How Do You Algebraically Solve for X in a Dividend Formula?

  • Thread starter EdLeafs
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    Algebra
In summary, to solve for x algebraically in the equation $30.85 = X/$1.15 + X/($1.15)^2, multiply both sides by 1.15 and then divide by 2.15. The result should be $18.98 when substituted for x.
  • #1
EdLeafs
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This isn't a homework question...just something I picked up on a finance book about dividends. Say you have something like this.

$30.85 = X/$1.15 + X/($1.15)^2

How do I solve for x step by step? They give me an answer of $18.98 and if I sub that # into x I get 30.85. How do I do it the other way...algebraically?

Thanks.
 
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  • #2
Welcome to PF!

Hi EdLeafs ! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)
EdLeafs said:
$30.85 = X/$1.15 + X/($1.15)^2

Easy … use brackets …

30.85 = X(1/1.15 + 1/1.152) … take it from there. :smile:​
 
  • #3
The first thing I would do is multiply both sides by 1.15 to get rid of the fractions!

(1.15)2(30.85)2= (1.15)X+ X= 2.15X

Now divide both sides by 2.15.

I'm wondering what X represents that dividing by "dollars" results in "dollars"! It has units of "dollars squared"?
 
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  • #4
tiny-tim said:
30.85 = X(1/1.15 + 1/1.152) … take it from there. :smile:​
HallsofIvy said:
The first thing I would do is multiply both sides by 1.15 to get rid of the fractions!

(1.15)(30.85)2= (1.15)X+ X= 2.15X

(erm)2 :redface: …
 
  • #5
Ahh...I see. thanks a lot.
 

FAQ: How Do You Algebraically Solve for X in a Dividend Formula?

How do you solve a simple algebraic equation?

To solve a simple algebraic equation, you need to isolate the variable on one side of the equation and simplify the other side. This can be done by using inverse operations, such as addition, subtraction, multiplication, and division.

What is the order of operations in algebra?

The order of operations in algebra is the same as in regular math: Parentheses, Exponents, Multiplication and Division (from left to right), and Addition and Subtraction (from left to right).

Can you show me an example of solving a simple algebraic equation?

For example, let's solve the equation 3x + 5 = 17. First, we subtract 5 from both sides to isolate the variable: 3x = 12. Then, we divide both sides by 3 to get the value of x: x = 4. Therefore, the solution to this equation is x = 4.

What is the purpose of solving algebraic equations?

The purpose of solving algebraic equations is to find the value of the unknown variable in the equation. This can be useful in real-life situations, such as calculating the cost of an item on sale or determining the amount of time needed to complete a task.

What should I do if I get stuck while solving an algebraic equation?

If you get stuck while solving an algebraic equation, take a step back and review the steps you have already taken. Make sure you have correctly applied the order of operations and check for any errors in your calculations. If you are still struggling, seek help from a teacher or tutor.

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