Solving An Equation With Rational Terms

In summary, the conversation is about solving an equation involving fractions and finding the LCD. The LCD is determined to be 21x and the process of multiplying both sides of the equation by 21x is discussed. The final step is solving for x by distributing, collecting like terms, and simplifying. The solution is x = 0.3956.
  • #1
Simon T
16
0
Hey guys, I am not to good in Math and I am having issues solving this equation.

2x-9/3x + 8 = 3/7x How do you solve this equation? The answer is supposed to be x = 0.3956?
 
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  • #2
Re: equation help

Okay, what you mean is:

(2x - 9)/(3x) + 8 = 3/(7x)

or, using $\LaTeX$:

\(\displaystyle \frac{2x-9}{3x}+8=\frac{3}{7x}\)

The first thing we could do is multiply through by the LCD...can you state what the LCD is in this problem?
 
  • #3
Re: equation help

it doesn't say what the LCD is, we have to figure that out. and yes, the second part of your post is correct
 
  • #4
Re: equation help

cyrus23 said:
it doesn't say what the LCD is, we have to figure that out. and yes, the second part of your post is correct

Yes, we need to come up with that on our own...can you find it? :D
 
  • #5
Re: equation help

would it be 21?
 
  • #6
Re: equation help

cyrus23 said:
would it be 21?

It would be 21 if the denominators were 3 and 7, but there is also $x$ as a factor in both, so the LCD is:

$21x$

So, what do you get when you multiply through by $21x$?
 
  • #7
Re: equation help

do you multiply both dominators by 21?
 
  • #8
Re: equation help

cyrus23 said:
do you multiply both dominators by 21?

No, you multiply everything by $21x$:

\(\displaystyle \frac{2x-9}{3x}\cdot21x+8\cdot21x=\frac{3}{7x}\cdot21x\)

After dividing out common factors, we then have:

\(\displaystyle (2x-9)7+8\cdot21x=3\cdot3\)

Now it is a matter of distributing, collecting like terms, and solving for $x$...what do you get?
 
  • #9
Re: equation help

where did you get 7 + 8 from?
 
  • #10
Re: equation help

cyrus23 said:
where did you get 7 + 8 from?

\(\displaystyle \frac{2x-9}{3x}\cdot21x+8\cdot21x=\frac{3}{7x}\cdot21x\)

\(\displaystyle \frac{2x-9}{3x}(21x)+8\cdot21x=\frac{3}{7x}(21x)\)

\(\displaystyle \frac{2x-9}{\cancel{3x}}(\cancel{3x}\cdot7)+8\cdot21x=\frac{3}{\cancel{7x}}(\cancel{7x}\cdot3)\)

\(\displaystyle (2x-9)(7)+8\cdot21x=3(3)\)

\(\displaystyle 7(2x-9)+8\cdot21x=3\cdot3\)
 
  • #11
Re: equation help

where did you get the 7? I am very confused :/
 
  • #12
Re: equation help

cyrus23 said:
where did you get the 7? I am very confused :/

\(\displaystyle \frac{2x-9}{3x}\cdot21x=(2x-9)\frac{21x}{3x}=(2x-9)\frac{3x\cdot7}{3x}=(2x-9)\frac{\cancel{3x}\cdot7}{\cancel{3x}}=(2x-9)7=7(2x-9)\)
 

FAQ: Solving An Equation With Rational Terms

What is an equation with rational terms?

An equation with rational terms is an equation where all the coefficients and constants are rational numbers. Rational numbers are numbers that can be expressed as a ratio of two integers, such as 1/2, -3/4, or 5/6.

How do you solve an equation with rational terms?

To solve an equation with rational terms, you can follow the same steps as you would for any other equation. First, combine like terms on both sides of the equation. Then, isolate the variable by using inverse operations. Finally, check your solution by plugging it back into the original equation.

What are the common methods for solving equations with rational terms?

The most common methods for solving equations with rational terms are the addition and subtraction method, the multiplication and division method, and the substitution method. These methods involve manipulating the equation using inverse operations to isolate the variable.

Can you solve equations with rational terms using the quadratic formula?

Yes, you can solve equations with rational terms using the quadratic formula. However, the polynomial must be in standard form, with the highest degree term having a coefficient of 1. If the polynomial is not in standard form, you will need to rearrange the equation before applying the quadratic formula.

Are there any special considerations when solving equations with rational terms?

One special consideration when solving equations with rational terms is to be aware of any restrictions on the variable. Since rational numbers cannot have a denominator of zero, any solution that would make the denominator zero must be excluded from the solution set.

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