Assume that the width of the frame is x, then
The length and the width of the picture with the frame will increase by 2x (x for each side), then
L = 6 + 2x
W = 4 + 2x
Since the total area is 80 in.^2, then
[tex]L\times W=80[/tex]Substitute the values of L and W
[tex](6+2x)(4+2x)=80[/tex]Now, we will solve the equation
[tex]\begin{gathered} (6+2x)(4+2x)=(6)(4)+(6)(2x)+(4)(2x)+(2x)(2x) \\ (6+2x)(4+2x)=24+12x+8x+4x^2 \end{gathered}[/tex]Add the like terms
[tex](6+2x)(4+2x)=24+20x+4x^2[/tex]Now, equate it by 80
[tex]4x^2+20x+24=80[/tex]Subtract 80 from both sides
[tex]\begin{gathered} 4x^2+20x+24-80=80-80 \\ 4x^2+20x-56=0 \end{gathered}[/tex]Divide all terms by 4 to simplify the equation
[tex]\begin{gathered} \frac{4x^2}{4}+\frac{20x}{4}-\frac{56}{4}=\frac{0}{4} \\ x^2+5x-14=0 \end{gathered}[/tex]The equation is
[tex]x^2+5x-14=0[/tex]Now, factorize it into 2 factors
[tex]\begin{gathered} x^2=(x)(x) \\ -14=(7)(-2) \\ 7x-2x=5x \\ (x+7)(x-2)=0 \end{gathered}[/tex]Equate each factor by 0 to find x
[tex]\begin{gathered} x+7=0 \\ x+7-7=0-7 \\ x=-7 \\ x-2=0 \\ x-2+2=0+2 \\ x=2 \end{gathered}[/tex]We will refuse x = -7 because the length must be a positive number, then
The width of the frame is 2 inches
