width with unknown variable

Algebra 2

width with unknown variable

Postby cmorton557 » Sat Jul 09, 2016 2:08 pm

The length of a rectangular label is 3 centimeters less than twice the width. The perimeter is 54 centimeters. Find the Width. Define your work with the unknown quantities in terms of variable. Please help. I don't even know where to begin.
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Re: width with unknown variable

Postby Guest » Tue Aug 28, 2018 12:35 am

Lets length-a,width-b. The two sentences turned into equations.

[tex]\begin{array}{|l} a+3=2b \\ 2(a+b)=54 /:2 \end{array}[/tex]

[tex]\begin{array}{|l} a=2b-3 (we- are- replacing)\\ a+b=27 \end{array}[/tex]

2b-3+b=27 [tex]\Leftrightarrow[/tex] b=10 ;we know a=2b-3 [tex]\Leftrightarrow[/tex] a=2.10-3=17
a=17 centimeters ;b=10 centimeters
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