Question

Please Solve This


Please
$\frac{y - 1}{3} \: - \: \frac{y - 2}{4} \: = \: 1$
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Answer 1

$\large\underline{ \underline{ \sf \maltese{ \: Question:- }}}$

$\frac{y - 1}{3} \: - \: \frac{y - 2}{4} \: = \: 1$

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

$\begin{gathered}\begin{gathered}\begin{gathered}: \implies \underline\blue{ \boxed{\displaystyle \sf \bold\orange{\: \frac{y - 1}{3} \: - \: \frac{y - 2}{4} \: = \: 1 }} }\\ \\\end{gathered}\end{gathered}\end{gathered}$

LCM of denominator 3 and 4 on L.H.S is 12

$\qquad \quad {:} \longrightarrow \sf{\sf{ \frac{y - 1}{3} \: - \: \frac{y - 2}{4} \: = \: 1 }}$

Multiplying both sides by 12 , we get

$\qquad \quad {:} \longrightarrow \sf{\sf{12 \times \bigg( \dfrac{y - 1}{3} \bigg) \: - \:12 \times \bigg(\dfrac{y - 2}{4} \bigg) \: = \: 12 \: \times \: 1}}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ \cancel{12} \times \dfrac{y - 1}{ \cancel{3}} \: - \: \cancel{12} \times \dfrac{y - 2}{ \cancel{4}}\: = \: 12 \: \times \: 1}}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ 4(y - 1) \: - 3(y - 2) \: = \: 12 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ 4y \: - \: 4 \: - 3y \: + \: 6\: = \: 12 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ 4y \: - \: 3y \: - \: 4 \: + \: 6\: = \: 12 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ \: y \: + \: 2\: = \: 12 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ \: y \: = \: 12 \: - \: 2 }}$

$\qquad\quad {:} \longrightarrow \underline \red{\boxed{\sf{y\: = \: 10 }}}$

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$\large\underline{ \underline{ \sf \maltese{ \: Verification:- }}}$

$\begin{gathered}\begin{gathered}\begin{gathered}: \implies \underline\blue{ \boxed{\displaystyle \sf \bold\orange{\: \frac{y - 1}{3} \: - \: \frac{y - 2}{4} \: = \: 1 }} }\\ \\\end{gathered}\end{gathered}\end{gathered}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ \frac{10- 1}{3} \: - \: \frac{10 - 2}{4} \: = \: 1 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ \frac{9}{3} \: - \: \frac{8}{4} \: = \: 1 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ \frac{9 \times 4}{3 \times 4} \: - \: \frac{8 \times 3}{4 \times 3} \: = \: 1 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ \frac{36 \: - \: 24}{12} = \: 1 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ \frac{12}{12} = \: 1 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ \cancel{\frac{12}{12} } = \: 1 }}$

$\qquad\quad {:} \longrightarrow \underline \red{\boxed{\sf{1\: = \: 1 }}}$

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$\bf \therefore \; y \; = 10$

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More For Knowledge :-

$\underbrace\red{\boxed{ \sf \green{ Rules \: to \:Solve \: a \: Linear \: Equation \: in \: 1 \: Variable}}}$

• Rule 1 :- Same quantity ( number ) can be added to both side of an equation without changing the equality.
• Rule 2 :- Same quantity can be subtracted from both sides of an equation without changing the quality
• Rule 3 :- Both sides of an equation may be multiplied by the same non zero number without changing the quality.
• Rule 4 :- Both sides of an equation may be divided by the same non zero number without changing the quality.

Note :-

It should be noted that some complicated equation can be solved by using two or more of these rules together.

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Answer 2

Given :

• $\rm{\dfrac{y - 1}{3} \: - \: \dfrac{y - 2}{4} \: = \: 1}$

To find :

• The value of y

Solution :

$\ \sf{\dfrac{y - 1}{3} \: - \: \dfrac{y - 2}{4} \: = \: 1}$

Taking out the LCM

${:} \implies \sf{ \dfrac{4y - 4 - 3y + 6 }{12} = 1}$

${:} \implies \sf{ \dfrac{y + 2 }{12} = 1}$

By applying cross multiplication rule

${:} \implies \sf{y + 2 = 12}$

By bringing 2 to RHS

${:} \implies \sf{y = 12 - 2}$

${:} \implies \boxed{ \frak \pink{y = 10}}$

Verification :

By putting the value of y in equation

${:}\implies \ \sf{\dfrac{10 - 1}{3} \: - \: \dfrac{10 - 2}{4} \: = \: 1}$

${:}\implies \ \sf{\dfrac{9}{3} \: - \: \dfrac{8}{4} \: = \: 1}$

${:}\implies \ \sf{ \dfrac{36 - 24}{12} \: = \: 1}$

${:}\implies \ \sf{ \dfrac{ \cancel{12}}{ \cancel{12}} \: = \: 1}$

${:}\implies \ \sf{ 1 = 1}$

LHS = RHS

Hence Verified!!