Question

Chapter = Linnear simulation equation. Please answer it fast whose answer is correct I give it brainlieast answer​

Answer 1

Answer.

(a) $\begin{cases} & x= \dfrac{1}{4}\\ & y= 6\end{cases}$

(b) $\begin{cases} & x= 2\\ & y= 3\end{cases}$

(c) $\begin{cases} & x= 1\\ & y= \dfrac{1}{2}\end{cases}$

(d) $\begin{cases} & x= \dfrac{1}{2}\\ & y= \dfrac{1}{5}\end{cases}$

Answer Keys.

System Equations

We eliminate one variable and solve for only one variable. The ways we can eliminate variables are DMAS(Division/Multiplication/Addition/Subtraction).

Equations

We can multiply, divide, add, subtract equal values on both sides because it keeps the equality. But you cannot divide by 0.

Solution.

Check whether we can eliminate one of the variables.

(a) $\begin{cases} & 4x+\dfrac{6}{y}= 2\\ & 15x-\dfrac{6}{y}= \dfrac{11}{4}\end{cases}$

By adding both equations,

$\therefore \boxed{x=\dfrac{1}{4}}$

Solving for $y$,

$\therefore \boxed{y=6}$

(b) $\begin{cases} & \dfrac{10}{x}+\dfrac{15}{y}=10\\ & \dfrac{10}{x}+\dfrac{20}{y}=\dfrac{70}{6} \end{cases}$

By subtracting both equations,

$\therefore \boxed{y=3}$

Solving for $x$,

$\therefore\boxed{x=2}$

(c) $\begin{cases} & \dfrac{1}{x}+\dfrac{1}{y}=3\\ & -\dfrac{1}{x}+\dfrac{1}{y}=1\end{cases}$

By adding both equations,

$\therefore \boxed{y=\dfrac{1}{2}}$

By subtracting both equations,

$\therefore\boxed{x=1}$

(d) $\begin{cases} & x+y=\dfrac{7}{10}\\ & x-y=\dfrac{3}{10}\end{cases}$

By adding both equations,

$\therefore\boxed{x=\dfrac{1}{2} }$

By subtracting both equations,

$\therefore\boxed{y=\dfrac{1}{5} }$

Answer 2

$\Large{\textsf{\textbf{\underline{\underline{Solution\::}}}}}$

❶ $\begin{cases} 2x \ + \ \dfrac{3}{y} \ = \ 1 ---(i) \\ \\ 5x \ - \ \dfrac{2}{y} \ = \ \dfrac{11}{12} ---(ii) \end{cases}$

Multiple 2 in equation (i) & 3 in equation (ii), we get,

$\dashrightarrow\:\tt{\begin{cases} 4x \ + \ \dfrac{6}{y} \ = \ 2 ---(i) \\ \\ 15x \ - \ \dfrac{6}{y} \ = \ \dfrac{11}{4} ---(ii) \end{cases}}$

Now adding both equation, we get

$\dashrightarrow\:\tt{\left( 4x \: + \: \dfrac{6}{y} \right) + \left( 15x \: - \: \dfrac{6}{y} \right) \: = \: 2 \: + \: \dfrac{11}{4} }$

$\dashrightarrow\:\tt{4x \: + \: \dfrac{6}{y} + 15x \: - \: \dfrac{6}{y} \: = \: 2 \: + \: \dfrac{11}{4} }$

$\dashrightarrow\:\tt{19x \: = \: \dfrac{19}{4} }$

$\dashrightarrow\:\tt{x \: = \: \dfrac{19}{4} \times \dfrac{1}{19} }$

$\dashrightarrow\:\bf\pink{x \: = \: \dfrac{1}{4} }$

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

$\dashrightarrow\:\bf\pink{y \: = \: 6 }$

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❷ $\begin{cases} \dfrac{2}{x} \ + \ \dfrac{3}{y} \ = \ 2 ---(i) \\ \\ \dfrac{5}{x} \ + \ \dfrac{10}{y} \ = \ 5\dfrac{5}{6} ---(ii) \end{cases}$

First of all solve equation (ii) as,

➠ $\tt{\dfrac{5}{x} \: + \: \dfrac{10}{y} \: = \: 5\dfrac{5}{6}}$

➠ $\tt{\dfrac{5}{x} \: + \: \dfrac{10}{y} \: = \: \dfrac{35}{6}}$

➠ $\tt{5\:\left(\dfrac{1}{x} \: + \: \dfrac{2}{y}\right) \: = \: \dfrac{35}{6}}$

➠ $\tt{\dfrac{1}{x} \: + \: \dfrac{2}{y} \: = \: \dfrac{7}{6}}$

Now, we write both equation as,

$\dashrightarrow\:\tt{\begin{cases} \dfrac{2}{x} \ + \ \dfrac{3}{y} \ = \ 2 ---(i) \\ \\ \dfrac{1}{x} \ + \ \dfrac{2}{y} \ = \ \dfrac{7}{6} ---(ii) \end{cases}}$

Multiple 2 in equation (ii), we get

$\dashrightarrow\:\tt{\begin{cases} \dfrac{2}{x} \ + \ \dfrac{3}{y} \ = \ 2 ---(i) \\ \\ \dfrac{2}{x} \ + \ \dfrac{4}{y} \ = \ \dfrac{7}{3} ---(ii) \end{cases} }$

Substracting equation (2) from equation (1), we get

$\dashrightarrow\:\tt{ \dfrac{2}{x} \: + \: \dfrac{3}{y} \: - \: \left( \dfrac{2}{x} \: + \: \dfrac{4}{y} \right) \: = \: 2 \: - \: \dfrac{7}{3} }$

$\dashrightarrow\:\tt{ \dfrac{2}{x} \: + \: \dfrac{3}{y} \: - \: \dfrac{2}{x} \: - \: \dfrac{4}{y} \: = \: 2 \: - \: \dfrac{7}{3} }$

$\dashrightarrow\:\tt{ \dfrac{3}{y} \: - \: \dfrac{4}{y} \: = \: \dfrac{6\:-\:7}{3} }$

$\dashrightarrow\:\tt{ - \: \dfrac{1}{y} \: = \: - \: \dfrac{1}{3} }$

$\dashrightarrow\:\bf\pink{ y \: = \: 3 }$

Putting the value of y in equation (i), we get

$\dashrightarrow\:\bf\pink{ x \: = \: 2 }$

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❸ $\begin{cases} \dfrac{x \ + \ y}{xy} \ = \ 3 ---(i) \\ \\ \dfrac{x \ - \ y}{xy} \ = \ 1 ---(ii) \end{cases}$

$\dashrightarrow\:\tt{\begin{cases} x \ + \ y \ = \ 3xy ---(i) \\ \\ x \ - \ y \ = \ xy ---(ii) \end{cases}}$

Adding both equation, we get

$\dashrightarrow\:\tt{x \: + \: y \: + \: x \: - \: y \: = \: 3xy \: + \: xy }$

$\dashrightarrow\:\tt{2x \: = \: 4xy }$

$\dashrightarrow\:\tt{2y \: = \: 1 }$

$\dashrightarrow\:\bf\pink{y \: = \: \dfrac{1}{2} }$

Putting the value of y in equation (i), we get

$\dashrightarrow\:\bf\pink{x \: = \: 1 }$

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❹ $\begin{cases} \dfrac{x \ + \ y}{x \ - \ y} \ = \ \dfrac{7}{3} ---(i) \\ \\ x \ + \ y \ = \ \dfrac{7}{10} ---(ii) \end{cases}$

Putting the value of equation (ii) in equation (i), we get

➠ $\tt { \dfrac{x \: + \: y}{x \: - \: y} \: = \: \dfrac{7}{3} \:}$

➠ $\tt { \dfrac{\frac{7}{10}}{x \: - \: y} \: = \: \dfrac{7}{3} \:}$

➠ $\tt { x \: - \: y \: = \: \dfrac{7}{10} \times \dfrac{3}{7} \:}$

➠ $\tt { x \: - \: y \: = \: \dfrac{3}{10} \:}$

Now, we write both equation as,

$\dashrightarrow\:\tt{\begin{cases} x \ - \ y \ = \ \dfrac{3}{10} ---(i) \\ \\ x \ + \ y \ = \ \dfrac{7}{10} ---(ii) \end{cases}}$

Adding both equation, we get

$\dashrightarrow\:\tt{x \: - \: y \: + \: x \: + \: y \: = \: \dfrac{3}{10} \: + \: \dfrac{7}{10} }$

$\dashrightarrow\:\tt{2x \: = \: \dfrac{10}{10} }$

$\dashrightarrow\:\tt{2x \: = \: 1 }$

$\dashrightarrow\:\bf\pink{x \: = \: \dfrac{1}{2} }$

Putting the value of x in equation (ii), we get

$\dashrightarrow\:\bf\pink{x \: = \: \dfrac{1}{5} }$

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