Question

Find the point of intersection of the ines 5a+ 7b = 3 and 2a – 3b =7.​

Answer 1

Answer:

a= 2 and b = - 1

Step-by-step explanation:

Let 5a+7b=3....(1)

. . . 2a-3b=7......(2)

(1)*2. (2)*5

(2)-(1)

a= 2. b= - 1

Answer 2

Answer:

Required intersection point of two given lines is (2,-1)

Step-by-step explanation:

Here given two lines are $5a + 7b = 3....(1)$ and $2a - 3b = 7.....(2)$

We want to find point of intersection of these two lines.

By property of straight line It is clear that (a,b) is point of intersection of given two lines.

From equation (1),

$5a = 3 - 7b \\ a = \frac{3 - 7b}{5} .....(3)$

and from equation (2),

$2a = 7 + 3b \\ a = \frac{7 + 3b}{2} ......(4)$

From equation (3) and (4),

$\frac{3 - 7b}{5} = \frac{7 + 3b}{2}$

By cross multiplication,

$2 \times (3 - 7b) = 5 \times (7 + 3b) \\ 6 - 14b = 35 + 15b \\ 15b + 14b = 6 - 35 \\ 29b = - 29 \\ b = \frac{ - 29}{29} \\ b = - 1$

We are putting b = (-1) in equation (3),

$a = \frac{3 - 7 \times ( - 1)}{5} \\ a = \frac{3 + 7}{5} \\ a = \frac{10}{5} \\ a = 2$

So,  value of a is 2 and value of b is (-1).

Therefore required intersection point of two given lines is (2,-1)

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