6th 7th 8th Grade Algebra | MathHelp.com - Free Educational videos for Students in K-12 | Lumos Learning

6th 7th 8th Grade Algebra | MathHelp.com - Free Educational videos for Students in k-12


6th 7th 8th Grade Algebra | MathHelp.com - By MathHelp.com



Transcript
00:0-1 in this problem we are asked to solve the following
00:02 equation and check your solution To a plus five equals
00:07 17 . To solve this equation for a , we
00:11 must first isolate the term containing A . Which in
00:15 this case is to A . Since five is being
00:18 added to to A . We need to subtract five
00:21 from both sides of the equation . Yeah , On
00:25 the left side of the equation , the positive five
00:28 and -5 cancel each other out . And we have
00:31 to a . And on the right side of the
00:34 equation 17 -5 is 12 . So we have to
00:38 a equals 12 . Now to get a by itself
00:42 , since A is being multiplied by two , we
00:45 need to divide by two on both sides of the
00:48 equation , On the left side of the equation ,
00:51 the two's cancel . And we have a . And
00:54 on the right side of the equation , 12 divided
00:57 by two is six , So a equals six which
01:01 is the solution to our equation . To check our
01:04 solution , we plug six back in for a in
01:07 the original equation . So we have two times six
01:12 plus five equals 17 , two times six is 12
01:18 , and we have 12 plus five equals 17 or
01:22 17 equals 17 , which is a true statement ,
01:27 so our answer checks .
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