Question

15. Find the values a and b for which the following pair of linear equations have an infinitely

number of solutions:

(i) ( a + b ) x – 2 b y = 5 a + 2b + 1, 3x – y = 14 (ii) ( 2 a -1 ) x + 3 y = 5 , 3 x + ( b – 1 ) y = 2​

Answer 1

Answer:

(i) a = 5

b=1

Step-by-step explanation:

Infinite no. of solutions are possible if and only if-

$\frac{a_{1} }{a_{2} } = \frac{b_{1} }{b_{2} } =\frac{c_{1} }{c_{2}}$

For (i)

$a_{1}$= (a+b)         $b_{1} = -2\\ b_{2} = -1$

$a_{2}$ = 3

and $c_{1}= -5a-2b-1\\c_{2} = -14$

so, $\frac{a_{1}}{a_{2}}$ = (a+b)/3

and $\frac{b_{1}}{b_{2}} =$ -2/-1= 2

Since $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}$

Therefore

(a+b)/3 = 2

so, a+b =6.................eq. (1)

Also, $\frac{b_{1}}{b_{2}} =\frac{c_{1}}{c_{2}}$

So, 2= (5a + 2b + 1)/14

or 5a + 2b = 27................eq (2)

Solving (1) and (2) , we get

a = 5

b=1

Similarly you can solve for (ii)

Hope this is useful. Please rate the answer!

Answer 2

Step-by-step explanation:

(a + b)x - 2by - (5a + 2b +1) = 0

3x - y - 14 = 0

Here, a¹= 3, b¹= -1, c¹= -14

a²= (a+ b), b²= (-2b) , c²= -(5a + 2b +1)

For equation to have infinite solutions, a¹/a²= b¹/b² =c¹/c²

Now put the values, you shall get 2 equations for each of a, b, c

Add or substract the equations and get result.