Video: Evaluating Algebraic Expressions by Factoring Out the Highest Common Factor

If 3𝑐 βˆ’ 2𝑏 = 5 and 2π‘Ž + 𝑑 = βˆ’3, what is the value of 6π‘Ž+3𝑑 βˆ’ 3𝑐 + 2𝑏?

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Video Transcript

If three 𝑐 minus two 𝑏 is equal to five and two π‘Ž plus 𝑑 is equal to negative three, what is the value of six π‘Ž plus three 𝑑 minus three 𝑐 plus two 𝑏?

In this question, we’ve been given the value of two algebraic expressions. What we’re absolutely not going to do is individually try and work out the value of π‘Ž, 𝑏, 𝑐, and 𝑑. There’s simply not enough information to do so. Instead, we’ll try to find a way to relate these two expressions to the third expression in our question. That’s six π‘Ž plus three 𝑑 minus three 𝑐 plus two 𝑏.

Now if we look carefully, we might notice that six π‘Ž plus three 𝑑 looks a little bit like the expression two π‘Ž plus 𝑑. And negative three 𝑐 plus two 𝑏 has something in common with the expression three 𝑐 minus two 𝑏. In fact, if we factor three from the expression six π‘Ž plus three 𝑑, we see we can write it as three times two π‘Ž plus 𝑑. But of course we already saw that two π‘Ž plus 𝑑 is equal to negative three. So we can write this as three times negative three, which is simply negative nine.

Now this next step isn’t so obvious. But we’re going to factor negative three 𝑐 plus two 𝑏. That’s the second half of our third expression. We’re going to factor negative one. When we divide negative three 𝑐 by negative one, we get three 𝑐. And when we divide two 𝑏 by negative one, we get negative two 𝑏.

But of course, the question tells us that three 𝑐 minus two 𝑏 is five. So this becomes negative one times five, which is negative five. We’ll rewrite the third expression just slightly. We’ll write it as six π‘Ž plus three 𝑑 plus negative three 𝑐 plus two 𝑏. We’ve just calculated the value of six π‘Ž plus three 𝑑. It’s negative nine. Similarly, we calculated the value of negative three 𝑐 plus two 𝑏. It’s negative five. So six π‘Ž plus three 𝑑 minus three 𝑐 plus two 𝑏 is the same as negative nine plus negative five, which is equal to negative 14.

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