Equation 1)
2b + 4d = 100 (Equation 2)
To solve this system, we can use either substitution or elimination method. Let's use the elimination method:
Multiply Equation 1 by 2 to make the coefficients of b the same:
2(b + d) = 2(40)
2b + 2d = 80 (Equation 3)
Now subtract Equation 3 from Equation 2:
(2b + 4d) - (2b + 2d) = 100 - 80
2d = 20
Divide both sides by 2 to solve for d:
2d / 2 = 20 / 2
d = 10
Now, substitute the value of d into Equation 1 to solve for b:
b + 10 = 40
b = 30
Therefore, there are 30 boys and 10 dogs at the playground.
Let x = boys and y = dogs
x + y = 40, then x = 40-y
2x + 4y = 100
Substitute 40-y for x in second equation and solve for y. Insert that value into the first equation and solve for x. Check by inserting both values into the second equation.