Question 622247
Hi, there--
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Here is one way to solve this problem.
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Let W be the width of the rectangle.
Let L be the length of the rectangle.
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We need to write two equations that use the information in the problem to model the situation. 
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The perimeter is 48 cm. Perimeter is the distance around the the rectangle: two widths and two lengths.
[two widths of the rectangle] + [two lengths of the rectangle] = [the perimeter]
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In algebra, we write
{{{2W+2L=48}}}
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We also know that the length of a rectangle is 6 cm longer than its width. 
[the length of the rectangle] = [the width of the rectangle] + [6 more centimeters]
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In algebra, we write,
{{{L=W+6}}}
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Now we have a system of two equations. We will solve it using the substitution method. Substitute W+6 for L in the perimeter equation.
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{{{2W+2L=48}}}
{{{2W+2(W+6)=48}}}
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Simplify and solve for W. Clear the parentheses.
{{{2W+2W+12=48}}}
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Combine like terms.
{{{4W+12=48}}}
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Subtract 12 form both sides of the equation.
{{{4W=48-12}}}
{{{4W=36}}}
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Divide both sides of the equation by 4.
{{{W=9}}}
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The width of the rectangle is 9 cm. Since the length of the rectangle is 6 more the the width, the length is 15 cm.
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The formula for the area of a rectangle is A=L*W. The area is 135 square cm since 9*15=135.
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The last step to to check our answer against the original problem. The perimeter should be 48 cm. We see that 2 widths is 18 cm, and 2 lengths is 30 cm; 18+30=48. Check!
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Please email me if I need to explain any part of this better. By the way, there is a way to solve this problem without using two variables and a system of equations. Let me know if you'd like to see that method.
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Ms.Figgy
math.in.the.vortex@gmail.com