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5 less than a number b is 16

Sagot :

Given :-

[tex] \begin{gathered}\\ \large\dashrightarrow\mathsf { \: \: 5 < b = 6 \: \: \: or \: \: \: b - 5 = 6 } \\ \end{gathered}[/tex]

Solution :-

[tex] \begin{gathered}\\ \large\dashrightarrow\mathsf { \: \: b - 5 = 6 } \\ \end{gathered}[/tex]

[tex] \begin{gathered}\\ \large\dashrightarrow\mathsf { \: \: b = 6 + 5 } \\ \end{gathered}[/tex]

[tex] \begin{gathered}\\ \large\dashrightarrow \: \: \underline{ \boxed{\mathbf \red{ \: \: b = 11 \: \: }}} \\ \end{gathered}[/tex]

Hence,

  • b is equal to 11 .

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Additional information

[tex]\begin{gathered}\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \small \color{blue}{ \underline{\boxed{ \begin{array}{cc} \small \underline{\underline{\bf{ \color{red}{{ \orange \bigstar \: MᴏʀE \: IᴅᴇɴᴛɪᴛɪᴇS \: \orange \bigstar}}}}} \\ \\ \: \frak{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \: \frak{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \: \frak{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \: \frak{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \: \frak{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \: \frak{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \: \frak{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \: \frak{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \\ \: \end{array} }}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered} \: \: [/tex]

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