Math, asked by princy5114, 1 month ago

from the pairs of linear equations for the following problem and find their solution by substitution method:a fract in becomes 4/5,if 1 is added to both the numnerator and denominator .if 5 is subtracted to both the numnerator and the denominator, it becomes 1/2 find the fraction​

Answers

Answered by mathdude500
5

\large\underline{\sf{Solution-}}

\begin{gathered}\begin{gathered}\bf\: Let \: -\begin{cases} &\sf{numerator \: of \: fraction = x}  \\ \\ &\sf{denominator \: of \: fraction = y} \end{cases}\end{gathered}\end{gathered}

\begin{gathered}\begin{gathered}\bf \: So - \begin{cases} &\rm{required \: fraction \:  =  \: \dfrac{x}{y} }\end{cases}\end{gathered}\end{gathered}

According to first condition

The fraction becomes 4/5, if 1 is added to both numerator and denominator.

\begin{gathered}\begin{gathered}\bf\: So \: -\begin{cases} &\sf{numerator \: of \: fraction = x + 1}  \\ \\ &\sf{denominator \: of \: fraction = y + 1} \end{cases}\end{gathered}\end{gathered}

\rm :\longmapsto\:\dfrac{x + 1}{y + 1}  = \dfrac{4}{5}

\rm :\longmapsto\:5x + 5 = 4y + 4

\rm \implies\:5x + 5 - 4 = 4y

\rm \implies\:5x + 1 = 4y

\rm \implies\:\boxed{ \tt{ \: y =  \frac{5x + 1}{4} \: }} -  -  -  - (1)

According to second condition

If 5 is subtracted from both the numnerator and the denominator, it becomes 1/2.

\begin{gathered}\begin{gathered}\bf\: So \: -\begin{cases} &\sf{numerator \: of \: fraction = x - 5}  \\ \\ &\sf{denominator \: of \: fraction = y - 5} \end{cases}\end{gathered}\end{gathered}

\rm :\longmapsto\:\dfrac{x - 5}{y - 5}  = \dfrac{1}{2}

\rm :\longmapsto\:2x - 10 = y - 5

\rm :\longmapsto\:2x - 10 + 5 = y

\rm :\longmapsto\:2x -5 = y

On substituting the value of y from equation (1), we get

\rm :\longmapsto\:2x -5 = \dfrac{5x + 1}{4}

\rm :\longmapsto\:8x - 20 = 5x + 1

\rm :\longmapsto\:8x - 5x = 20 + 1

\rm :\longmapsto\:3x = 21

\rm \implies\:\boxed{ \tt{ \: x = 7 \: }} -  -  - (2)

On substituting the value of x in equation (1), we get

\rm :\longmapsto\:y = \dfrac{5 \times 7 + 1}{4}

\rm :\longmapsto\:y = \dfrac{35 + 1}{4}

\rm :\longmapsto\:y = \dfrac{36}{4}

\rm \implies\:\boxed{ \tt{ \: y = 9 \: }} -  -  - (3)

\begin{gathered}\begin{gathered}\bf\: So \: -\begin{cases} &\sf{numerator \: of \: fraction = x = 7}  \\ \\ &\sf{denominator \: of \: fraction = y = 9} \end{cases}\end{gathered}\end{gathered}

\begin{gathered}\begin{gathered}\bf \: So - \begin{cases} &\rm{required \: fraction \:  =  \: \dfrac{7}{9}}\end{cases}\end{gathered}\end{gathered}

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Basic Concept Used :-

There are 4 methods to solve this type of pair of linear equations.

1. Method of Substitution

2. Method of Eliminations

3. Method of Cross Multiplication

4. Graphical Method

We prefer here Method of Substitution

To solve systems using substitution, follow this procedure:

  • Select one equation and solve it to get one variable in terms of second variables.

  • In the second equation, substitute the value of variable evaluated in Step 1 to reduce the equation to one variable.

  • Solve the new equation to get the value of one variable.

  • Substitute the value found in to any one of two equations involving both variables and solve for the other variable.

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