两个数之和是 7，它们的差是 1。找出这些数

17 世纪的印度数学家 Brahmagupta 是“算术之父”，Carl Friedrich Gauss 在 1801 年提供了数论的基本原理。算术是数学的基础，主要关注古代的数字、数制、它们的性质和它们的运算。该术语源自希腊语“arithmos”，仅表示数字。传统的算术运算方法是加法、差法、乘法和除法的组成部分。几个世纪以来，这些行动都是为了人类社会和经济发展而进行的。

The branch of mathematics that specifically deals with the study of numbers and properties of traditional operations like addition, subtraction, multiplication, and division is called as Arithmetic.

除了传统的加减乘除运算外，还包括百分比、对数、取幂、平方根等高级计算方法。

算术中的基本运算类型

下面讨论算术的四种基本运算：

加法(+)

加法或也称为求和是将两个或多个值或数字组合成单个值的操作。将 n 个值相加的过程称为求和。

将 0 添加到任何值都会得到相同的结果。因此，0 被称为加法的单位元。例如：如果我们将 0 与 1 相加，结果仍然是 1。

0 + 1 = 1

然而，当我们添加相同数字的相反值时，我们称其为逆元素。逆元素相加的结果是一个为 0 的单位元素。例如，如果我们将 1 与它的相反值 -1 相加，那么结果将是

1 + (-1) = 0

减法(-)

减法是加法的逆运算。确定两个值之差的算术运算（即被减数减去减数）是减法。在被减数大于减数的情况下，差值是正数。

5 – 2 = 3

同时，如果减数大于被减数，则它们之间的差异将为负数。

2 – 5 = -3

乘法 (×)

将两个值组合成一个乘积的运算称为乘法。乘法运算中涉及的两个值称为被乘数和乘数。

假设两个值的乘积为 p，q 以 pq 或 p×q 形式表示。

除法 (÷)

除法是乘法的逆运算。它是计算两个值的商的操作。其中涉及的两个值被除数称为红利。如果商大于 1，如果被除数大于除数，则结果将为正数。

9/3 = 3

现在，让我们跳到这个问题上，

哪两个数之和为 7，差为 1？

回答：

Let the two numbers be x and y.

According to the question,

x + y = 7……(i)

x – y = 1…….(ii)

Now, to know the value of one of the variables we need to solve one of the equations given above and substitute the value in the other equation.

Now, from equation (i) we can obtain the value of x by isolating it.

=>x = 7 – y

Then, substituting the value of x in equation (ii)

=>7 – y – y = 1

=>7 – 2y = 1

=>2y = 7 – 1

=>y = 6/2 = 3

Substituting the value of y = 3 in x = 7 – y

=>x = 7 – 3 = 4

Hence, the two variables are x = 4 and y = 3.

类似问题

问题1：两个数之和为8，差为2。求变量。

解决方案：

Let the two numbers be x and y.

According to the question,

x + y = 8……(i)

x – y = 2…….(ii)

Now, to know the value of one of the variables we need to solve one of the equations given above and substitute the value in the other equation.

Now, from equation (i) we can obtain the value of x by isolating it.

=>x = 8 – y

Then, substituting the value of x in equation (ii)

=>8 – y – y = 2

=>8 – 2y = 2

=>2y = 8 – 2

=>y = 6/2 = 3

Substituting the value of y = 3 in x = 8 – y

=>x = 8 – 3 = 5

Hence, the two variables are x = 5 and y = 3.

问题2：让我们取两个数字，它们的和是11，它们的乘积是30。求数字的值。

解决方案：

Let the two numbers be x and y.

Now, according to the question,

x + y = 11……(i)

x.y = 30…….(ii)

Now, from equation (ii) we can obtain the value of x by isolating it.

=>x = 30/y

Then, substituting the value of x in equation (i)

=>30/y + y = 11

=>30 + y² = 11y

=>y² – 11y + 30 = 0

=>y² – 6y – 5y + 30 = 0

=>y(y – 6) – 5(y – 6) = 0

=>(y – 6)(y – 5) = 0

=>y = (5, 6)

=>x = (6, 5)

问题 3：两个数之和是 8，它们的倒数之和是 8/15。找到数字。

解决方案：

Let the two numbers be x and y.

According to the question,

x + y = 8…….(i)

1/x + 1/y = 8/15…….(ii)

Now, to know the value of one of the variables we need to solve one of the equations given above and substitute the value in the other equation.

Now, from equation,

=>8/xy = 8/15

=>xy = 15

=>x = 15/y

Then, substituting the value of x in equation (i)

=>15/y + y = 8

=>15 + y² = 8y

=>y² – 8y + 15 = 0

=>(y – 5)(y – 3) = 0

=>y = (5, 3)

=>x = (3, 5)

问题 4：如果两个非零数的和是 8，它们的差是 7。求数的值。

解决方案：

Let the two numbers be x and y.

According to the question,

x + y = 25……(i)

x – y = 7…..(ii)

Now, to know the value of one of the variables we need to solve one of the equations given above and substitute the value in the other equation.

Now, from equation (i) we can obtain the value of x by isolating it.

=>x = 25 – y

Then, substituting the value of x in equation (ii)

=>25 – y – y = 7

=>25 – 2y = 7

=>2y = 25 – 7

=>2y = 18

=>y = 18/2 = 9

Substituting the value of y = 9 in x = 25 – y

=>x = 25 – 9 = 16

Hence, the two variables are x = 16 and y = 9.