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In the following exercises, determine whether each given value is a solution to the equation.$20 h-5=15 h+35$(a) $h=6 \quad$ ( b $h=8$

a. Nob. Yes

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Chapter 2

The Language of Algebra

Section 3

Solving Equations Using the Subtraction and Addition Properties of Equality

01:14

In the following exercises…

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00:27

01:46

Solve each equation and ch…

00:54

00:52

Solve each equation. Then …

00:42

Solve each equation, and c…

00:41

00:23

Either solve the given equ…

Yeah. The next problem of 20 H minus five goes 15 h plus 35. A couple possible answers. We need to check here, but most are minute substitute. Every time I see h, I'm gonna put either six or eight, depending on when I'm looking at. So for a who now 20 time six minus five equals 15 time six plus 35. And therefore be we could have 20 times eight minus five equals 15 times a plus 35. And then we need to check both of these because we don't actually know if these air equal yet. So we do the math. So for a 20 times, six is 1 20 minus five is 100 and 15 and then on the right side, 15 times six is 90 plus 35 is 1 25 So these are not equal. So it's a no for a ah never be 20 times eight is 1 60 minus five is 1 55 up 15 times AIDS would be 90. No, not 90. Let's see about 15 times eight zero for this 1 20 Sorry. So 1 20 votes 35 is 1 55 That's right, cause if faes wrong be has to be right, So say yes, Herbie. And there you go

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