Posted by **Blue** on Monday, October 7, 2013 at 9:04pm.

1.)A picture 8 in. by 12 in. is placed in a frame which has a uniform width. if the area of the frame is equal to the area of the picture, find the dimension of the frame

(8+2x)(12+2x)-96 = 96

(8+2x)(12+2x)=96+96

96+4x(2)=192

4x(2)=192-96

4x(2)=96

x(2)=96/4

x/2=24/2

x=12

is it correct that the answer is 12? i'm not confident with my answer,please help!

Arman has 1160 annual income from bonds bearing 3% and 5 % interest. then he added 25% more of the 3% bonds and 40% more of the 5 % bonds, thereby increasing his annual income by 410. find his initial investment in each type

of bond

x=original 3% investment

y=original 5% investment

.03x+.05y=1160

.03(1.25x)+.05(1.4y)=1570

3.75x+7y = 1570

i don't know what to do next,please help!

thank you

- 12th grade MATH -
**Steve**, Tuesday, October 8, 2013 at 12:16am
Not even close. Not only is the algebra wrong, but x is not the answer to the question.

(8+2x)(12+2x)=96+96

96+40x+4x^2 = 96+96

4x^2+40x-96 = 0

x^2+10x-24 = 0

(x-2)(x+12) = 0

x = 2 or -12

Naturally, -12 is not a possibility, so x=2

The frame is 2" wide, making the dimensions of the framed picture 12x16

Note that 12x16 = 96+96

on the 2nd,

first, note that .05x1.4 is NOT 7. It is 0.07

.03x + .05y = 1160

.0375x + .07y=1570

7(1st)-5(2nd):

.21x + .35y = 8120

.1875x + .35y = 7850

.0225x = 270

x = 12000

y = 16000

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