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For example, in the simple equation 3 + 2y = 8y, both sides actually contain 2y (because 8y is the same as 2y + 6y). Therefore, the 2y on both sides can be cancelled out, leaving 3 = 6y, or y = 0.5. This is equivalent to subtracting 2y from both sides. At times, cancelling out can introduce limited changes or extra solutions to an equation.
In the diagram, the shaded part represents the difference between the areas of the two squares, i.e. . The area of the shaded part can be found by adding the areas of the two rectangles; () + (), which can be factorized to (+) ().
The plus–minus sign, ±, is used as a shorthand notation for two expressions written as one, representing one expression with a plus sign, the other with a minus sign. For example, y = x ± 1 represents the two equations y = x + 1 and y = x − 1. Sometimes, it is used for denoting a positive-or-negative term such as ±x.
To solve this kind of equation, the technique is add, subtract, multiply, or divide both sides of the equation by the same number in order to isolate the variable on one side of the equation. Once the variable is isolated, the other side of the equation is the value of the variable. [ 37 ]
However, if one searches for real solutions, there are two solutions, √ 2 and – √ 2; in other words, the solution set is {√ 2, − √ 2}. When an equation contains several unknowns, and when one has several equations with more unknowns than equations, the solution set is often infinite. In this case, the solutions cannot be listed.
[2] In a sense, subtraction is the inverse of addition. That is, c = a − b if and only if c + b = a. In words: the difference of two numbers is the number that gives the first one when added to the second one. Subtraction follows several important patterns. It is anticommutative, meaning that changing the order changes the sign of the answer.
In the first equation, solve for one of the variables in terms of the others. Substitute this expression into the remaining equations. This yields a system of equations with one fewer equation and unknown. Repeat steps 1 and 2 until the system is reduced to a single linear equation.
Divide the highest term of the remainder by the highest term of the divisor (x 2 ÷ x = x). Place the result (+x) below the bar. x 2 has been divided leaving no remainder, and can therefore be marked as used. The result x is then multiplied by the second term in the divisor −3 = −3x. Determine the partial remainder by subtracting 0x − ...