Create New Account
Create New Account

Sign in to Innings2

or

New Account     Reset Password     

Simple Equations

    More Equations

    What Have We Discussed ?

Powered by Innings 2

Reset Progress

This will delete your progress and chat data for all chapters in this course, and cannot be undone!

Glossary

Select one of the keywords on the left…

7th class > Simple Equations > More Equations

More Equations

Let us practise solving some more equations. While solving these equations, we shall learn about transposing a number, i.e., moving it from one side to the other. We can transpose a number instead of adding or subtracting it from both sides of the equation.

Here's a simple example to illustrate this

Instruction

3x+4=10

  • We need to 4 from both sides
  • This gives us: =
  • We then, 3 from both sides
  • Upon solving we get: x =
  • We have found the solution.

Example 6: Solve 12p-5 = 25

To isolate the variable, we first:

5 on both sides of the equations.

i.e. 12p - 5 + 5 = 25 + 5 (or) =

Dividing both sides by 12,

12p12 = 3012 or p =

Putting p = 52 in the LHS of the given equation:

LHS = 12 × 52 - 5 = 6 × 5 - 5 =

Thus, p = 52 is the solution.

NOTE: Transposing in mathematics means moving a term from one side of an equation to the other while changing its sign. This technique helps isolate the variable, making it easier to solve the equation.

When transposing terms in an equation:

  • Additions become subtractions and vice versa.

  • Multiplications become divisions and vice versa

Let us take two more examples of transposing.

3p−10=5 , 5x+12=27

Adding or Subtracting on both sidesTransposing
(i) 3p – 10 = 5(i) 3p – 10 = 5
Add 10 to both sidesTranspose (–10) from LHS to RHS
3p – 10 + 10 = 5 + 10(On transposing – 10 becomes + 10)
or 3p = 3p = 5 + 10 or 3p =
(ii) 5x + 12 = 27(ii) 5x + 12 = 27
Subtract 12 from both sidesTransposing + 12
5x + 12 – 12 = 27 – 125x = 27 – 12
or 5x = or 5x =

Changing side is called transposing. While transposing a number, we change i.e. reverse the operation sign.

We shall now solve two more equations. As you can see they involve brackets, which have to be solved before proceeding.

Example 7: Solve (a) 4(m + 3) = 18 by transposing

Instruction

4m+3=18

  • When transposing, we need to isolate x - we can 4 on the RHS.
  • This gives us the equation: =
  • Now, transpose 3 to the other side by 3 on RHS.
  • This gives us: m =
  • We have found the value of m. Now, let's validate it. Substitute m = 32 in the equation.
  • We get the resulting value to be .
  • We see that LHS = RHS
  • Thus, m = 32 is the solution.

(b) –2(x + 3) = 8 by transposing

Instruction

−2x+3=8

  • When transposing, we need to isolate m - we can −2 on the RHS.
  • This gives us the equation: =
  • Now, transpose 3 to the other side by 3 on RHS.
  • This gives us: x =
  • We have found the value of x. Now, let's validate it. Substitute x = -7 in the equation.
  • We get the resulting value to be .
  • We see that LHS = RHS
  • Thus, x = −7 is the solution.