How do you solve x^3 + 2y^3x3+2y3 for x = 7 & y = 1?

2 Answers
Apr 5, 2018

345345

Explanation:

Just by replaicing the values in xx and yy:

x^3+2y^3x3+2y3

7^3+2·1^3=34573+213=345

Apr 5, 2018

345345

Here's how I did it:

Explanation:

Your expression is x^3 + 2y^3x3+2y3 and you know that x = 7x=7 and y = 1y=1. We just plug in these values for xx and yy, and then we simplify:

(7)^3 + 2(1)^3(7)3+2(1)3

7^373 is the same thing as (7)(7)(7)(7)(7)(7), or 343343.

Now, for 2(1)^32(1)3, we have to look this order of operations called PEMDAS. This stands for:
P arentheses
E xponents
MD Multiplication/Division
AS Addition/Subtraction

In your expression 2(1)^32(1)3, the first thing we need to do is take care of the exponent. (1)^3 is the same thing as (1)(1)(1)(1)(1)(1), which is just 11.

So now we do 2(1)2(1), which is 22.


Finally we add 22 with 343343 (from earlier) to get the final simplification of 345345.