Question
Prove that there is a value of $(\text{c}\neq0)$ for which the system has infinitely many solutions. Find this value.
$6x + 3y = c - 3$
$12x + cy = c$
✓
Answer
The given system of equations may be written as,
$6x + 3y - (c - 3) = 0$
$12x + cy - c = 0$
This is of the form
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where, $a_1 = 6, b_1 = 3, c_1 = -(c - 3)$
And, $a_2 = 12, b_2 = c, c_2 = -c$
For infinitely many solutions, we must have
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{6}{12}=\frac{3}{\text{c}}=\frac{-(\text{c}-3)}{-\text{c}}$
$\Rightarrow\frac{6}{12}=\frac{3}{\text{c}}$ and $$$\frac{3}{\text{c}}=\frac{\text{c}-3}{\text{c}}$
$\Rightarrow6\text{c}=12\times3$ and $3=(\text{c}-3)$
$\Rightarrow\text{c}=\frac{36}{6}$ and $\text{c}-3=3$
$\Rightarrow\text{c}=6$ and $\text{c}=6$
Now,
$\frac{\text{a}_1}{\text{a}_2}=\frac{6}{12}=\frac{1}{2}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{3}{6}=\frac{1}{2}$
$\frac{\text{c}_1}{\text{c}_2}=\frac{-(6-3)}{-6}=\frac{1}{2}$
$\therefore\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_1}$
Clearly, for this value of c, we have $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
Hence, the given system of equations has infinitely many solutions, if $c = 6$.
$6x + 3y - (c - 3) = 0$
$12x + cy - c = 0$
This is of the form
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where, $a_1 = 6, b_1 = 3, c_1 = -(c - 3)$
And, $a_2 = 12, b_2 = c, c_2 = -c$
For infinitely many solutions, we must have
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{6}{12}=\frac{3}{\text{c}}=\frac{-(\text{c}-3)}{-\text{c}}$
$\Rightarrow\frac{6}{12}=\frac{3}{\text{c}}$ and $$$\frac{3}{\text{c}}=\frac{\text{c}-3}{\text{c}}$
$\Rightarrow6\text{c}=12\times3$ and $3=(\text{c}-3)$
$\Rightarrow\text{c}=\frac{36}{6}$ and $\text{c}-3=3$
$\Rightarrow\text{c}=6$ and $\text{c}=6$
Now,
$\frac{\text{a}_1}{\text{a}_2}=\frac{6}{12}=\frac{1}{2}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{3}{6}=\frac{1}{2}$
$\frac{\text{c}_1}{\text{c}_2}=\frac{-(6-3)}{-6}=\frac{1}{2}$
$\therefore\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_1}$
Clearly, for this value of c, we have $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
Hence, the given system of equations has infinitely many solutions, if $c = 6$.
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