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  • For what value of k, will the following pair of equations have infinitely many solutions :             2x + 3y = 7 and (k + 2) x – 3 (1 – k) y = 5k + 1

For what value of k, will the following pair of equations have infinitely many solutions :
            2x + 3y = 7 and (k + 2) x – 3 (1 – k) y = 5k + 1

 

 

 
 
 
 
 

Answers (2)

Given pair of equations are

    2x + 3y = 7 and

    (k + 2) x – 3 (1 – k) y = 5k + 1

For a pair of linear equations to have infinitely many solutions:

        \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \qquad -(1)

From the given equtions,

        \begin{matrix} a_1 = 2 & & b_1 = 3 & & c_1 = 7 \\ a_2 = k+2 & & b_2 = -3(1- k) & & c_2 = 5k+1 \end{matrix}

Put above values in equation (1)

        \frac{2}{k+2} = \frac{3}{-3(1- k)} = \frac{7}{5k+1}

\Rightarrow \frac{2}{k+2} = \frac{\cancel3}{-\cancel{3}(1- k)}

\Rightarrow -2(1-k) = k + 2

\Rightarrow -2 + 2k = k +2

\Rightarrow k = 4

Posted by

Safeer PP

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Given pair of equations are

    2x + 3y = 7 and

    (k + 2) x – 3 (1 – k) y = 5k + 1

For a pair of linear equations to have infinitely many solutions:

        \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \qquad -(1)

From the given equtions,

        \begin{matrix} a_1 = 2 & & b_1 = 3 & & c_1 = 7 \\ a_2 = k+2 & & b_2 = -3(1- k) & & c_2 = 5k+1 \end{matrix}

Put above values in equation (1)

        \frac{2}{k+2} = \frac{3}{-3(1- k)} = \frac{7}{5k+1}

\Rightarrow \frac{2}{k+2} = \frac{\cancel3}{-\cancel{3}(1- k)}

\Rightarrow -2(1-k) = k + 2

\Rightarrow -2 + 2k = k +2

\Rightarrow k = 4

Posted by

Shriddhi Buddhpriya

View full answer

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