Solve for x: 1/((a + b + x)) = 1/a + 1/b + 1/x
[a ≠ 0, b ≠ 0, x ≠ 0, x ≠ –(a + b)]
CBSE Class 10 Sample Paper for 2019 Boards
CBSE Class 10 Sample Paper for 2019 Boards
Last updated at July 14, 2026 by Teachoo
Transcript
Question 23 (OR 2nd question) Solve for x: 1/((𝑎 + 𝑏 + 𝑥)) = 1/𝑎 + 1/𝑏 + 1/𝑥 [a ≠ 0, b ≠ 0, x ≠ 0, x ≠ –(a + b)] 1/((𝑎 + 𝑏 + 𝑥)) = 1/𝑎 + 1/𝑏 + 1/𝑥 1/((𝑎 + 𝑏 + 𝑥)) – 1/𝑥 = 1/𝑎 + 1/𝑏 (𝑥 − (𝑎 + 𝑏 + 𝑥))/(𝑎 + 𝑏 + 𝑥)𝑥 = 1/𝑎 + 1/𝑏 (𝑥 − 𝑎 − 𝑏 − 𝑥)/(𝑎 + 𝑏 + 𝑥)𝑥 = (𝑏 + 𝑎)/𝑎𝑏 (−(𝑎 + 𝑏))/(𝑎 + 𝑏 + 𝑥)𝑥 = ((𝑎 + 𝑏))/𝑎𝑏 (−1)/(𝑎 + 𝑏 + 𝑥)𝑥 = 1/𝑎𝑏 –ab = x(a + b + x) –ab = x(a + b) + x2 0 = x2 + x(a + b) + ab x2 + x(a + b) + ab = 0 Comparing with Ax2 + BX + C = 0 A = 1, B = (a + b), C = ab x = (−𝐵 ± √(𝐵^(2 )− 4𝐴𝐶) )/2𝐴 x = (−(𝑎 + 𝑏) ± √(〖(𝑎 + 𝑏)〗^(2 )− 4×1×𝑎𝑏) )/(2 × 1) x = (−(𝑎 + 𝑏) ± √(𝑎^2 + 𝑏^2 + 2𝑎𝑏 − 4𝑎𝑏) )/2 x = (−(𝑎 + 𝑏) ± √(𝑎^2 + 𝑏^2 − 2𝑎𝑏) )/2 x = (−(𝑎 + 𝑏) ± √((𝑎 − 𝑏)^2 ) )/2 x = (−(𝑎 + 𝑏) ± (𝑎 − 𝑏) )/2 So, x = (−(𝑎 + 𝑏) + (𝑎 − 𝑏) )/2 x = (−𝑎 − 𝑏 + 𝑎 − 𝑏)/2 x = (−2𝑏)/2 x = –b x = (−(𝑎 + 𝑏) − (𝑎 − 𝑏) )/2 x = (−𝑎 − 𝑏 − 𝑎+ 𝑏)/2 x = (−2𝑎)/2 x = –a