Say you are at a bike shop. They sell tricycles, bicycles and unicycles. If there is a total of 150 wheels and 70 seats. If there are twice as many bicycles as tricycles how many of each type of bike do you have?
T + B + U = 70
B = 2T
U + 2B + 3T = 150
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Solve those equations simultaneously.
2T + B = 80
2B = 80
B = 40
T = B/2 = 20
U = 70 - 40 - 20 = 10
To find the number of each type of bike, we can set up a system of equations based on the given information.
Let's represent the number of tricycles as "t", bicycles as "b", and unicycles as "u".
From the information given, we know that:
1) The total number of wheels is 150: 3t + 2b + 1u = 150.
2) The total number of seats is 70: t + b + u = 70.
3) There are twice as many bicycles as tricycles: b = 2t.
We can now solve these equations simultaneously.
Substituting the value of b from equation 3 into equations 1 and 2, we have:
3t + 2(2t) + 1u = 150,
t + 2t + u = 70.
Simplifying these equations, we get:
7t + u = 150, and
3t + u = 70.
Now, we can solve this system of equations using substitution or elimination.
Let's use elimination:
Multiplying the second equation by 3, we have:
(3t + u) × 3 = 70 × 3,
which becomes:
9t + 3u = 210.
Now, subtract the resulting equation from the first equation:
(7t + u) - (9t + 3u) = 150 - 210,
which simplifies to:
-2t - 2u = -60.
Rearranging this equation, we have:
2t + 2u = 60.
Divide this equation by 2:
t + u = 30.
Now, we have a system of equations:
t + u = 30, and
3t + u = 70.
Subtracting the equations, we get:
(3t + u) - (t + u) = 70 - 30,
which simplifies to:
2t = 40.
Solving for t, we find:
t = 20.
Since b = 2t, we have:
b = 2 × 20 = 40.
Now that we know the number of tricycles and bicycles, we can find the number of unicycles by substituting the values into equation 2:
20 + 40 + u = 70,
which simplifies to:
60 + u = 70.
Solving for u, we find:
u = 10.
So, there are 20 tricycles, 40 bicycles, and 10 unicycles at the bike shop.