Question

Solve and check your result
a) 3x – 1 = 11
b) 7 + 5a = 5

Answer 1

The above question can be written as, ⇒[(×(3−1)−(×(x))/35=3          [Taking L.C.M of 5 and 7]

⇒((21−7−(5x))/3=3

⇒(21−7−5x)/3=3

⇒(16x−7)/35=3...[Multiplying both side by 35]

⇒[(16x−7)/35]×35=3×35

⇒16x−7=105

⇒16x=105+7

⇒16x=105+7

⇒16x=112...[Multiplying both side by 1/16]

⇒16x×(1/16)=112×(1/16)

⇒x=7

By substituting 7 in the place of x in given equation, we get

LHS,

=(16x−7)/35

=((16×7)−7)/35

(112−7)/35

=(105/35)

=3

RHS,

=3

By comparing LHS and RHS

⇒(4/3)=(4/3)

∴LHS=RHS

Hence, the result is verified.

Answer 2

Question:-

i) 3x - 1 = 11

ii) 7 + 5a = 5

Solution:-

i) 3x - 1 = 11

= 3x = 11 + 1

= 3x = 12

= x = 12/3

= x = 4

ii) 7 + 5a = 5

= 5a = 5 - 7

= 5a = -2

= a = -2/5

= a = -0.4

Verification:-

i) 3x - 1 = 11

= (3 × 4) - 1 = 11 (replace the value of x)

= 12 - 1 = 11

hence, verified

ii) 7 + 5a = 5

= 7 + (5 × -0.4) = 5 (replace the value of x)

= 7 + (-2) = 5

= 7 - 2 = 5 ( + - = -)