Question

#Quality Question.
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Q. Solve This Linear Equation in two Variables.
$\frac{2}{x } + \frac{5}{y} = 1 \\ \frac{60}{x} + \frac{40}{y} = 19$
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Answer 1

Answer:

$\huge \huge \bf {\: \pmb{{Question}}}$

Q. Solve This Linear Equation in two Variables.

$\frac{2}{x } + \frac{5}{y} = 1 \\ \frac{60}{x} + \frac{40}{y} = 19$

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given that ,

$\frac{2}{x } + \frac{5}{y} = 1 \\ \frac{60}{x} + \frac{40}{y} = 19$

is a linear equation

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$\huge \huge \bf {\: \pmb{{Answer}}}$

$\sf \dashrightarrow\frac{2}{x } + \frac{5}{y} = 1 \: can \: be \: written \: as \\ \sf 2a \: + 5b \: = 1 \: \\ \sf \: by \: taking \: \frac{1}{x} as \: \color{purple}a \color{black}\: and \: \frac{1}{y} as \:\color{purple} b \: \\ \sf \dashrightarrow\frac{60}{x} + \frac{40}{y} = 19 \: \: can \: be \: written \: as \\ \sf 60a \: + 40b \: = 19 \: \\ \sf \: by \: taking \: \frac{1}{x} as \: \color{purple}a \color{black}\: and \: \frac{1}{y} as \:\color{purple} b$

$\sf \: now \: we \: have \: 2a + 5b \: = 1 \: and \\ \sf \: 60a \: + 40b \: = 19 \\ \sf \: to \: slove \: we \: need \: to \: make \: into \: expression \: \\ \sf \dashrightarrow \color{teal}2a + 5b - 1 = 0\: \color{tel} - - (1)\\ \sf\dashrightarrow \:\color{teal} 60a \: + 40b \: - 19 = 0 \color{tal}- - (2)$

⇢By elimination method

we need to make the first number of equation equal means, multiple 30 in equation (1)

30 ( 2a + 5b - 1 )

= 60a + 150b -30 --- (3)

$\sf 60a + 150b - 30 = 0\: \\ \sf 60a \: + 40b \: - 19 = 0 \ \\ \sf( - ) \: \: \: \: \: \: \: ( - ) \: \: \: \: \: \: ( + ) \\ \sf - - - - - - - - - - - - - \\ \sf \: 0a + 110b \: - 11 \\ \sf \: b \: = \cancel \frac{11}{110} \\ \sf b \: = \color{blue} \frac{1}{10}$

now, substituting value of b in equation (1)

$\sf \dashrightarrow2a \: + 5b \: - 1 \\ \sf \: 2a \: + 5( \frac{1}{10} ) = 1 \\ \sf 2a \: = 1 - \frac{5}{10} \\ \sf \sf \: 2a = 1 - \frac{1}{2} \\ \sf \: a \: = \frac{1}{2 \times 2} = \color{blue} \frac{1}{4}$

now we know that value of a = $\frac{1}{x}$

so, $\frac{1}{4}$ = $\frac{1}{x}$

by comparison ,

x = 4

we know that value of b = $\frac{1}{y}$

so, $\frac{1}{10}$ = $\frac{1}{y}$

by comparison ,

y = 10

hence solved ☑