If a number is subtracted from the numerator of the fraction 3/5 and thrice that number is added to the denominator, the fraction becomes 1/4. Find the number.
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If a number is subtracted from the numerator of the fraction 3/5 and thrice that number is added to the denominator, the fraction becomes 1/4. Find the number.
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Verified answer
[tex]\huge{\sf\color{green}\underline{Solution}}[/tex]
[tex] \sf\color{yellow}\underline{Let \: }[/tex]
[tex] \sf\color{cyan}\underline{A/Q \: }[/tex]
[tex] \sf\pink{:\longmapsto \: \: \: \: \: \: \: \: \:{\dfrac{3 - x}{5 + 3x} = \dfrac{1}{4}}} \: [/tex]
Cross Multiplying
[tex]\sf{:\longmapsto} \: \: \: \: \: \: \: \: \: {4(3 - x) = 5 + 3x}[/tex]
[tex]\sf{:\longmapsto} \: \: \: \: \: \: \: \: \: {12 - 4x = 5 + 3x}[/tex]
Transposing 4x to RHS and 5 to LHS
[tex]\sf{:\longmapsto} \: \: \: \: \: \: \: \: \: {12 - 5 = 4x + 3x}[/tex]
[tex]\sf{:\longmapsto} \: \: \: \: \: \: \: \: \:{7 = 7x}[/tex]
[tex]\sf{:\longmapsto} \: \: \: \: \: \: \: \: \:{7x = 7}[/tex]
Dividing
[tex]\sf{:\longmapsto} \: \: \: \: \: \: \: \: \:{x = \cancel\dfrac{7}{7}}[/tex]
[tex]\sf\blue{:\longmapsto} \: \: \: \: \: \: \: \: \:{\red{\underline{\boxed{\bf{{x = 1}}}}}}[/tex]
[tex] \sf\large{\orange{\therefore{\underline{The \: required \: number \: is \: 1. \: }}}}[/tex]
[tex]\red{\rule{200pt}{2pt}}[/tex]
[tex] \sf{\large\color{yellow}\underline{Verification}}[/tex]
Substituting x = 1 on the LHS, we get
[tex]\sf{\dfrac{3 - 1}{5 + 3(1)}}[/tex]
[tex]\sf=\cancel{\dfrac{2}{8}}[/tex]
[tex]\sf=\dfrac{1}{4}.[/tex]
RHS = [tex]\sf\dfrac{1}{4}.[/tex]
LHS = RHS
1/4 = 1/4
Hence, verified! ✓.
Steps for solving linear equations based word problems
Steps for solving linear equations
[tex]\space\space\space\orange{\bigstar}{\color{yellow}{\rule{150pt}{4pt}}\orange{\bigstar}}[/tex]
Given :
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
To Find :
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
Solution :
~ According to the Question :
➳ If a number is subtracted from the numerator of the fraction 3/5 and thrice that number is added to the denominator, the fraction becomes 1/4. Let the no. be y .Hence,
[tex]\large{\qquad{\gray{\bigstar}} \: \: {\underline{\underline{\red{\sf{ \dfrac{3 - y}{5 + 3y} = \color{cyan}{\dfrac{1}{4}} }}}}}}[/tex]
[tex]\qquad{━━━━━━━━━━━━━━━━━━━━━━}[/tex]
~ Now Cross Multiplication :
[tex]{\implies{\qquad{\sf{ \dfrac{3 - y}{5 + 3y} = \dfrac{1}{4}}}}} \\ \\ \ {\implies{\qquad{\sf{ 4(3 - y) = 1(5 + 3y) }}}} \\ \\ \ {\implies{\qquad{\sf{ 12 - 4y = 5 + 3y}}}} \\ \\ \ {\implies{\qquad{\sf{ 12 - 5 = 4y + 3y }}}} \\ \\ \ {\implies{\qquad{\sf{ 7 = 7y}}}} \\ \\ \ {\implies{\qquad{\sf{ y = \cancel\dfrac{7}{7} }}}} \\ \\ \ {\qquad{\sf{ Value \: of \: y \: = {\color{maroon}{\sf{ 1}}}}}}[/tex]
[tex]\qquad{━━━━━━━━━━━━━━━━━━━━━━}[/tex]
Therefore :
[tex]\large{\red{\dashrightarrow{\orange{\underline{\underline{\green{\sf{ Required \: Number = 1}}}}}}}}[/tex]
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Verification :
[tex]{\leadsto{\sf{ \dfrac{3 - y}{5 + 3y} = \dfrac{1}{4}}}} \\ \\ {\leadsto{\sf{ \dfrac{3 - 1}{5 + 3 \times 1 } = \dfrac{1}{4}}}} \\ \\ {\leadsto{\sf{ \dfrac{2 }{5 + 3} = \dfrac{1}{4}}}} \\ \\ {\leadsto{\sf{ \cancel\dfrac{2}{8} = \dfrac{1}{4}}}} \\ \\ {\leadsto{\red{\underline{\sf{ \dfrac{1}{4} = \dfrac{1}{4}}}}}} \\ \\ {\red{\underline{\sf{ LHS = RHS}}}}[/tex]
Hence, Verified.