Question

Q1. Solve the equation 2(3x-1)-5x= -7/2 - 2 ( 2x-7)​

Answer 1

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

How to solve :

Step 1: First step is to open the brackets and Multiply with each and every term.

Step2:Take LCM of the term to both sides.

Step3:Now , simplify the term and cross Multiply to both sides.

Step4:Make constant term together and Variable term together.

Step5:Shifts all the constant term to constant side and let the variable alone.

Step6: Evaluate it and hence it is your required answer.

$\sf \longmapsto2(3x - 1) - 5x = - \dfrac{7}{2} - 2(2x - 7)$

$\sf \longmapsto6x - 2 - 5x = - \dfrac{7}{2} - 4x + 14$

$\sf \longmapsto \: x - 2 = - \dfrac{7}{2} + 14 - 4x$

$\sf \longmapsto \: x - 2 = - \dfrac{7}{2} + 14 - 4x$

Taking LCM to right side :

$\sf \longmapsto \: x - 2 = \dfrac{ - 7 + 28 - 8x}{2}$

Now,cross multiply to both sides:

$\sf \longmapsto21 - 8x = 2(x - 2)$

$\sf \longmapsto21 - 8x = 2x - 4$

$\sf \longmapsto21 + 4 = 2x + 8x$

$\sf \longmapsto25 = 10x$

$\sf \longmapsto \bold{x = \dfrac{25}{10} = \dfrac{5}{2} }$

Check:-

$\sf \longmapsto21 - 8x = 2x - 4$

$\sf \longmapsto21 - 8 \times \dfrac{5}{2}$

$\sf \longmapsto21 - 20 = 1$

Taking RHS

$\sf \longmapsto2 \times \dfrac{5}{2} - 4$

$\sf \longmapsto5 - 4 = 1$