Question

Find the value of t : 4 (t + 1 ) + ( t + 2) = 6 ( t – 3 ) – 1​

Answer 1

$\large\sf\underline{Given\::}$

• $\sf\:4(t+1) +(t+2) =6(t-3) -1$

‎

‎

$\large\sf\underline{To\:find\::}$

• Value of x

‎

‎

$\large\sf\underline{Solution\::}$

$\sf\:4(t+1) +(t+2) =6(t-3) -1$

‎

• Let's multiply the terms and remove the brackets

‎

$\sf\implies\:4t + 4 + t + 2 = 6t- 18 -1$

‎

• Let's now arrange the like terms together

‎

$\sf\implies\:4t + t + 4 + 2 = 6t- 18 -1$

‎

• Adding the like terms

‎

$\sf\implies\:5t + 6 = 6t- 19$

‎

• Transposing 6t to the other side it becomes -6t

‎

$\sf\implies\:5t - 6t + 6 = - 19$

$\sf\implies\:-t + 6 = - 19$

‎

• Transposing 6 to other side it becomes -6

‎

$\sf\implies\:-t = - 19-6$

$\sf\implies\:\cancel{-}t = \cancel{-} 25$

$\small\fbox\red{⟹\:\:t\:\:=\:\:25}$ ⍟

‎

_____________________________

‎

$\large\sf\underline{Verifying\::}$

$\sf\:4(t+1) +(t+2) =6(t-3) -1$

‎

• Let's substitute the value of t as 25

‎

$\sf\to\:4(25+1) +(25+2) =6(25-3) -1$

‎

• Simplifying the terms inside brackets

‎

$\sf\to\:4(26) +(27) =6(22) -1$

‎

• Removing brackets

‎

$\sf\to\:4 \times 26 +27 =6 \times 22 -1$

‎

• Multiplying the terms

‎

$\sf\to\:104+27 =132 -1$

‎

• Adding in [ LHS ] and Subtracting [ RHS ]

‎

$\sf\to\:131 =131$

${\sf{{\orange{[Hence\:Verified]}}}}$ ☑

‎

_____________________________

!! Hope it helps !!