how to find the value of y by substituting the value of x=b1c2-b2c1/a1b2-a2b1 in equation a1x+b1y+c=0
Answers
Answer:
Let’s assume that we have to find solution for a1x + b1y + c1 = 0 (i) and a2x + b2y + c2 = 0 (ii)
Step 1: Multiply Equation (i) by b2 and Equation (ii) by b1, to get
b2a1x + b2b1y + b2c1 = 0 ---(iii)
b1a2x + b1b2 y + b1c2 = 0 ----(iv)
Step 2: Subtracting Equation (4) from (3), we get:
(b2a1 – b1a2) x + (b2b1 – b1b2) y + (b2c1– b1c2) = 0
Or (b2a1 – b1a2) x + (b2c1– b1c2) = 0
Or x = (b1c2– b2c1)/ (a1b2 – a2b1)
Step 3: Substituting this value of x in (i) or (ii), we get
y = (c1a2 – c2a1)/ (a1b2 – a2b1)
Step 4: Calculate value of a1b2 – a2b1
Step 5: if a1b2 – a2b1 ≠ 0 or a1/a2 ≠ b1/b2, then equation has definite solution.
Step 6: if a1b2 –a2b1 =0 or a1/a2 = b1/b2, then there are two possibilities
When a1/a2 = b1/b2 = c1/c2 , then there are infinitely many solutions.
When a1/a2 = b1/b2 ≠ c1/c2, then there is no solution.