So if y equals 2, then 3×2=6, and 2×2=4 and 6+4=10.

Your homework, what I scanned of it is correct.

Math is a beautiful subject because if you know your basic operations, you can always check your work by plugging in a number for each variable and see if it works out right.

2 and 3 are correct, but 4 is wrong. For number 5 and 9, you can add the numbers without variable to further simplify, for example, in number 5, you can add the 7 and 2 and you get 9z+9. 6, 7, and 8 are correct.

For the second set (10-15) try to do the same, but treat a variable to a power as a different variable. so #10 would just be 10a+2a(to the second power) and #12 is 3x(to the third power)+2x (to the second power)

For 13-15, you need to mulitply it all out beofre adding the like terms. #13 is 3y+6+6y, then add the like terms to get 9y+6

If you get confused, just thinkit through and take it one step at a time.

GL!

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4. You have to double check your work… it doesn’t make sense the way you typed it…

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5. You can “combine” (add) the 2 and 7…. 2+7 = ?

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9. Again… just as in Problem 5., here you can “combine” the numbers… so what is 2+9? 2+9 = ?

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10. 4a+6a+2a (to the 2nd power) = … you can combine the 4a and the 6a…. but the 2a^2 stays the same…. so you have ___a + 2a^2…. you just have to fill in the blank…

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11. 5z+2z+6z (to the 2nd power)

Same as in 10. You can combine the 5z and the 2z… but the 6z^2 stays the same… so you have ___z + 6z^2…. you just have to fill in the blank…

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These, you first have to take care of the ()’s…. for example… in 13. you have to multiply each of the terms in the parentheses (the “y” and the +2) each by the 3 in front… so 3(y+2) = (3)(y) + (3)(+2) = 3y + 6… but don’t forget to add the 6y…. because your given problem is 3(y+2) + 6y…

This is how your work should look for Problem 13. Watch closely…

13. 3(y+2)+6y = (3)(y) + (3)(2) + 6y = 3y + 6 + 6y…

But you aren’t done yet… because you can combine “like” terms…. The “+6” stays by itself… but you can combine 3y and 6y…. so your final answer should be “9y + 6”

Okay… so I’m leaving 14. and 15. for you to do…

first expand the #(_ + _) …. by multiplying each term inside the ()’s by the # in front… In problem 14. you first expand 8(z+1) by multiplying “z” and “+1” each by 8…. you do the same with 2(z+4)…. after you have done that…. add all the “like” terms together… so add the 8z and the 2z….. and combine the numbers together also…. you should be adding an 8 and another 8…. then you will have your final answer…

14. 8(z+1)+2(z+4)

Same thing … expand #(_ + _ ) first….. then combine “like” terms… you know you are doing the math correctly if when you are combining your terms, you are adding an “ab” term and a “3ab” term and if you also have a “2a” term and a “+4” in your answer…

15. a(b+2)+3ab+4

#2, #3 are right, don’t know what you are doing with #4, maybe a typo?

#5 not finished because you still have 7+2

#10, #11 is the whole thing to the second power?

#13, #14, #15 take care of the () first

example #13 3(y+2)+6y =3y+6+6y=9y+6

Hope this helps

8z+10=8z+10

8z-8z=10-10

0=0

now do the samething for the other one

10) (4a+6a+2a) second power

12(a)second power

13) 3(y+2)+6y

3y+6+6y

9y+6

now do the others.