When two values of a quantity are compared to each other, it is a ratio of values. For example, comparing the weight of loads as 150 kg and 50 kg, then the ratio will be [br][br][math]\dfrac{150kg}{50kg}=3.[/math][br][br]In this case, it is said that load 1 is 3 times heavier than load 2. [br][br][color=#0000ff]Proportions are equations, both sides of which are ratios[/color]. Familiar examples of using a comparison are scale (drawing, map, etc.) and roof slope.

Consider a cooking situation, where the recipe has the following instructions for us: for 2,5 cups of flour, add 2 tablespoons of oil. If we added 10 cups of flour, how much oil would we need? [br][br]This problem is a typical example of a [b]proportion equation[/b]. In the equation, we have two quantities, where the [b]proportion[/b] (ratio) of the quantities is known. If the amount of one of the quantities is known, we can solve the amount of the other quantity. The unknown quantity is denoted by [i]x[/i] and it is solved from the equation. [br][br]You can start by making a table:[br][br][table][tr][td][b]flour[/b][/td][td][b]oil[/b][/td][td][/td][/tr][tr][td]2,5 cups [/td][td]2 Tbsp[/td][td][/td][/tr][tr][td]10 cups[/td][td]x[/td][td][/td][/tr][/table][br]Proportion is easily done by writing the columns as fractions:[br][br][math]\large \frac{\text{2,5 cups flour}}{\text{10 cups flour}}=\frac{\text{2 Tbsp oil}}{x}[/math]

Simple proportion equations have an algorithm that can be used for [br]obtaining the solution of the equation:[br][br]1.[color=#0000ff] Cross multiplication[/color]: multiply the numerator of the first ratio with [br]the denominator of the second ratio, and also multiply the denominator [br]of the first ratio with the numerator of the second ratio. Both products[br] yield the same value, creating a new equation that is equivalent to the[br] original equation.[br][br][math]\large \text{2,5 cups flour}\cdot \text{x}=\text{10 cups flour}\cdot \text{2 Tbsp oil}[/math][br][br]2. [color=#0000ff]Divide[/color] the equation by the coefficient of the unknown variable.[br][br][math]\large x =\frac{\text{10 \cancel{cups\; flour}}\cdot \text{2 Tbsp oil}}{\text{2,5 \cancel{cups \;flour}}}\vspace{10mm}\\[br]\large x[br]=\frac{\cancel {10}^{\Large 4} \cdot \text{2 Tbsp oil}}{\cancel{2,5}}=4\cdot \text{2 Tbsp oil}=\text{8 Tbsp oil} [/math]