Question

Solve the following systems of equations:
7(y + 3) – 2(x + 2) = 14
4(y – 2) + 3(x – 3) = 2

Answer 1

Point of intersection will be (5, 1)

Step-by-step explanation:

7(y + 3) – 2(x + 2) = 14

7y + 21 - 2x - 4 = 14

-2x + 7y = -3 -------------(1)

4(y – 2) + 3(x – 3) = 2

4y - 8 + 3x - 9 = 2

3x + 4y = 19 -------------(2)

 Adding equations (1)*3 & (2)*2, we get: 29y = +29. So y = 29/29 = 1

Substituting y = 1 in equation 1, we get: -2x + 7 = -3

-2x = - 10

x = 5

Hence solved. x = 5 and y = 1.

If you draw a graph for the equations, you will get two straight lines. The co-ordinates of the point of intersection of the two lines will be the solution. The point of intersection will be (5, 1)

Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Given,

7(y + 3) - 2(x + 2) = 14 .............(1)

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

Now,

From (1) we have,

7y + 21 - 2x - 4 = 14

7y = 14 + 4 - 21 + 2x

7y = 18 - 21 + 2x

7y = 2x - 3

y = (2x - 3)/7 ........... (3)

Also,

From (2) we have,

4y - 8 + 3x - 9 = 2

4y + 3x - 17 - 2 = 0

4y + 3x - 19 = 0 …........(4)

Putting value of y from (3) in (4) we get :-

4(2x - 3)/7 + 3x - 19 = 0

(8x - 12)/7 + 3x = 19

(8x - 12 + 7 × 3x)/7 = 19

(8x - 12 + 21x) = 19 × 7

29x - 12 = 133

29x = 133 + 12

29x = 145

x = 145/29

x = 5

Now,

Substituting value of x in (3),

y = (2x - 3)/7

y = (2 × 5 - 3)/7

y = (10 - 3)/7

y = 7/7

y = 1

Therefore,

x = 5 and y = 1