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Question

Answers

$

\left( {{\text{k - 3}}} \right){\text{x + 3y = k}} \\

{\text{kx + ky = 12}} \\

$

Answer
Verified

Hint:- For the system of equations to have infinitely many solutions the ratios of coefficients of x ,y and constant term should be equal.

Given,

$\left( {{\text{k - 3}}} \right){\text{x + 3y = k and kx + ky = 12}}$ .

Let

$

\left( {{\text{k - 3}}} \right){\text{x + 3y = k }} \cdots \left( 1 \right) \\

{\text{kx + ky = 12 }} \cdots \left( 2 \right) \\

$

For a general system of equations of two variables, let the equations be

$

{{\text{a}}_1}{\text{x + }}{{\text{b}}_1}{\text{y = }}{{\text{c}}_1}{\text{ }} \cdots \left( 3 \right) \\

{{\text{a}}_2}{\text{x + }}{{\text{b}}_2}{\text{y = }}{{\text{c}}_2}{\text{ }} \cdots \left( 4 \right) \\

$

The equations will have infinite solution if and only if,

$\dfrac{{{{\text{a}}_1}}}{{{{\text{a}}_2}}} = \dfrac{{{{\text{b}}_1}}}{{{{\text{b}}_2}}} = \dfrac{{{{\text{c}}_1}}}{{{{\text{c}}_2}}}$

On comparing the coefficients of equation (1) and (3) we get,

${{\text{a}}_1}{\text{ = k - 3, }}{{\text{b}}_1}{\text{ = 3 and }}{{\text{c}}_1}{\text{ = k }} \cdots \left( 5 \right)$

On comparing the coefficients of equation (2) and (4) we get,

${{\text{a}}_2}{\text{ = k, }}{{\text{b}}_2}{\text{ = k and }}{{\text{c}}_2}{\text{ = 12 }} \cdots \left( 6 \right)$

Now, dividing the equation (5) and (6), we get

$\dfrac{{{\text{k - 3}}}}{{\text{k}}} = \dfrac{3}{{\text{k}}}{\text{ and }}\dfrac{3}{{\text{k}}} = \dfrac{{\text{k}}}{{12}}$

Solving above equations, we get

$

\left( {{\text{k - 3}}} \right){\text{k = 3k and }}{{\text{k}}^2}{\text{ = 36}} \\

{\text{k - 3 = 3 and k = }}\sqrt {36} \\

{\text{k = 6 }} \\

$

Both the equations will satisfy for k =6. Hence , the required answer is 6.

The equations will be 3x + 3y=6 and 6x + 6y =12.

Note:- The system of equations having infinite solutions is consistent and dependent. Equations of two variables must have the same slope and same y-intercept for having infinite solutions.

Given,

$\left( {{\text{k - 3}}} \right){\text{x + 3y = k and kx + ky = 12}}$ .

Let

$

\left( {{\text{k - 3}}} \right){\text{x + 3y = k }} \cdots \left( 1 \right) \\

{\text{kx + ky = 12 }} \cdots \left( 2 \right) \\

$

For a general system of equations of two variables, let the equations be

$

{{\text{a}}_1}{\text{x + }}{{\text{b}}_1}{\text{y = }}{{\text{c}}_1}{\text{ }} \cdots \left( 3 \right) \\

{{\text{a}}_2}{\text{x + }}{{\text{b}}_2}{\text{y = }}{{\text{c}}_2}{\text{ }} \cdots \left( 4 \right) \\

$

The equations will have infinite solution if and only if,

$\dfrac{{{{\text{a}}_1}}}{{{{\text{a}}_2}}} = \dfrac{{{{\text{b}}_1}}}{{{{\text{b}}_2}}} = \dfrac{{{{\text{c}}_1}}}{{{{\text{c}}_2}}}$

On comparing the coefficients of equation (1) and (3) we get,

${{\text{a}}_1}{\text{ = k - 3, }}{{\text{b}}_1}{\text{ = 3 and }}{{\text{c}}_1}{\text{ = k }} \cdots \left( 5 \right)$

On comparing the coefficients of equation (2) and (4) we get,

${{\text{a}}_2}{\text{ = k, }}{{\text{b}}_2}{\text{ = k and }}{{\text{c}}_2}{\text{ = 12 }} \cdots \left( 6 \right)$

Now, dividing the equation (5) and (6), we get

$\dfrac{{{\text{k - 3}}}}{{\text{k}}} = \dfrac{3}{{\text{k}}}{\text{ and }}\dfrac{3}{{\text{k}}} = \dfrac{{\text{k}}}{{12}}$

Solving above equations, we get

$

\left( {{\text{k - 3}}} \right){\text{k = 3k and }}{{\text{k}}^2}{\text{ = 36}} \\

{\text{k - 3 = 3 and k = }}\sqrt {36} \\

{\text{k = 6 }} \\

$

Both the equations will satisfy for k =6. Hence , the required answer is 6.

The equations will be 3x + 3y=6 and 6x + 6y =12.

Note:- The system of equations having infinite solutions is consistent and dependent. Equations of two variables must have the same slope and same y-intercept for having infinite solutions.

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