Question

please solve this with explanation.​

Answer 1

Answer:

2x + 3y = 4

Let X = 0

2+3y = 4

3y = 2

y = 2/3

Let y = 0

2x+3 = 4

2x = 1

x = 1/2

x+y = 2/3+1/2

= 1 ( 1 approx ).

This above is the proof that the question is correct.

NOW SOL;

2x-3y + x+y = 4+1

2x+x-3y+y = 5

3x-2y = 5

It means your answer can be:-

3x-2y and 5.

NONE OF THESE. (D)

Answer 2

Answer:

$\huge\underline\bold\red{AnswEr}$

The required line is parallel to line x + y = 1 and passes through point (1, 1).

For parallel lines, slope is equal.

x + y = 1

y = -x + 1 => slope is -1

Line passing though point (1, 1) with slope = -1 is given by y - 1 = -(x - 1)

y - 1 = -x + 1

y + x = 1 + 1 = 2

i.e. Equation of the required lin is x + y = 2

Required distance is the distance between point (1, 1) and the point of intersection of lines x + y = 2 and 2x - 3y = 4.

x + y = 2 . . . (1)

2x - 3y = 4 . . . (2)

(1) x 2 => 2x + 2y = 4 . . . (3)

(2) - (3) => -5y = 0 => y = 0

From (1), x + 0 = 2 => x = 2.

i.e. the point of intersection of lines x + y = 2 and 2x - 3y = 4 is (2, 0).

Required distance is

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ =\sqrt{(0 - 1)^2+(2-1)^2}=\sqrt{(-1)^2+1^2} \\ =\sqrt{1+1} \\ = \sqrt{2}$