负数减法的规则是什么？

代数是处理算术运算及其相关符号的数学分支。这些符号被称为变量，当受到不同约束时可能取不同的值。变量大多表示为 x、y、z、p 或 q，可以通过加法、减法、乘法和除法的不同算术运算进行操作，以计算值。

负数

负数由前面加减号的整数表示。例如，-4、-2 是负数。负数位于数轴的左侧，与正数相隔0。可以说负数是正数的补数。通过使用两个负操作数，可以轻松地添加或减去负数。让我们学习如何在适当的情况下专门减去负数，

负数减法的规则是什么？

解决方案：

Rule 1: Subtracting a negative number from a negative number (-) a minus sign followed by a negative sign, turns the two signs into a plus sign.

Subtraction of a negative number from another negative number is simply an addition of negative and positive numbers. This is because, according to the known rule, – (-4) becomes +4. The resultant operation becomes positive in nature. The final operation may be positive or negative in nature. However, the magnitude of the final output is greater than both the operands, in case none of the operands is 0. In the case of subtracting negative numbers, the following scenarios may arise where we are subtracting the second operand from the first operand: 

• Second operand > First operand 
  In case the magnitude of the second operand is greater than the first operand, the final output has a positive sign associated with it. For example, we have, -2 – (-4). This equation is equivalent to -2 + 4, which boils down to the addition of 4 to -2. On the number line, it starts at -2.

Then we move forward with 4 units: +4.

The answer is -2 – (-4) = 2.

• Second operand < First operand 
  In case the magnitude of the second operand is greater than the first operand, the final output has a negative sign associated with it. For example, we have, -4 – (-2). This equation is equivalent to -4 + 2, which boils down to the addition of 2 to -4. On the number line, it starts at -4. On addition of 2, the result becomes -2.
• Second operand = First operand
  In case the magnitude of second operand is equal to the first operand, the final output is 0. For example, we have, -2 – (-2). This equation is equivalent to -2 + 2, which boils down to the addition of 2 to -2 and produces 0.

示例问题

问题 1：评估 -4 – (-10) – 2 – (-25)。

解决方案：

 -4 – (-10) – 2 – (-25)

• First open the brackets.

= -4 + 10 – 2 + 25 

• Add the positive and negative integers separately.

= -4 – 2 + 10 + 25

= -6 + 35

= 29

问题 2：求解： (2 × 2) – (3 × 3) – (4 × 4)

解决方案：

(2 × 2) – (3 × 3) – (4 × 4)

• First solve the brackets.

= (4) – (9) – (16)

• Now open the brackets.

= 4 – 9 – 16

• Add the positive and negative integers separately.

= 4 – 25

= -21

问题 3：从 (4x – 5y) ^{2 中减去 (2x + 3y) 2 。}

解决方案：

(4x – 5y)² – (2x + 3y)²

• Solve the brackets.

Using algebraic identity,

(x + y)² = x² + y² + 2xy

= (16x² + 25y² – 40xy) – (4x² + 9y² + 12xy)

• Now open the brackets

= 16x² + 25y² – 40xy – 4x² – 9y² – 12xy

• Now add or subtract the like terms

= 16x² – 4x² + 25y² – 9y² – 40xy – 12xy 

= 12x² + 16y² – 52xy

问题 4：从 2x ² – 4y ² – 12xy 中减去 (6x – 8y) ²

解决方案：

2x² – 4y² – 12xy – (6x – 8y)² 

• Solve the bracket.

Using algebraic identity,

(x + y)² = x² + y² + 2xy

= 2x² – 4y² – 12xy – (36x² + 64y² – 96xy)

• Open the bracket.

= 2x² – 4y² – 12xy – 36x² – 64y² + 96xy

• Add or subtract like terms.

= 2x² – 36x² – 4y² – 64y² – 12xy + 96xy

= -34x² – 68y² + 84xy