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

sandyk109

New member
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.

Thanks,

Sandy[/code][/list]

galactus

Super Moderator
Staff member
You have it correct. Must be an algebra mistake.

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

Solve for w.

Subhotosh Khan

Super Moderator
Staff member
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.

Thanks,

Sandy[/code][/list]

Denis

Senior Member
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?