![(-1)/(3)(6x-21)=-5(x+1)](https://img.qammunity.org/2023/formulas/mathematics/college/b0ydl9fqtnugpfribw58ikcxszk08reiub.png)
Open the bracket on LHS by multiplying with -1/3 and RHS by multiplying with -5 with quantity inside the bracket.
![\begin{gathered} (-1)/(3)*6x-((-1)/(3)*21)=-5* x+(-5)*1 \\ -2x-(-7)=-5x-5 \\ -2x+7=-5x-5 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/otito51wrjqq2a68lp982lge7d0d76l88j.png)
Add '5x' to LHS (Left-hand side) and RHS (Right-hand side) of the above expression to eliminate the 5x in the RHS.
![\begin{gathered} -2x+7+(5x)=-5x-5+(5x) \\ (-2x+5x)+7=(-5x+5x)-5 \\ 3x+7=0x-5 \\ 3x+7=-5 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/uayy5i6giwtmf9hznixvoxkxrzmbfy9mqf.png)
Substract '7' from both RHS and LHS of the above expression.
![\begin{gathered} 3x+7-(7)=-5-(7) \\ 3x+0=-12 \\ 3x=-12 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/q597cdgbyvo972tdjhxcerxx0z6m9mss1o.png)
Divide '3' from the RHS and LHS of the above expression.
![\begin{gathered} (3x)/(3)=(-12)/(3) \\ x=-4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/o9xe3r67donhfnqez2lr1ylwaiak8x62an.png)
Thus, the value of x is -4.