哪两个数的和为20，差为6？

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。数字系统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 是左边的表达式，5 是右边的表达式，用 '=' 符号连接。

主要有3种方程：

• 线性方程
• 二次方程
• 多项式方程

在这里，我们将研究线性方程组。一个变量的线性方程是写成 ax + b = 0 的方程，其中 a 和 b 是两个整数，x 是一个变量，并且只有一个解。例如，3x+2=5 是一个只有一个变量的线性方程。因此，这个方程只有一个解，即 x = 1。另一方面，一个有两个变量的线性方程有两个解。

A one-variable linear equation is one with a maximum of one variable of order one. The formula is ax + b = 0, using x as the variable.

这个方程只有一个解。这里有一些例子：

• 4x = 8
• 5x + 10 = -20
• 1 + 6x = 11

一个变量中的线性方程以标准形式写成：

ax + b = 0

Here,

• The numbers ‘a’ and ‘b’ are real.
• Neither ‘a’ nor ‘b’ are equal to zero.

求解一个变量中的线性方程

求解只有一个变量的方程的步骤如下：

步骤 1：如果有任何分数，请使用 LCM 将其删除。

第 2 步：等式两边都应该简化。

第 3 步：从方程中删除变量。

第 4 步：确保您的回答是正确的。

哪两个数的和为20，差为6？

解决方案：

Let both numbers be first and second.

According to the problem statement:
first + second = 20 (Consider this as 1st equation)
first – second = 6  (Consider this as 2nd equation)

Add both equations:

first + second + first – second = 20 + 6
2 * first = 26
first = 26 / 2
first = 13

So from this we get first = 13, put this value in any equation i.e.

first + second = 20 (Put the value of first in this equation)
13 + second = 20
second = 20 – 13
second = 7

So, the numbers are 13 and 7.

If we consider the case i.e. second – first = 6 then the solution will be same and the first number will become 7 and second number will become 13.
 

类似问题

问题1：三个数之和是30，这三个数中前两个数之和是19。任务是找到第三个数。

解决方案：

Let the numbers be first, second and third.

According to the problem statement:
first + second + third = 30 (Consider this as 1st equation)
first + second = 19  (Consider this as 2nd equation)

So, put the value of 2nd equation in 1st equation i.e.

first + second +third = 30 (Put the value of first+second in this equation)
19 + third = 30
third = 30-19
third = 11

So, the third number is 11.

问题2：哪两个数的和为34，差为6？

解决方案：

Let both numbers be first and second.According to the problem statement:
first + second = 34 (Consider this as 1st equation)
first – second = 6  (Consider this as 2nd equation)

Add both equations:

first + second + first – second = 34 + 6
2 * first = 40
first = 40 / 2
first = 20

So, from this we get first = 20, put this value in any equation i.e.

first + second = 34 (Put the value of first in this equation)
20 + second = 34
second = 34 – 20
second = 14

So, the numbers are 20 and 14.

If we consider the case i.e. second – first = 6 then the solution will be same and the first number will become 14 and second number will become 20.