The length of a rectangle is 3 meters more than the width. The perimeter of the rectangle is 26 meters. If the width is b, which equation represents the situation? How many solutions will this equation have?

A. (b+3)(b) = 26, which has one solution
B. 2(b+3) + 2b = 26 which has infinitely many solutions
C. 2[(b+3) + b] = 26 which has one solution
D. 2[b+3) + b] = 26 which has no solution

L=b + 3 and 2L+2b=26(this equation gives you the perimeter)
you now substitute L=b + 3 into the equation to get:
2(b+3) + 2b = 26 (this is the perimeter)
so your answer is C

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cerverusdante cerverusdante · Answer 2 · 2018-11-24T04:59:36+01:00

Answer

C. 2[(b+3) + b] = 26 which has one solution

Explanation

The perimeter of a rectangle is given by the formula:

[tex]p=2(l+w)[/tex]

where

[tex]p[/tex] is the perimeter of the rectangle

[tex]l[/tex] is the length of the rectangle

[tex]w[/tex] is the width of the rectangle

We know form our problem that the width of our rectangle is b, so [tex]w=b[/tex]. We also now that the length of a rectangle is 3 meters more than the width, so [tex]l=b+3[/tex]. Let's replace those values in our formula:

[tex]p=2(l+w)[/tex]

[tex]p=2((b+3)+b)[/tex]

[tex]p=2[(b+3)+b][/tex]

But we also know that the perimeter of our rectangle is 26 meters, so [tex]p=26[/tex]. Let's replace that value as well:

[tex]26=2[(b+3)+b][/tex]

[tex]2[(b+3)+b]=26[/tex]

Now that we have the equation for the perimeter of our rectangle we can solve for [tex]b[/tex] to check how many solutions it has:

[tex]2[(b+3)+b]=26[/tex]

[tex]2b+6+2b]=26[/tex]

[tex]4b]=20[/tex]

[tex]b=\frac{20}{4}[/tex]

[tex]b=5[/tex]

Since our equation has one solution only, we can conclude that the correct answer is: C. 2[(b+3) + b] = 26 which has one solution.