Answer: 5x-6y-9=0. (1 equation) 3x+4y-25=0. (2 equation) Multiplying equation 2 by 5 and equation 1 by 3 we get, 15x-18y-27=0. (3 equation) 15x+20y-125=0. (4 equation) Now, Solving equations 3 & 4 by substitution method Substracting equation 4 from 3, we get, 15x-18y-27=0 -15x-20y+125=0 x portion will be cancelled out and remaining is -38y+97=0 38y=97 y=2.55 Substituting the value of y in equation 1 we get, 5x-6×2.55-9=0 5x-24.3=0 5x=24.3 x=4.86 Reply
Answer:
5x-6y-9=0. (1 equation)
3x+4y-25=0. (2 equation)
Multiplying equation 2 by 5 and equation 1 by 3 we get,
15x-18y-27=0. (3 equation)
15x+20y-125=0. (4 equation)
Now, Solving equations 3 & 4 by substitution method
Substracting equation 4 from 3, we get,
15x-18y-27=0
-15x-20y+125=0
x portion will be cancelled out and remaining is
-38y+97=0
38y=97
y=2.55
Substituting the value of y in equation 1 we get,
5x-6×2.55-9=0
5x-24.3=0
5x=24.3
x=4.86