Question

plzz answer fast 1st question ka 4th prt​

Answer 1

AnSwER :

$\underline{\underline{ \sf \qquad Solution \: (1) : \qquad}}$

$\longrightarrow\:\:\sf x - 2 = 7 \qquad \bigg \lgroup Given \: equation\bigg \rgroup$

$\longrightarrow\:\:\sf x = 7 + 2 \qquad \bigg \lgroup By \: transposing \: 2 \: towards \: RHS\bigg \rgroup$

$\longrightarrow\:\: \underline{ \boxed{\sf x = 9}}$

$\therefore\:\underline{\textsf{Here the value of x is \textbf{9}}}.$

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$\underline{\underline{ \sf \qquad Solution \: (2) : \qquad}}$

$\longrightarrow\:\:\sf \dfrac{3}{7} + x = \dfrac{17}{7} \qquad \bigg \lgroup Given \: equation\bigg \rgroup$

$\longrightarrow\:\:\sf x = \dfrac{17}{7} - \dfrac{3}{7} \qquad \bigg \lgroup By \: transposing \: \dfrac{3}{7} \: towards \: RHS\bigg \rgroup$

$\longrightarrow\:\:\sf x = \dfrac{17 - 3}{7} \qquad \bigg \lgroup By \: substracting \: 3 \: from \: 17 \bigg \rgroup$

$\longrightarrow\:\:\sf x = \dfrac{14}{7}\qquad \bigg \lgroup By \: Dividing \: 14 \: by \: 7 \: we \: get \: answer \: as \: 2 \bigg \rgroup$

$\longrightarrow\:\: \underline{ \boxed{\sf x = 2}}$

$\therefore\:\underline{\textsf{Here the value of x is \textbf{2}}}.$

Answer 2

(1)x -2 = 7

According to the question:

$\sf \implies \: x - 2 = 7 \\ \sf \implies \: x = 7 + 2 \\ \sf \implies \: x = 9$

Hence, x is equal to 9.

(2) 3/7 +x = 17/7

According to the question:

$\sf \implies \: \frac{3}{7} + x = \frac{17}{7} \\ \sf \implies \: x = \frac{17}{7} - \frac{3}{7} \\ \sf \implies \: x = \frac{17 - 3}{7} \\ \sf \implies \: x = \frac{14}{7} \\ \sf \implies \: x = 2$

Hence, x is equal to 2.