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Question Video: Subtracting Algebraic Expressions Mathematics • 7th Grade

Subtract 6π‘₯⁡ βˆ’ 3𝑦³ + 3𝑧² from 8π‘₯⁡ βˆ’ 5𝑦³ βˆ’ 2𝑧².

04:30

Video Transcript

Subtract six π‘₯ to the fifth power minus three 𝑦 cubed plus three 𝑧 squared from eight π‘₯ to the fifth power minus five 𝑦 cubed minus two 𝑧 squared.

To solve this problem, we need to think carefully about how to write out this subtraction. We’re subtracting six π‘₯ to the fifth power minus three 𝑦 cubed plus three 𝑧 squared from eight π‘₯ to the fifth power minus five 𝑦 cubed minus two 𝑧 squared. And this means that our first expression will be the eight π‘₯ to the fifth power minus five 𝑦 cubed minus two 𝑧 squared. And our second term will be six π‘₯ to the fifth power minus three 𝑦 cubed plus three 𝑧 squared. As we are subtracting the entire second expression from the first expression, we must put that second expression in parentheses and then distribute the subtraction across every term in the expression.

In this step, we pay close attention to the sign of every term. Six π‘₯ to the fifth power is positive. And we are going to subtract six π‘₯ to the fifth power. But the three 𝑦 cubed is negative. And if we want to subtract negative three 𝑦 cubed, we rewrite that as adding three 𝑦 cubed. The three 𝑧 squared is positive. We’re subtracting positive three 𝑧 squared, which we can rewrite as minus three 𝑧 squared. And in order to do any subtraction here, we need to see if we have like terms. Those are terms who have variables taken to the same exponent.

We have one term, that is, π‘₯ to the fifth power, and another π‘₯ to the fifth power term. We then have two terms with the variable 𝑦 cubed and two terms with the variable 𝑧 cubed. If you want, you can regroup so that the like terms are next to each other in the expression. As you’re doing this, pay close attention to the signs. And then we remember that, to combine like terms, we combine their coefficients. Eight π‘₯ to the fifth power minus six π‘₯ to the fifth power will be eight minus six π‘₯ to the fifth power, which is two π‘₯ to the fifth power.

For the second terms, we have negative five 𝑦 cubed plus three 𝑦 cubed. That means we need to combine negative five and positive three, which is negative two. And for our 𝑧 squared term, we’re combining the coefficients, negative two and negative three. Negative two minus three is negative five. Putting all this together, we get two π‘₯ to the fifth minus two 𝑦 cubed minus five 𝑧 squared.

Before we leave this question, let’s look at one other way you could solve, starting with our first expression. Instead of writing the second expression horizontally, you could vertically line the second expression up underneath the first. Notice that we’ve grouped the π‘₯ to the fifth terms, the 𝑦 cubed terms, and the 𝑧 squared terms. But again, our biggest challenge here is to make sure we don’t make any sign mistakes. For the first term, we’re saying eight π‘₯ to the fifth minus six π‘₯ to the fifth. So we subtract six from eight. And we get two π‘₯ to the fifth.

But our second set of like terms are not as straightforward. We have negative five 𝑦 cubed. And we are subtracting negative three 𝑦 cubed, negative five 𝑦 cubed plus three 𝑦 cubed. This is doing that distribution. Negative five plus three is negative two. And the variable is 𝑦 cubed. Then we have negative two 𝑧 squared minus positive three 𝑧 squared, which means we’ll have negative two 𝑧 squared minus three 𝑧 squared. Negative two minus three is negative five. And our variable is 𝑧 squared. And we see here that both methods will yield the same result.

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