The length of a rectangle is $3x+10$ feet and its width is $x+12$ feet.  If the perimeter of the rectangle is 76 feet, how many square feet are in the area of the rectangle?
To find the area of the rectangle, I need to know the length and the width. The perimeter of the rectangle is 76 feet, so I can use that to find $x$, which is the unknown part of the length and the width. Substituting the given values, I get $76=2(3x+10)+2(x+12)$. Simplifying, I get $76=6x+20+2x+24$. Combining like terms, I get $76=8x+44$. Subtracting 44 from both sides, I get $32=8x$. Dividing both sides by 8, I get $x=4$. Now that I have $x$, I can plug it into the expressions for the length and the width. The length is $3x+10$, so when $x=4$, the length is $3(4)+10=22$ feet. The width is $x+12$, so when $x=4$, the width is $4+12=16$ feet. Substituting the values of the length and the width, I get $A=22(16)$. Multiplying, I get $A=352$ square feet. 