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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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