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Use systems to solve :The length of a rectangle is 2 cm more than itswidth. If the perimeter is 52 cm, find the width.

Sagot :

ANSWER

The width is 12 cm

EXPLANATION

The length L of the rectangle is 2 cm more than its width W. With this we have one equation:

[tex]L=W+2[/tex]

Then the perimeter is 52cm, which is the sum of the sides of the rectangle:

[tex]P=W+W+L+L=2W+2L[/tex]

Therefore the system to solve is:

[tex]\begin{cases}L=W+2 \\ 52=2W+2L\end{cases}[/tex]

Using the substitution method we can solve just for W. Replace L in the second equation by its value in terms of W from the first equation:

[tex]52=2W+2(W+2)[/tex]

Use the distributive property to eliminate the parenthesis:

[tex]52=2W+2W+4[/tex]

Add like terms:

[tex]52=4W+4[/tex]

And solve for W:

[tex]\begin{gathered} 4W=52-4 \\ 4W=48 \\ W=\frac{48}{4} \\ W=12 \end{gathered}[/tex]

Therefore, the width of the rectangle is 12cm

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