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Ontario Tech acknowledges the lands and people of the Mississaugas of Scugog Island First Nation.

We are thankful to be welcome on these lands in friendship. The lands we are situated on are covered by the Williams Treaties and are the traditional territory of the Mississaugas, a branch of the greater Anishinaabeg Nation, including Algonquin, Ojibway, Odawa and Pottawatomi. These lands remain home to many Indigenous nations and peoples.

We acknowledge this land out of respect for the Indigenous nations who have cared for Turtle Island, also called North America, from before the arrival of settler peoples until this day. Most importantly, we acknowledge that the history of these lands has been tainted by poor treatment and a lack of friendship with the First Nations who call them home.

This history is something we are all affected by because we are all treaty people in Canada. We all have a shared history to reflect on, and each of us is affected by this history in different ways. Our past defines our present, but if we move forward as friends and allies, then it does not have to define our future.

Learn more about Indigenous Education and Cultural Services

Rational Equations

A rational expression is a quotient whose numerator and denominator are polynomials, where the denominator cannot equal zero. For example:

$$\frac{2x^2 +4x -7}{x^2 -3x+8}$$

A rational equation is one that involves only a rational expression. For example:

$$\frac{2x^2 +4x -7}{x^2 -3x+8}=0$$

Solving By Factoring

Factoring is often an important step in solving rational equations. To solve the following rational equation, the numerator must be factored. The solutions of the equation are the solutions that result when the numerator has been set to zero. The denominator is not set equal to zero, because if you remember, one of the conditions of a rational expression is that the denominator cannot equal zero. Once the numerator has been factored, set each of the factors to zero and solve for x. 

Example: Solve for x 



\frac{x^2-5x+6}{x^2+3x+2} & =0 \\
x^2-5x+6 & =0 \\
(x-3)(x-2) & =0 \\
$$\smash{x-3=0 \qquad \text{or} \qquad  x -2 = 0}$$
$$\smash{x=3     \qquad \qquad \qquad        x =2}$$

Therefore, the above rational equation has two solutions, one when x = 3 and the other when x = 2.

A Review of the Lowest Common Denominator

Finding a common denominator is a math technique that is often required to solve rational equations. To find the lowest common denominator of a rational expression, you basically just multiply all of the unique factors in the denominator together. To change each term in the expression to have that denominator, figure out what you multiplied it's denominator by to make it become the lowest common denominator, and then multiply the numerator by that same factor (otherwise, you're changing the expression). For example, if you need to multiply the denominator by (x - 1), then the numerator must also be multiplied by (x -1). Be sure to collect the like terms of the numerator for your final answer. 

Example: Combine  \(\frac{2}{x+4}+\frac{3}{x-1}\) into a single rational expression.


\frac{2}{x+4}+\frac{3}{x-1}& = \frac{2}{x+4} \cdot \frac{x-1}{x-1} + \frac{3}{x-1}\cdot\frac{x+4}{x+4}  \\
& = \frac{2(x-1)+3(x+4)}{(x+4)(x-1)} \\
& = \frac{2x-2+3x+12}{(x+4)(x-1)} \\
& = \frac{5x+10}{(x+4)(x-1)}
\end{align} $$

Solving By Multiplying By a Common Denominator

When solving a rational equation, a common denominator for all terms may have to be found. The terms must then be rearranged through addition and subtraction so that only zero remains on one side of the equation. Once this has been done, the numerator can be set equal to zero and it becomes easy to solve. 

Example: Solve \(\frac{5x}{x+2}=7\) for x.


\frac{5x}{x+2}&=7 \\
\frac{5x}{x+2} &=\frac{7(x+2)}{x+2} \\
\frac{5x}{x+2} - \frac{7(x+2)}{x+2} &=0 \\
\frac{5x-7x-14}{x+2} &= 0 \\
5x-7x-14 &=0 \\
-2x &= 14 \\
x &= -7 

Another way to solve the equation above is to simply multiply both sides of the equation by the common denominator. 

\frac{5x}{x+2}&=7 \\
5x &=7(x+2)\\
5x &= 7x+14 \\
5x-7x &=14 \\
-2x &= 14 \\
x &= -7 


Example (Harder):