Confused about how to do this equation

[sharpsbc]

Find the value(s) of a for which the equation is an identity.
10x-35+3ax = 5ax-7a

From the Lesson, the book focuses a bit on equations that "are an identity" thus the solution for such equations is all real numbers. An eg of such would be
2x+10=2(x+5) which becomes 2x+10=2x+10. This statement is true for all values of x. So, the equation is an identity, and the solution is all real numbers.

With that background, I cannot figure out how to isolate a or x so that I can figure this out.
Here is my work:
10x-2ax+7a=35
10x-a(2x-7)=35

10x/2x-7 - a = 35/2x-7
Now what do I do? Or have I gone in the wrong direction?

 

[Deleted member 4993]

Find the value(s) of a for which the equation is an identity.
10x-35+3ax = 5ax-7a

From the Lesson, the book focuses a bit on equations that "are an identity" thus the solution for such equations is all real numbers. An eg of such would be
2x+10=2(x+5) which becomes 2x+10=2x+10. This statement is true for all values of x. So, the equation is an identity, and the solution is all real numbers.

With that background, I cannot figure out how to isolate a or x so that I can figure this out.
Here is my work:
10x-2ax+7a=35

10x-a(2x-7)=35

10x/2x-7 - a = 35/2x-7
Now what do I do? Or have I gone in the wrong direction?

Here you have to Isolate 'a'

10x-2ax+7a=35

10x - 35 = 2ax - 7a

5 * (2x - 7) = a * (2x - 7)

Now continue....

 

[HallsofIvy]

I would NOT "isolate" a. Instead try to solve for x ("isolate" x). Since you want this to be an identity, true for all x, you should NOT be able to solve for a specific x! You should be able to get to, say, mx= n for number m and n (depending on a).

If \(\displaystyle a\ne 0\), you can divide both sides by m to get x= n/m as the only x for which this is true. But we don't want that so we must have m= 0. Of course, if [itex]n\ne 0[/tex] then 0x= n is not true for any x. So we must also have n= 0- in that case, the equation is 0x= 0 which is true for all x. Is there a value of a that makes those both equal to 0.

 

[JeffM]

Find the value(s) of a for which the equation is an identity.
10x-35+3ax = 5ax-7a

From the Lesson, the book focuses a bit on equations that "are an identity" thus the solution for such equations is all real numbers. An eg of such would be
2x+10=2(x+5) which becomes 2x+10=2x+10. This statement is true for all values of x. So, the equation is an identity, and the solution is all real numbers.

With that background, I cannot figure out how to isolate a or x so that I can figure this out.
Here is my work:
10x-2ax+7a=35
10x-a(2x-7)=35

10x/2x-7 - a = 35/2x-7
Now what do I do? Or have I gone in the wrong direction?

In my opinion, the easiest way to do this kind of problem is to take advantage of the fact that an identity is true for any x. Zero looks as though it would be a nice number to work with on this example.

\(\displaystyle 10(0) - 35 + 3a(0) = 5a(0) - 7a \implies - 35 = - 7a \implies a = 5.\)

Let's check

\(\displaystyle 10x - 35 + (3)(5)x = (5)(5)x - 7(5) \implies 10x - 35 + 15x = 25x - 35 \implies 25x - 35 = 25x - 35,\) which sure looks like an identity to me.

Edit: Of course, 0 was a guess intended to minimize work because I am lazy. If the relationship represents an identity, any choice would have worked but the arithmetic might have been a bit messier.

\(\displaystyle 10(100) - 35 - 3a(100) = 5a(100) - 7a \implies 1000 - 35 + 300a = 500a - 7a \implies 965 = 500a - 300a - 7a = 193a \implies a = 5.\)