document.write( "Question 27655: A rectangular picture is 13 by 15 inches. A frame of the same width all the way around has an area of 93 square inches. What is the width of the frame.
\n" ); document.write( "I tried but I was getting nowhere I asked my dad but he couldn't help either. Can you? \n" ); document.write( "
Algebra.Com's Answer #15033 by bmauger(101)\"\" \"About 
You can put this solution on YOUR website!
The area of the frame must be the area of the outer dimentions minus the area of the picture. So you have two rectangles, and need to find the difference between the two. The inner rectangle is given as 13\"x15\" which is 195 square inches.\r
\n" ); document.write( "\n" ); document.write( "Because the width of the frame is consistant all the way around, we'll call that value \"x\". Because it's being added to all four sides (twice to the width and twice to the lenght) So the area of the rectangle created by the outside of the frame is \"%282x%2B15%29%282x%2B13%29=4x%5E2%2B56x%2B195\"\r
\n" ); document.write( "\n" ); document.write( "The problem states that frame's area (i.e. the difference between the outside and inside of the frame) is 93 in^2. So we take our outside subtract our inside, and set it equal to 93 like:
\n" ); document.write( "\"4x%5E2%2B56x%2B195-195=93\" We now have a quadratic equation we can solve:
\n" ); document.write( "\"4x%5E2%2B56x-93=0\"Which factors to:
\n" ); document.write( "\"%282x-3%29%282x%2B31%29=0\" 2x+31 will solve for a negative and nonsensical answer, so we'll look at 2x-3:
\n" ); document.write( "\"2x-3=0\"
\n" ); document.write( "\"2x=3\"
\n" ); document.write( "\"x=3%2F2\"
\n" ); document.write( "The border is 1.5\"
\n" ); document.write( "Double checking:
\n" ); document.write( "\"%2815%2B1.5%2A2%29%2813%2B1.5%2A2%29-195\"
\n" ); document.write( "\"%2818%29%2816%29-195\"
\n" ); document.write( "\"288-195=93\"Check.
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