1. Problem 1.4 Say It With Symbols

4. You can find the area of the two separate rectangles and add the two areas together or you can find the area of one large rectangle. A B
Rectangle A had dimensions of 30 by x. The area of A is 30x.
Rectangle B had dimensions of 30 by 10. The area of B is 300.
The combined area is 30x
If I look at the pool as one big rectangle the side lengths will be 30 by (x+10).
I can use the distributive property to multiply 30(x+10)
The combined area is 30x

5. You can find the area of the two separate rectangles and add the two areas together or you can find the area of one large rectangle.
Rectangle A had dimensions of 25 by x. The area of A is 25x.
Rectangle B had dimensions of x by x. The area of B is x2.
The combined area is 25x + x2. A B
If I look at the pool as one big rectangle the side lengths will be x by (x+25).
I can use the distributive property to multiply x(x+25)
The combined area is x2 + 25x.

6. This one can be done in more than two ways
This one can be done in more than two ways. I can split it up into 4 smaller rectangles. Or I can make two tall rectangles. Or I can make two wide rectangles. Or I can write it as finding the area of one large rectangle.
Rectangle A had dimensions of x by x. The area of A is x2.
Rectangle B had dimensions of 3 by x. The area of B is 3x.
Rectangle C had dimensions of 2 by x. The area of C is 2x.
Rectangle D had dimensions of 3 by 2. The area of D is 6.
The combined area is x2+ 5x +6 A B C D
Rectangle A had dimensions of x by x+3. The area of A is x(x+3) = x2+3x
Rectangle B had dimensions of 2 by x+3
The area of B is 2(x+3) = 2x+6
The combined area is x2+ 5x +6 A B

7. This one can be done in more than two ways
This one can be done in more than two ways. I can split it up into 4 smaller rectangles. Or I can make two tall rectangles. Or I can make two wide rectangles. Or I can write it as finding the area of one large rectangle.
As one large rectangle the side lengths would be (x+2) and (x+3)
The combined area is x2+ 5x +6.
To get that area, I would have done one of the other methods to still get the area of the parts. A
Rectangle A had dimensions of x by x+2. The area of A is x(x+2) = x2+2x
Rectangle B had dimensions of 3 by x+2
The area of B is 3(x+2) = 3x+6
The combined area is x2+ 5x +6 A B

8. You can find the area of the two separate rectangles and add the two areas together or you can find the area of one large rectangle.
Rectangle X had dimensions of a by b. The area of X is ab.
Rectangle Y had dimensions of a by c. The area of Y is ac.
The combined area is ab+ac. X Y
If I look at the pool as one big rectangle the side lengths will be a by (b+c).
I can use the distributive property to multiply a(b+c)
The combined area is ab+ac.

9. = 6x - 20 = 3x + 15 3 2 x 5 3x -10 = x2 +7x +10 = 2x2 + 10 2x x 5 5 x
aka Area as SUM
= 3x + 15
= 6x - 20 x 3 3x 2
= 2x2 + 10
= x2 +7x +10 x 2 5 x 2x

10. = 12(2x + 1) = 3x+6 = 3(x + 2) 2x 1 12 = x(x + 3) = (x+1)(x + 3) x 1 3
aka Area as PRODUCT
= 12(2x + 1)
= 3x+6
= 3(x + 2) 2x 12
= x(x + 3)
= (x+1)(x + 3) x 1 3 x x

11. = 4s+4 multiply the 4 by the S and by the 1
= 4s+4 Add the S terms together
= 2s+2s+4 = 4s + 4
multiply the 2 by the S and by the 2 THEN add 2s
= 4s = 4s + 4
multiply the 4 by the S and by the 2 THEN subtract 4
= (s+2)(s+2) = s2+2s+2s+4 – s2 = 4s+4
multiply the (s+2) times itself then subtract the extra s2 term

12. I can combine the like terms of 2x and -12x to get -10x
I can combine the like terms of 2x and -12x to get -10x. My simplified answer is -10x + 10 or x
I can combine the like terms of 12x and -2x to get 10x. My simplified answer is 10x + 10 or x
Not equal
This is already simplified. They are not like terms. They cannot be combined and there is nothing to multiply. I can rearrange terms to have -10x+10 if I wanted.
I can distribute the 10 on the outside and multiply it by each term on the inside. I’ll do 10·1 and 10(-x) to get 10-10x or -10x +10.

13. 6(p+2) -2p = 6p+12-2p = p+12
Six times the quantity of p plus 2 minus 2p
Six times the sum of p and 2 minus 2p
6p+(2 - 2)p = 6p+0p = p
Six p plus the quantity of 2 minus 2 times p
Six p plus the difference of 2 and 2 times p