Angie Applegate has a photograph for which the length is 2 inches longer than the width. If she increases each dimension by 4 inches, the area is increased by 88 square inches. How do you find the original dimensions?

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Answer 1 · 2017-08-03T06:31:07

$8 in \times 10 in$ $8"in" xx 10"in"$

At first you have $x (x + 2) = a$ $x(x+2)=a$ , and after, you have $(x + 4) (x + 6) = a + 88$ $(x+4)(x+6)=a+88$

By expanding the brackets you get $x^{2} + 10 x + 24 = a + 88$ $x^2+10x+24=a+88$

By putting all terms onto one side, you get $x^{2} + 10 x - 64 = a$ $x^2+10x-64=a$

$x^{2} + 10 x - 64 = a$ $x^2+10x-64=a$
$x^{2} + 2 x = a$ $x^2+2x=a$

$x^{2} + 10 x - 64 = x^{2} + 2 x$ $x^2+10x-64=x^2+2x$

$8 x - 64 = 0$ $8x-64=0$

$8 x = 64$ $8x=64$

$x = \frac{64}{8} = 8$ $x=64/8=8$

$8 (8 + 2) = 8 \cdot 10 = 80$ $8(8+2)=8*10=80$
$(8 + 4) (8 + 6) = 12 \cdot 14 = 168$ $(8+4)(8+6)=12*14=168$
$168 - 80 = 88$ $168-80=88$

$x = 8 in$ $x=8"in"$
$x + 2 = 8 + 2 = 10 in$ $x+2=8+2=10"in"$

$8 in \times 10 in$ $8"in" xx 10"in"$