If the frame is 8 square inches, how big is the picture?

sandyk109

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A picture is 1 in. longer than it is wide. It is put into a frame 1/2 in. wide. If the area of the frame intself is 8 in.2, how big is the picture?

(l + 2) (l + 1) -(l + 1) (l) = 8in.2

LSquared + l +2L -LSquared -l =*inSquared

2l = 8in Squared

l = 4

l + 1= 5

This is the wrong answer. the correct answer not 4 by 5 but 3 by 4.

Please show me how to get the correct answer.

Thanks,

Sandy[/code][/list]
 
You have it correct. Must be an algebra mistake.

\(\displaystyle \L\\(w+1)(w+2)-w(w+1)=8\)

Solve for w.
 
Re: Beginning Algebra

sandyk109 said:
A picture is 1 in. longer than it is wide. It is put into a frame 1/2 in. wide. If the area of the frame intself is 8 in.2, how big is the picture?

(l + 2) (l + 1) -(l + 1) (l) = 8in.2

LSquared + l +2L -LSquared -l =*inSquared ... what happened to 2*1 - 'L' of FOIL

Also you are dealing with width call it 'W' like galactus did - just to avoid confusion

2l = 8in Squared

l = 4

l + 1= 5

This is the wrong answer. the correct answer not 4 by 5 but 3 by 4.

Please show me how to get the correct answer.

Thanks,

Sandy[/code][/list]
 
Sandy, (a + x)(a + y) = a^2 + ay + ax + xy = a^2 + a(x + y) +xy;
you didn't know that?

Also, 3 by 4 is NOT the solution: who told you that?
 
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