Trying to learn Expanded Brackets to teach my Granchild during half term.

[cdaley]

Hi all,

I'm a Gran mother in her 60's and I'm trying to help my Gran son whilst he's with me during half term with Extended Brackets, I've watched a few youtube videos and I think I'm beginning to understand extended brackets.

Please can you help me to see if I'm heading in the right direction.

Find the value of each expression when p=3 and q=4.
a. 3p(q-2)

3x4 =12 12x4= 48 48-2= 46

b. (2p +q)^2

2x3 = 6 6+4 = 10 10^2 = 100

c. q(4p-7)

4 x (4x3) = 48 48-7 = 41

Can you let me know if I'm on the right track please.

Thank you ever so much

 

[stapel]

...I think I'm beginning to understand extended brackets.

Please can you help me to see if I'm heading in the right direction.

Find the value of each expression when p=3 and q=4.
a. 3p(q-2)

3x4 =12 12x4= 48 48-2= 46

I think you mean the above to indicate the following:

. . . . .\(\displaystyle 3p\, (q\, -\, 2)\)

. . . . .\(\displaystyle 3(3)\, \bigg((4)\, -\, 2\bigg)\)

. . . . .\(\displaystyle 9\, \bigg(4\, -\, 2\bigg)\)

...but this already doesn't appear to match what you've posted, so something is wrong...?

b. (2p +q)^2

2x3 = 6 6+4 = 10 10^2 = 100

I think your steps were as follows:

. . . . .\(\displaystyle (2p\, +\, q)^2\)

. . . . .\(\displaystyle \bigg( 2(3)\, +\, (4) \bigg)^2\)

. . . . .\(\displaystyle \bigg( 6\, +\, 4\bigg)^2\)

. . . . .\(\displaystyle \bigg(10\bigg)^2\)

. . . . .\(\displaystyle 10^2\)

. . . . .\(\displaystyle 100\)

If so, then I agree with your result.

c. q(4p-7)

4 x (4x3) = 48 48-7 = 41

I think your work was along these lines:

. . . . .\(\displaystyle q\, (4p\, -\, 7)\)

. . . . .\(\displaystyle (4)\, \bigg( 4(3)\, -\, 7\bigg)\)

. . . . .\(\displaystyle 4\, \bigg(12\, -\, 7\bigg)\)
But this again doesn't match what you've posted, so something is wrong.

Are you familiar with what is called "the order of operations"? (here) This information may prove helpful in doing this sort of computation.

Please reply with corrections and clarification, showing your steps clearly, as above. Thank you!

 

[HallsofIvy]

There is a reason for the parentheses (or braces). a(b+ c) is not the same as ab+ c. The reason b+ c is set in parentheses is that it is to be done as a separate calculation. if a= 3, b= 4, and c= 5, then b+ c= 9 so a(b+ c)= 3(9)= 27. But ab+ c= (3)(4)+ 5=12+ 5= 1 7.

At a "higher" level of algebra, you can use the "distributive law": a(b+ c)= ab+ ac. With a= 3, b= 4, and c= 5, as before, that is 3(4)+ 3(5)= 12+ 15= 27 as before.