哪两个数之和为 19，差为 5？

引入了一个系统来定义从负无穷到正无穷的数字。该系统被称为数字系统。数系很容易在数轴上表示，整数、整数、自然数都可以在数轴上定义。数轴包含正数、负数和零。

方程是一种数学语句，它用“=”符号连接两个相等值的代数表达式。例如：在等式 3x+2 = 5 中，3x+ 2 是左侧表达式，5 是与“=”符号连接的右侧表达式。

主要有3种方程：

• 线性方程
• 二次方程
• 多项式方程

这里给出了两个由两个变量组成的方程，要找到数字，我们必须将这两个方程相加。
因为，一个方程包含正值，第二个方程包含正值和负值，通过将两个方程相加，一个变量被取消，我们得到另一个变量的值。
然后使用第二个变量的值，我们可以通过使用任何给定的方程来获取第一个变量的值。

因此，无法求解由两个变量组成的单个方程。

Just adding few points related to numbers:

• Sum of two odd numbers will always be even.
• Sum of two even numbers will always be even.
• Sum of one odd and one even number will always be odd.

两个数字的和是 19，它们的差是 5。找出这些数字。

解决方案：

Let the both numbers be first and second.

According to the problem statement:
first + second = 19 (Consider this as 1st equation)
first – second = 5  (Consider this as 2nd equation)

Add both equations:

first + second + first – second = 19 + 5
2 * first = 24
first = 24 / 2
first = 12

So from this we get first = 12, put this value in any equation i.e.

first + second = 19 (Put the value of first in this equation)
12+ second = 19
second = 19 – 12
second = 7

So, the numbers are 12 and 7.

If we consider the case i.e. second – first = 5, then the solution will be same and the first number will become 7 and second number will become 12.

类似问题

问题1：两个数之和是50，两个数之差是30。任务是找出这些数。

解决方案：

Let the both numbers be first and second.

According to the problem statement:
first + second = 50 (Consider this as 1st equation)
first – second = 30  (Consider this as 2nd equation)

Add both equations:

first + second + first – second = 50 + 30
2 * first = 80
first = 80 / 2
first = 40

So from this we get first = 40, put this value in any equation i.e.

first + second = 50 (Put the value of first in this equation)
40+ second = 50
second = 50 – 40
second = 10

So, the numbers are 40 and 10.

If we consider the case i.e. second – first = 30, then the solution will be same and the first number will become 10 and second number will become 40.

问题2：两个数之和是33，两个数之差是25。任务是找到这两个数。

解决方案：

Let both numbers be first and second.

According to the problem statement:
first + second = 33 (Consider this as 1st equation)
first – second = 25  (Consider this as 2nd equation)

Add both equations:

first + second + first – second = 33 + 25
2 * first = 58
first = 58 / 2
first = 29

So from this we get first = 29, put this value in any equation i.e.

first + second = 33 (Put the value of first in this equation)
29+ second = 33
second = 33 – 29
second = 4

So, the numbers are 29 and 4.

If we consider the case i.e. second – first = 25, then the solution will be same and the first number will become 4 and second number will become 29.