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

1.	Find the value of t : 4 (t + 1 ) + ( t + 2) = 6 ( t – 3 ) – 1
Answer» ALIGN="absmiddle" ALT="\large\sf\underline{Given\::}" class="latex-formula" id="TexFormula1" src="https://tex.z-dn.net/?f=%5Clarge%5Csf%5Cunderline%7BGiven%5C%3A%3A%7D" title="\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 !!

Answer» ALIGN="absmiddle" ALT="\large\sf\underline{Given\::}" class="latex-formula" id="TexFormula1" src="https://tex.z-dn.net/?f=%5Clarge%5Csf%5Cunderline%7BGiven%5C%3A%3A%7D" title="\large\sf\underline{Given\::}">