Solve ##\dfrac{3x-6}{5-x}+\dfrac{11-2x}{10-4x}=3\dfrac{1}{2}##

  • Thread starter RChristenk
  • Start date
  • Tags
    Precalculus
In summary, the equation \(\dfrac{3x-6}{5-x}+\dfrac{11-2x}{10-4x}=3\dfrac{1}{2}\) can be solved by first rewriting \(3\dfrac{1}{2}\) as \(\dfrac{7}{2}\), then finding a common denominator for the fractions on the left side, simplifying the expression, and solving for \(x\). The solution involves isolating \(x\) and checking for any extraneous solutions that may arise from the denominators.
  • #1
RChristenk
64
9
Homework Statement
Solve ##\dfrac{3x-6}{5-x}+\dfrac{11-2x}{10-4x}=3\dfrac{1}{2}##
Relevant Equations
Algebraic manipulation
I've multiplied everything out on paper and got ##x=2, \dfrac{15}{4}##, which is correct. However multiplying directly is tedious and from observing this problem I suspect there is a simplification or trick that I missed.

##3\cdot\dfrac{x-2}{5-x}+\dfrac{1}{2}\cdot\dfrac{11-2x}{5-2x}=\dfrac{7}{2}##

Multiply both sides by ##2##:

##\dfrac{6(x-2)}{5-x}+\dfrac{11-2x}{5-2x}=7##

And then I'm stuck. But the denominators ##5-x, 5-2x## are tantalizingly close to each other, but I just can't figure out how to simplify/substitute/manipulate it to process this problem. Besides just multiplying it out of course.
 
Physics news on Phys.org
  • #2
I don't see a way to simplify things. The best you could do is:
$$6(x-2)(5-2x) + (11-2x)(5-x) = 7(5-x)(5-2x)$$$$6(x-2)(5-2x) + (5-x)(11-2x - 35 + 14x) = 0$$$$6(x-2)(5-2x) + (5-x)(12x-24) = 0$$$$6(x-2)(5-2x) +6(x-2)(10 - 2x) = 0$$$$6(x-2)(15 - 4x) = 0$$
 
  • Like
Likes MatinSAR, chwala, SammyS and 2 others
  • #3
[tex]
-3-\frac{9}{x-5}+\frac{1}{2}-\frac{3}{2x-5}=3+\frac{1}{2}[/tex]
[tex]\frac{3}{x-5}+\frac{1}{2x-5}=-2[/tex]
[tex]4x^2-23x+30=0[/tex]
[tex](x-2)(4x-15)=0[/tex]
 
Last edited:
  • Like
Likes MatinSAR, chwala, RChristenk and 1 other person
  • #4
RChristenk said:
Homework Statement: Solve ##\dfrac{3x-6}{5-x}+\dfrac{11-2x}{10-4x}=3\dfrac{1}{2}##
Relevant Equations: Algebraic manipulation

I've multiplied everything out on paper and got ##x=2, \dfrac{15}{4}##, which is correct. However multiplying directly is tedious and from observing this problem I suspect there is a simplification or trick that I missed.

##3\cdot\dfrac{x-2}{5-x}+\dfrac{1}{2}\cdot\dfrac{11-2x}{5-2x}=\dfrac{7}{2}##

Multiply both sides by ##2##:

##\dfrac{6(x-2)}{5-x}+\dfrac{11-2x}{5-2x}=7##

And then I'm stuck. But the denominators ##5-x, 5-2x## are tantalizingly close to each other, but I just can't figure out how to simplify/substitute/manipulate it to process this problem. Besides just multiplying it out of course.
Here is my two cents.
I much prefer methodical approaches that use basic principles to reliably solve the vast majority of problems. I don't really care if they require a little more work. If you look for cute and clever tricks, you might learn a million of them and still not have a basic understanding.
 
  • Like
Likes MatinSAR, chwala, docnet and 2 others

FAQ: Solve ##\dfrac{3x-6}{5-x}+\dfrac{11-2x}{10-4x}=3\dfrac{1}{2}##

What are the first steps to solve the equation?

The first steps involve simplifying the equation. Convert the mixed number on the right-hand side to an improper fraction, so \(3 \frac{1}{2}\) becomes \(\frac{7}{2}\). Then, find a common denominator for the fractions on the left-hand side.

How do you find a common denominator for the fractions?

The common denominator for the fractions \(\dfrac{3x-6}{5-x}\) and \(\dfrac{11-2x}{10-4x}\) can be found by factoring and simplifying. Notice that \(10-4x = 2(5-x)\), so the common denominator is \(2(5-x)\).

What do you do after finding the common denominator?

Rewrite each fraction with the common denominator and combine them. This involves multiplying the numerator and denominator of each fraction by the necessary factors to achieve the common denominator, and then adding the fractions together.

How do you solve the equation after combining the fractions?

Once the fractions are combined, set the resulting equation equal to \(\frac{7}{2}\). Cross-multiply to eliminate the denominator, resulting in a polynomial equation that can be simplified and solved for \(x\).

What is the final solution for \(x\)?

After solving the polynomial equation, you find the value of \(x\). Ensure to check that this solution does not make any denominators zero, as that would make the original equation undefined. The final solution is \(x = -1\).

Back
Top