In primary school mathematics textbooks, simple equations can be divided into the following two situations.

(1) A simple equation that only needs one step to solve.

① Find the unknown addend.

Solution: Subtract the known addend from the sum.

Example solution equation x+36 = 97

Answer: 97 is the sum of two numbers and 36 is a known addend. therefore

x+36=97

x=97-36

x=6 1

② Find the unknown minuend

Solution: Add the difference to the known subtraction.

Example solution equation X-55 = 48

Solution: 48 is the difference and 55 is the subtraction. therefore

x-55=48

x=48+55

x= 103

③ Find the unknown subtraction.

Solution: Subtract the difference from the minuend.

Example solution equation 200-x = 95

Answer: 200 is the minuend and 95 is the difference. therefore

200-x=95

x=200-95

x= 105

④ Find out the unknown factors.

Solution: Divide the product by a known factor.

Example solution equation 7x = 9 1

The solution 9 1 is a product and 7 is a known factor. therefore

7x=9 1

x=9 1÷7

x= 13

⑤ Seek an unknown dividend.

Solution: Multiply the quotient by the divisor.

Example solution equation x ÷ 29 = 75.

Solution: 75 is the quotient and 29 is the divisor. therefore

x÷29=75

x=75×29

x=2 175

③ Find the unknown divisor.

Solution: Divide the dividend by the quotient.

Example solution equation 432 ÷ x = 27

Solution: 432 is the dividend and 27 is the quotient. therefore

432÷x=27

x=432÷27

x= 16