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Solve for the variable and check. Each solution is an integer.$(3 a+7)-(a-1)=14$
$a=3$
Algebra
Chapter 1
THE INTEGERS
Section 5
Multiplying Polynomials
The Integers
Equations and Inequalities
Polynomials
Campbell University
Baylor University
University of Michigan - Ann Arbor
Idaho State University
Lectures
01:32
In mathematics, the absolu…
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02:04
Solve for the variable and…
05:24
Solve each equation. Check…
01:26
Solve each equation. $$a+7…
01:22
In $3-14,$ write the solut…
00:22
Find the solution set for …
00:41
Solve each equation and ch…
01:29
Solve each equation, and c…
02:05
welcome. We're going to be solving for a in this equation. And so the question we're giving us three a minus seven minus a minus one is 0 to 14. Our first step is going to be to distribute this negative sign. So we get 3897 minus a negative, and a negative makes a positive. So get a plus one is equal to 14 will combine like firms. So three a minus a is to a and negative seven plus one is negative. Six is equal to 14. Add six to both sides to get to a is equal to 20 and divide both sides by two. And so to a divided by two is a and 20 divided by two is 10. So our answer is A is equal to 10. But there's more of that this problem because it wants us to check our work. And so the way we could check our work is by substituting this value of a 0 to 10 right back into this equation on dhe. Um, and if we substituted into this equation and get off the value of 14 we know that is indeed equal to 10. Otherwise, we know we've made an error somewhere, so let's substitute this. So three times 10 minus seven, minus 10 minus one. And so this is pretty much just where they used to be. And now there's 10 because that's the value we found for it. We'll evaluate this further, to get 30 minus seven minus 10 minus one, which is the same thing as 23 minus nine. Uh, 23 minus nine is indeed 14 and 14 is also the value that we were actually looking for. And so now we know that a is indeed equal 10 and we've checked our work.
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