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1. simplify the following expression

(a) 2(3r + 1) + 2r
- for this i multiplied out of the brackets to give 6r + 1 + 2r, then collect like terms to give 8r + 1 - that is my final answer. my book says the correct answer is 8r + 2. can someone explain to me what i have done wrong here.

can someone check my answer for another question please as it is not in my book!

the question is:

solve these simultaneous equations by the method of elimination

4d – 3c = 10
2d + 3c = 23

i know that the two lhs added will equal the two rhs added, so i have done this:

4d - 3c + 2d + 3c = 10 + 23

the c is cancelled out because of the status sign and this leaves

4d + 2d = 33
(6d = 33)

6d/6 = 33/6

.:. d = 5½

then i use this and plug it into the second simultaneous equation

11 + 3c = 23

11 + 3c = 23
-11 -11

3c = 12

3c/3 = 12/3

.:. c = 4

hence the solution is d = 5½, c = 4. can someone tell me if this is correct? i think it might be, because it works when i plug them into the other equation. my book which i am learning from is crap and does not tell me some answers. i am starting gcses next year and i wish to have a head start!
2. a)

2(3r + 1) + 2r
6r + 2 + 2r
8r + 2

you didn't multiply the 1 in brackets by 2.

b)

4d – 3c = 10
2d + 3c = 23

so:

3c = 4d - 10
3c = 23 - 2d

4d - 10 = 23 - 2d
6d = 33
d = 5.5

plugging in...

2(5.5) + 3c = 23
3c = 23 - 11
3c = 12
c = 4

so yeah, you're right. well done.
3. (Original post by einsci)
(a) 2(3r + 1) + 2r
- for this i multiplied out of the brackets to give 6r + 1 + 2r, then collect like terms to give 8r + 1 - that is my final answer. my book says the correct answer is 8r + 2. can someone explain to me what i have done wrong here.
2(3r + 1) + 2r = 6r + 2 + 2r = 8r + 2
4. (Original post by Solomon)
2(3r + 1) + 2r = 6r + 2 + 2r = 8r + 2
ahh yews, careless mistake, thankyou.
5. (Original post by chewwy)
a)

2(3r + 1) + 2r
6r + 2 + 2r
8r + 2

you didn't multiply the 1 in brackets by 2.

b)

4d – 3c = 10
2d + 3c = 23

so:

3c = 4d - 10
3c = 23 - 2d

4d - 10 = 23 - 2d
6d = 33
d = 5.5

plugging in...

2(5.5) + 3c = 23
3c = 23 - 11
3c = 12
c = 4

so yeah, you're right. well done.
thanks mates, i like the helpfullness of some members here
6. (Original post by einsci)
simplify the following expression

(a) 2(3r + 1) + 2r
- for this i multiplied out of the brackets to give 6r + 1 + 2r, then collect like terms to give 8r + 1 - that is my final answer. my book says the correct answer is 8r + 2. can someone explain to me what i have done wrong here.

can someone check my answer for another question please as it is not in my book!

the question is:

solve these simultaneous equations by the method of elimination

4d – 3c = 10
2d + 3c = 23

i know that the two lhs added will equal the two rhs added, so i have done this:

4d - 3c + 2d + 3c = 10 + 23

the c is cancelled out because of the status sign and this leaves

4d + 2d = 33
(6d = 33)

6d/6 = 33/6

.:. d = 5½

then i use this and plug it into the second simultaneous equation

11 + 3c = 23

11 + 3c = 23
-11 -11

3c = 12

3c/3 = 12/3

.:. c = 4

hence the solution is d = 5½, c = 4. can someone tell me if this is correct? i think it might be, because it works when i plug them into the other equation. my book which i am learning from is crap and does not tell me some answers. i am starting gcses next year and i wish to have a head start!
need help with gcse maths? add [email protected] to your msn, hopefully i can provide yet another year of help

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