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Linear Equations in One Variable

    Reducing Equations to Simpler Form

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8th class > Linear Equations in One Variable > Reducing Equations to Simpler Form

Reducing Equations to Simpler Form

Example 3: Solve6x+13+1 = x−36

6x+13+1=x−36

  • Multiplying both sides of the equation by 6
  • On simplifying the divisions
  • Solving for terms within the brackets
  • Solving the constant terms on LHS.
  • Move x to one side of the equation to simplify.
  • Simplify for the x vaiable
  • Now, simplify the constant terms
  • Solve the constant terms on the RHS
  • We get: x =
  • Thus, x = is the solution.

Now, that we have found the solution, let's check if the LHS and RHS of the equation are equal or not.

Instructions

Now check the LHS and RHS of the above expression by substituting the obtained x value.

LHS = 6(-1) + 13 + 1 = -6+13 + 1 = −53 + 33 = -5+33 =
RHS = (-1)-36 = -46 =
LHS = RHS. (as required)

Example 4: Solve 5x – 2 (2x – 7) = 2 (3x – 1) + 72

5x−22x−7=23x−1+72

  • Solving the terms with brackets
  • This can be further simplified
  • Solving for the constant terms on RHS.
  • Transposing the constant to the LHS and x to RHS
  • Solving we get
  • Now, we have
  • We get: x =
  • Therefore, required solution is x =

Check: LHS = 5 × 52-2(52× 2-7)

= 252-2(5-7) = 252 - 2 (-2) =252 + 4 = 25 + 82 =

RHS = 2(52 × 3-1) + 72 = 2 (152-22) + 72 = 2 × 132 + 72

=26+72 = = LHS (as required)