Glencoe Math: Course 2, Volume 2
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Glencoe Math: Course 2, Volume 2 View details
7. Subtract Linear Expressions
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Exercise 25 Page 409

To find the additive inverse of a linear expression, you can multiply each term of the expression by -1.

4x-15

Practice makes perfect
We want to find the following difference. 5(x-3)-x First, let's distribute 2 into the first set of parentheses.
5(x-3)-x
(5 * x-5 * 3)-x
(5x - 15) -x
Now, let's note that adding 0 to an expression does not change the value of the expression. For this reason, we can rewrite our expression as a difference of two linear expression. Let's do it! (5x - 15) -x = (5x - 15)-(x+ 0) Next, we will calculate this difference. First, let's find the additive inverse of the linear expression within the second set of parentheses. To do so, we can distribute -1 into the second set of parentheses. When we do so, we multiply each term of that linear expression by -1.
(5x - 15)-(x+0)
(5x - 15)+( (-1) * x+ (-1) * 0)
(5x - 15)+( - x+ (-1) * 0)
(5x - 15)+( - x+ 0)
Now, let's identify which, if any, terms can be combined. Remember, only like terms — constant terms or terms with the same variable — can be combined. ( 5x - 15)+( - x+ 0) In this case, we have two x-terms and two constants. Let's arrange the like terms in columns. Then, let's perform the addition. 5 x -15& (+) - x +0& 4 x - 15& Therefore, we get the following result. 5(x-3)-x = 4x-15