Solving systems of equations using substitution y=4x+2 7x-4y=-8
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Answer:
The crib is 30 inches wide and 23 inches long.
Step-by-step explanation:
Assuming that the crib is rectangular, then the crib has a length L and a width W.
And remember that for a rectangle of width W and length L, the perimeter is:
P = 2*L + 2*W = 2*(L + W)
And the area is:
A = L*W
In this case, the perimeter is 106 in, then:
P = 106in = 2*(L + W)
And we also know that the area is 690 in^2
Then:
A = 690in^2 = L*W
Then we have a system of equations:
106in = 2*(L + W)
690in^2 = L*W
To solve this, first, we need to isolate one of the variables in one of the equations.
Let's isolate L in the first one:
106in/2 = L + W
53 in = L + W
53in - W = L
Now we can replace this in the other equation:
690in^2 = L*W = (53in - W)*W = 53in*W - W^2
690in^2 = 53in*W - W^2
Now we need to solve this for W.
This is a quadratic equation:
W^2 - 53in*W + 690 in^2 = 0
The solutions of this equation are given by Bhaskara's formula:
Then the two possible solutions of W are:
W = (53in + 7in)/2 = 30in
W = (53in - 7in)/2 = 23in
We can choose any one of these, so let's choose W = 30in
If we replace this in the equation: "53in - W = L"
We can find the value of L:
53in - 30in = 23in = L
Then we have:
W = 30in
L = 23in