Question

2(x-3)+3(y-5)=0;5(x-1)+4(y-4)=0. solve it by substitution method​

Answer 1

Answer:

x = -3 and y = 9

Step-by-step explanation:

2(x - 3) + 3(y - 5) = 0

2x - 6 + 3y - 15 = 0

2x + 3y - 21 = 0

2x + 3y = 21 -------(1)

5(x - 1) + 4(y - 4) = 0

5x - 5 + 4y - 16 = 0

5x + 4y - 21 = 0

5x + 4y = 21 -------(2)

fron eq 2

5x = 21 - 4y

x = (21 - 4y)/5

put x in eq 1

2x + 3y = 21

2[(21 - 4y)/5] + 3y = 21

[(42 - 8y)/5] + 3y = 21

[42 - 8y + 15y]/5 = 21

42 + 7y = 21 × 5

42 + 7y = 105

7y = 105 - 42

7y = 63

y = 63/7

y = 9

put y in eq 2

5x + 4y = 21

5x + 4(9) = 21

5x + 36 = 21

5x = 21 - 36

5x = -15

x = -15/5

x = -3

Answer 2

Answer:

,

Step-by-step explanation:

hey buddy

here is ur answer....