Question

find the value of x in this algebraic expresion 2(x-3)+4(x-3)=20​

Answer 1

$\large{\underline{\underline{\mathfrak{\sf{Answer-}}}}}$

$\sf{\dfrac{19}{3}}$

$\large{\underline{\underline{\mathfrak{\sf{Explanation-}}}}}$

Given expression :

• $\sf{2(x-3)+4(x-3)=20}$

To find :

• Its value

Solution :

$\sf{2(x-3)+4(x-3)=20}$

After multiplying, we get,

$\implies$ $\sf{2x-6+4x-12=20}$

$\implies$ $\sf{6x-18=20}$

$\implies$ $\sf{6x=20+18}$

$\implies$ $\sf{6x=38}$

$\implies$ $\sf{x=\cancel{\dfrac{38}{6}}}$

$\implies$ $\sf{\dfrac{19}{3}}$

Answer 2

Answer: 19/3

Step by step explanation:

Given:

Expression = 2(x-3)+4(x-3)=20

To find: We have to find the value of x of the Algebraic expression.

Solution:

= 2(x-3)+4(x-3)=20

= 2x - 6 + 4x - 12 = 20

= 4x + 2x - 6 - 12 = 20

= 6x + 18 = 20

= 6x = 20+18

= 6x = 38

= x = 38/6

= x = 19/3

Value of x = 19/3

_________________________________________________

$2(x - 3) + 4(x - 3) \\\ 2x - 6 + 4x - 12 = 20 \\ 4x + 2x - 6 - 12 = 20 \\ 6x = 20 + 18 \\ 6x = 38 \\ x = \frac{38}{6} \\ x = \frac{19}{3}$