How to Solve Rational Equations

How to Solve Rational Equations

An equation that consists of at least one Rational expression is a Rational equation, and in this article, we will teach you how to solve this type of equation using two methods.

Related Topics

A step-by-step guide to solve Rational Equations

For solving rational equations, we can use following methods:

  • Converting to a common denominator: In this method, you need to get a common denominator for both sides of the equation. Then, make numerators equal and solve for the variable.
  • Cross-multiplying: This method is useful when there is only one fraction on each side of the equation. Simply multiply the first numerator by the second denominator and make the result equal to the product of the second numerator and the first denominator.

Examples

Rational Equations – Example 1:

Solve. \(\frac{x – 2}{x + 1 }=\frac{x + 4}{x – 2}\)

Solution:

Use cross multiply method: if \(\frac{a}{b}=\frac{c}{d}\), then: \(a×d=b×c \)
\(\frac{x – 2}{x + 1 }=\frac{x + 4}{x – 2}→(x-2)(x-2)=(x+4)(x+1)\)
Expand: \((x-2)^2=x^2-4x+4\) and \((x+4)(x+1)=x^2+5x+4\), Then:
\( x^2-4x+4=x^2+5x+4\), Now, simplify: \(x^2-4x=x^2+5x\), subtract both sides \((x^2+5x)\), Then: \(x^2-4x-(x^2+5x)=x^2+5x-(x^2+5x)→ -9x=0→x=0\)

Rational Equations – Example 2:

Solve. \(\frac{x – 3}{x + 1 }=\frac{x + 5}{x – 2}\)

Solution:

Use cross multiply method: if \(\frac{a}{b}=\frac{c}{d}\), then: \(a×d=b×c\)
Then: \((x-3)(x-2)=(x+5)(x+1)\)
Expand: \((x – 3)(x-2)=x^2-5x+6\)
Expand: \((x+5)(x+1)=x^2+6x+5\), Then: \(x^2-5x+6=x^2+6x+5\), Simplify: \(x^2-5x=x^2+6x-1\)
Subtract both sides \(x^2+6x ,Then: -11x=-1→x=\frac{1}{11}\)

Rational Equations – Example 3:

Solve. \(\frac{x +3}{x + 6 }=\frac{x + 2}{x – 4}\)

Solution:

Use cross multiply method: if \(\frac{a}{b}=\frac{c}{d}\), then: \(a×d=b×c \)
\(\frac{x+3}{x +6 }=\frac{x + 2}{x – 4}→(x+3)(x-4)=(x+2)(x+6)\)
Expand: \((x + 3)(x-4)=x^2-x-12\)
Expand: \((x+2)(x+6)=x^2+8x+12\), Then: \(x^2-x-12=x^2+8x+12\), Simplify: \(x^2-x=x^2+8x+24\)
Subtract both sides \(x^2+8x ,Then: -9x=24→x=-\frac{24}{9}=-\frac{8}{3}\)

Rational Equations – Example 4:

Solve. \(\frac{x +5}{x + 2 }=\frac{x -5}{x +3}\)

Solution:

Use cross multiply method: if \(\frac{a}{b}=\frac{c}{d}\), then: \(a×d=b×c \)
\(\frac{x+5}{x +2 }=\frac{x -5}{x+3}→(x+5)(x+3)=(x-5)(x+2)\)
Expand: \((x + 5)(x+3)=x^2+8x+15\)
Expand: \((x-5)(x+2)=x^2-3x-10\), Then: \(x^2+8x+15=x^2-3x-10\), Simplify: \(x^2+8x=x^2-3x-25\)
Subtract both sides \(x^2-3x ,Then: 11x=-25→x=-\frac{25}{11}\)

Exercises for Rational Equations

Solve Rational Equations.

  1. \(\color{blue}{\frac{10}{x+4}=\frac{15}{4x+4}}\)
  2. \(\color{blue}{\frac{x+4}{x+1}=\frac{x-6}{x-1}}\)
  3. \(\color{blue}{\frac{2x}{x+3}=\frac{x-6}{x+4}}\)
  4. \(\color{blue}{\frac{1}{x+5}-1=\frac{1}{1+x}}\)
  5. \(\color{blue}{\frac{1}{5x^2}-\frac{1}{x}=\frac{2}{x}}\)
  6. \(\color{blue}{\frac{2x}{2x-2}-\frac{2}{x}=\frac{1}{x-1}}\)
This image has an empty alt attribute; its file name is answers.png
  1. \(\color{blue}{x=\frac{4}{5}}\)
  2. \(\color{blue}{x=-\frac{1}{4}}\)
  3. \(\color{blue}{x=-9}\) or \(\color{blue}{x=-2}\)
  4. \(\color{blue}{x=-3}\)
  5. \(\color{blue}{x=\frac{1}{15}}\)
  6. \(\color{blue}{x=2}\)

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