1/a+b+c=1/a+1/b+1/c;a+b+x is not =0,a,b,x is not=0
ANS is -a,-b
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Correct Question is as below:
1/(a + b + x) = (1/a) + (1/b) + (1/x)
Solve for x?
1/(a + b + x) - (1/x) = (1/a) + (1/b)
((x - a - b - x)/(a + b + x)*x) = (a + b)/ab
(-(a + b)/(a + b + x)*x) = (a + b)/ab
Cancelling on (a + b) on both sides, we get
(-1/(a + b + x)*x) = 1/ab
cross multiplying, we get
-ab = (a + b + x)*x
-ab = ax + bx + x^2
Rearranging the above,
x^2 + ax + bx + ab = 0
x(x + a) + b(x + a) = 0
(x + a)(x + b) = 0
x + a = 0 ; x + b = 0
x = -a, x = -b ——> Answer
1/(a + b + x) = (1/a) + (1/b) + (1/x)
Solve for x?
1/(a + b + x) - (1/x) = (1/a) + (1/b)
((x - a - b - x)/(a + b + x)*x) = (a + b)/ab
(-(a + b)/(a + b + x)*x) = (a + b)/ab
Cancelling on (a + b) on both sides, we get
(-1/(a + b + x)*x) = 1/ab
cross multiplying, we get
-ab = (a + b + x)*x
-ab = ax + bx + x^2
Rearranging the above,
x^2 + ax + bx + ab = 0
x(x + a) + b(x + a) = 0
(x + a)(x + b) = 0
x + a = 0 ; x + b = 0
x = -a, x = -b ——> Answer
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