SOLVING ONE STEP EQUATIONS

About "Solving one step equations"

"Solving one step equations" is nothing but the initial stuff of learning algebra in math.

A one-step equation is as straightforward as it sounds. We just have to perform one step in order to solve the equation.

We have to isolate the variable which comes in the equation.

That is, we have to get rid of the number which is added to the variable or subtracted from the variable or multiplied by the variable or divides the variable.

Solving one step equations - Examples

Example 1 :

Solve : 5 + x = 3

Solution :

Here 5 is added to the variable "x". To get rid of 5, we have to take "negative 5" on both sides and solve the equation as explained below.

Hence, the vale of "x" is "-2"

Let us look at the next example on "Solving one step equations"

Example 2 :

Solve : p - 7 = 3

Solution :

Here 7 is subtracted from the variable "p". To get rid of "-7", we have to take "positive 7" on both sides and solve the equation as explained below.

Hence, the vale of "p" is "10"

Let us look at the next example on "Solving one step equations"

Example 3 :

Solve : 2r = 6

Solution :

Here "r" is multiplied by 2. To get rid of 2, we have to divide by 2 on both sides and solve the equation as explained below.

Hence, the vale of "r" is "3"

Let us look at the next example on "Solving one step equations"

Example 4 :

Solve : (1/4)m = 3

Solution :

Here "m" is divided by 4. To get rid of 4, we have to divide by 4 on both sides and solve the equation as explained below.

Hence, the vale of "m" is "12"

Example 5 :

Solve : 8 - p = 12

Solution :

Here "p" is having negative sign.

In this problem, first we have to make "p" to be positive.

For that, we have to add "p" on both sides. When we do so, we will have "12 + p" on the right side of the equation.

Then, to get rid of "12" on the right side, we have to subtract 12 on both sides.

Therefore, we have to add "p" and subtract "12" on both sides and solve the equation as explained below.

Hence, the vale of "p" is "-4"

Solving one step equations - Practice problems

Problem 1 :

Solve : m - 10 = -15

Solution :

Here "10" is subtracted from "m". To get rid of 10, we have to add 10 on both sides and solve the equation as explained below.

(m - 10) + 10 = (-15) + 10

m = -5

Hence, the value of "m" is "-5

Problem 2 :

Solve : v + 5 / 3 = - 1/ 3

Solution :

Here "5/3" is added to "v". To get rid of 5/3, we have to subtract 5/3 on both sides and solve the equation as explained below.

(v + 5/3) - 5/3 = (-1/3) - 5/3

v = -2

Hence, the value of "v" is "-2"

Let us look at the next problem on "Solving one step equations"

Problem 3 :

Solve : 38 - m = -44

Solution :

Here "m" is having negative sign.

In this problem, first we have to make "m" to be positive.

For that, we have to add "m" on both sides. When we do so, we will have "- 44 + m" on the right side of the equation.

Then, to get rid of "-44" on the right side, we have to add 44 on both sides.

Therefore, we have to add "m" and "44" on both sides and solve the equation as explained below.

(38 - m) + m + 44 = (-44) + m + 44

82 = m

Hence, the value of "m" is "82"

Problem 4 :

Solve : 11.7 = 1.1 + n

Solution :

Here "1.1" is added to "n". To get rid of 1.1, we have to subtract 1.1 on both sides and solve the equation as explained below.

(11.7) -1.1 = (1.1 + n) - 1.1

10.6 = n

Hence, the value of "n" is "10.6"

Let us look at the next problem on "Solving one step equations"

Problem 5 :

Solve : -25.7 - v = -40.3

Solution :

Here "v" is having negative sign.

In this problem, first we have to make "v" to be positive.

For that, we have to add "m" on both sides. When we do so, we will have "- 40.3 + v" on the right side of the equation.

Then, to get rid of "-40.3" on the right side, we have to add 40.3 on both sides.

Therefore, we have to add "v" and "40.3" on both sides and solve the equation as explained below.

(-25.7 - v) + v + 40.3 = (-40.3) + v + 40.3

14.6 = v

Hence, the value of "v" is "40.3"

Solving one step equations - Word problems

Problem 1 :

When 7 is added to a number, we get 25. Find the number.

Solution :

Let "x' be the required number.

According to the question, we have

x + 7 = 25

Here "7" is added to "x". To get rid of 7, we have to subtract 7 on both sides and solve the equation as explained below.

(x + 7) - 7 = (25) - 7

x = 18

Hence, the required number is "18".

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Problem 2 :

When we multiply a number by 4, we get 124. Find the number.

Solution :

Let "x' be the required number.

According to the question, we have

4x = 124

Here "x" is multiplied by "4". To get rid of 4, we have to divide by 4 on both sides and solve the equation as explained below.

4x / 4 = 124 / 4

x = 31

Hence, the required number is "31".

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Problem 3 :

When we divide a number by 7, we get 14. Find the number.

Solution :

Let "m' be the required number.

According to the question, we have

m / 7 = 14

Here "m" is divided by "7". To get rid of 7, we have to multiply by 7 on both sides and solve the equation as explained below.

(m/7) x 7 = 14 x 7

m = 98

Hence, the required number is "98".

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Problem 4 :

John had some candies. He gave 5 candies to his friend and now he has 18 candies. How many candies did John initially have ?

Solution :

Let "m' be the no. of candies that John initially had.

According to the question, we have

m - 5 = 18

Here "5" is subtracted from "m". To get rid of 5, we have to add 5 on both sides and solve the equation as explained below.

(m - 5) + 5 = 18 + 5

m = 23

Hence, John initially had 23 candies.

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Problem 5 :

Alex borrowed some money from Jose. After 3 years, Alex returned 2 times of borrowed money to Jose. If the returned money is $226, how much money did Alex borrow from Jose ?

Solution :

Let "x' be the borrowed money.

According to the question, we have

2x = 226

Here "x" is multiplied by 2. To get rid of 2, we have to divide by 2 on both sides and solve the equation as explained below.

2x / 2 = 226 / 2

x = 113

Hence, the borrowed money is $113.