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MATHMAA

# Linear Equation Two Variable3

Linear Equation Two Variable3 is the continuation of previous page.

7. For what value of 'k' the following pair has infinite number of solutions.

(k-3)x + 3y = k
k(x+y) = 12

Solution:-

If the lines $$a_1x + b_1y + c_1$$ = 0, $$a_2x + b_2y + c_2$$ = 0 has infinite number of solution then it must satisfy the condition -

$$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$

We have the lines,

(k-3)x + 3y = k
kx + ky = 12
Here,

$$a_1 = k-3, b_1 = 3, c_1 = k$$
$$a_2 = k, b_2 = k, c_2 = 12$$

Therefore,

$$\frac{k-3}{k} = \frac{3}{k} = \frac{k}{12}$$
$$\frac{k-3}{k} = \frac{k}{12}$$
12(k-3) = k(k)
$$12k -36 = k^2$$
$$k^2 - 12k + 36 = 0$$
$$(k-6)^2 = 0$$
k - 6 = 0
k = 6

Therefore the value of k is 6.

8. What is the condition so that $$a_1x + b_1y = c_1$$ , $$a_2x + b_2y = c_2$$ have unique solution.

Solution:-

The condition so that $$a_1x + b_1y = c_1$$ , $$a_2x + b_2y = c_2$$ have unique solution is

$$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$$.

What We Learn In
Linear Equation Two Variable3:

• In previous page we learned that if the pair of linear equation in two variable has infinitely many solutions then it is called dependent pair of linear equations.Now, in this page we learned about the case in which the given pair of linear equations have infinite number of solutions. The case is :
If
$$a_1x + b_1y = c_1$$ , $$a_2x + b_2y = c_2$$ are pair of linear equation in two variable, they have infinitely many solution when
$$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$.
• The conditions for which $$a_1x + b_1y = c_1$$ , $$a_2x + b_2y = c_2$$ have unique solution is
$$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$$.