Algebra problem on final step

[Restless]

Homework Statement

8x^2-7a^2=19a^2-5ax

Homework Equations

None

The Attempt at a Solution

8x^2-7a^2=19a^2-5ax
13ax^2-7a^2=19a^2
13ax^2=26a^2
Now I'm not sure what I should do next...
Maybe divide both sides by 2 to get rid of the squares
13ax=26a
Then say that ax = 13a ?
I am confused.

 

[jedishrfu]

Homework Statement

8x^2-7a^2=19a^2-5ax

Homework Equations

None

The Attempt at a Solution

8x^2-7a^2=19a^2-5ax
13ax^2-7a^2=19a^2
13ax^2=26a^2
Now I'm not sure what I should do next...
Maybe divide both sides by 2 to get rid of the squares
13ax=26a
Then say that ax = 13a ?
I am confused.

In your attempt at a solution where did the -5ax term go?

It seems you just dropped it. Oh I see no you can't add the x^2 and x terms together to get 13ax^2.

The strategy should be to group terms so that the right hand side of the equation is a quadratic in x and the left hand side is 0.

then find the roots of the quadratic.

 

[Mark44]

Homework Statement

8x^2-7a^2=19a^2-5ax

Is there something you're supposed to do with this, like, say, solve for x?

Homework Equations

None

The Attempt at a Solution

8x^2-7a^2=19a^2-5ax
13ax^2-7a^2=19a^2

You have a mistake above. Apparently you added 5ax to 8x² and got 13ax². They are not like terms, so can't be combined.

13ax^2=26a^2
Now I'm not sure what I should do next...
Maybe divide both sides by 2 to get rid of the squares

?
Dividing by 2 doesn't get rid of exponents.

13ax=26a
Then say that ax = 13a ?

There's an error above, as well. You divided the left side by 13, but divided the right side by 2. You can't do that.

I am confused.

 

[Restless]

Ok so I thought about the problem again and now I have

8x^2-7a^2=19a^2-5ax

8x^2=26a^2-5ax

But now I don't see any like terms or anything I can reduce or eliminate

 

[Hysteria X]

Ok so I thought about the problem again and now I have

8x^2-7a^2=19a^2-5ax

8x^2=26a^2-5ax

But now I don't see any like terms or anything I can reduce or eliminate

Bring it to the format

and
just use the quadratic root equation

in your case it would be 8 x^2 - 26a^2 + 5ax=0

 

[Restless]

I don't understand how to enter my equation into that formula

 

[Hysteria X]

I don't understand how to enter my equation into that formula

in you equation x=x
a= 8
b= +5a
c= -26a ^ 2

 

[Mark44]

Bring it to the format

and
just use the quadratic root equation

in your case it would be 8 x^2 - 26a^2 + 5ax=0

It would be better as 8x² + 5ax - 26a² = 0

 

[lendav_rott]

Does it tell you to solve for X, it's a bit confusing, since you have left it unmentioned or I am missing something. Anywho, what you cannot do for certain is divide the sides of the equal sign with x or a. If it were x you were solving for, if you divide it like that, then one of your solutions will disappear, the number of the dimension of the variable tells you how many solutions there will have to be.

You can treat both of the numbers a and x as variables here. 1 time, solve for x and then solve again for a, since you haven't said what needs to be done. I have a hunch one needs to solve for x here.