MATH - this is easy, middle school level - but I forgot how to do this stuff! Can you walk me thru it?

Question

I did this problem with my daughter and came up with the answer 126; but the book says the answer is 78. Can you walk me thru how u come up with the answer 78 please?

If x=3, y=4, and z=5

2[x+4(y=z)]

Answer 1

use the order of operations Parentheses,Exponents,Multiplication,Division,Addition, and Subtraction. First you plug in the values and you should get
2[3+4(4+5)] then you add the 4 and 5 = 9 2[3+4(9)] then you get
2[3+36] then 2(39) and the answer is 78.=)

Answer 2

Ok I understand where you went wrong, you did the two sums x+4 and y+z which make 7 and 9, you multiplied these together (63) and then by 2 (126). In fact the question is asking you to add x (3) to the sum of 4(4+5) which makes 36 + 3 = 39, multiplied by 2 makes 78.

Answer 3

You have copied the problem wrongly,it should be 2[x+4(y+z) 2[x+4(y+z)] = 2[3 +4(4+5)] = 2[3 +4X 9] = 2[ 3+36] =2 X39=78

Answer 4

2[3+4(4+5)]
2[3+4(9)]
2[3+36]
2*39
=78

Answer 5

I think the question is to find 2[x+4(y+z)]
y+z = 4+5 = 9
4(y+z) = 4x9 = 36
x+4(y+z) = 3+36 = 39
2[x+4(y+z)] = 2x39 = 78
GOT IT??

Answer 6

you typed something wrong, (y=z) it must be either (y-z or y+z). it is easy stuff though, type it again and il walk you through it. Remember your order of operations (BEDMAS= Brackets, exponents, division, multiplication, addition, subtraction) do what it in the brackets first then move on to multiplication. another potential source for error iz not distributing the term at the front of the brackets. each term in the brackets seperated by a + or - needs to be multiplied by the term in front of the brackets.