By Jeffrey Bergen, Susan Montgomery

This impressive reference covers subject matters akin to quantum teams, Hopf Galois concept, activities and coactions of Hopf algebras, damage and crossed items, and the constitution of cosemisimple Hopf algebras.

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**Example text**

Subtract 3x from both sides of the equation. Simplify both sides of the equation. Your common sense tells you that 0 will never equal 15. This means there is no value of the variable that will make the equation true because 0 will never equal 15. Since there is no value of x that will ever make the equation true, there is no solution. When there is no solution, it is called an empty set. This notation л is used for the empty set. 5x + 3 = 5(x – 1) + 8 5x + 3 = 5x – 5 + 8 5x + 3 = 5x + 3 Case 2 Use the distributive property.

In this example, you can see how the parentheses are used to indicate multiplication. 2(x + y) + 3(x – y) Use the distributive property. 2x + 2y + 3x – 3y Combine like terms. = 5x – y Example: 2(2x + y) – 3(x + 2y) Use the distributive property to get rid of the parentheses. The subtraction sign in front of the 3 is the same as multiplying (–3)(x) and (–3)(2y). Use the distributive property. 2(2x + y) – 3(x + 2y) Combine like terms. 4x + 2y – 3x – 6y = x – 4y Practice Use the distributive property to simplify the expressions.

What do you need to do to get the variable on a side by itself? You need to get rid of the 5. The equation says to subtract 5, so what undoes subtraction? If you said addition, you are right! In this problem, you need to add 5 to both sides of the equation to get rid of the 5. Example: x – 5 = 9 Add 5 to both sides of the equation. Simplify both sides of the equation. Add 0 to x. x–5+5=9+5 x + 0 = 14 x = 14 Example: a + 6 = 7 Subtract 6 from both sides of the equation. Simplify both sides of the equation.