Use the addition-subtraction method to solve each of the following systems of equations.

1. 4x + 3y = 17 and 5x – 3y = 1

2. –5m + 6n = –8 and –5m + 8n = 6

3. 2x – 7y = –9 and 6x + 7y = 57

4. 5p – 4q = 3 and –5p – 4q = 13

5. 6c + d = 12 and c – d = 2

the best answer will receive brainliest .

1. x=2, y=3

2. n=7, m=-6.8

3. x=6, y=3

4. q=-2, p=-1

5. c=2, d=0

Step-by-step explanation:

4x + 3y = 17

5x – 3y = 1

first make it so either x or y can cancel out in both equations, then combine the equations, giving you 9x=18. solve for x, x=2. substitute 2 for x in either equation, 4(2) + 3y = 17. solve for y, 8+3y=17, 3y=9, y=3

–5m + 6n = –8

–5m + 8n = 6

first make it so either m or n can cancel out in both equations, multiply –5m + 6n = –8 by -1, 5m - 6n = 8, then combine the equations, giving you 2n=14. solve for n, n=7. substitute 7 for n in either equation, 5m - 6(7) = 8. solve for m, 5m - 42 = 8, 5m = -34, m = -6.8

2x – 7y = –9

6x + 7y = 57

first make it so either x or y can cancel out in both equations, then combine the equations, giving you 8x=48. solve for x, x=6. substitute 6 for x in either equation, 2(6) – 7y = –9. solve for y, 12 – 7y = –9, –7y = –21, y = 3

5p – 4q = 3

–5p – 4q = 13

first make it so either p or q can cancel out in both equations, then combine the equations, giving you -8q=16. solve for q, q=-2. substitute -2 for q in either equation, 5p – 4(-2) = 3. solve for p, 5p + 8 = 3, 5p = -5, p = -1

6c + d = 12

c – d = 2

first make it so either c or d can cancel out in both equations, then combine the equations, giving you 7c=14. solve for c, c=2. substitute 2 for c in either equation, 6(2) + d = 12. solve for d, 12 + d = 12, d = 0

[EDIT] question 2 is impossible.

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐