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Solve 
(a)/(x+a)+(b)/(x-b)=(a+b)/(x+c) for 
x

Solve ax+a+bxb=a+bx+c \frac{a}{x+a}+\frac{b}{x-b}=\frac{a+b}{x+c} for x x

Full solution

Q. Solve ax+a+bxb=a+bx+c \frac{a}{x+a}+\frac{b}{x-b}=\frac{a+b}{x+c} for x x
  1. Rewrite with common denominator: Now, rewrite each fraction with the common denominator: a(xb)(x+a)(xb)+b(x+a)(x+a)(xb)=a+bx+c\frac{a(x-b)}{(x+a)(x-b)} + \frac{b(x+a)}{(x+a)(x-b)} = \frac{a+b}{x+c}.
  2. Combine fractions: Combine the fractions on the left side: (axab)+(bx+ab)(x+a)(xb)=a+bx+c\frac{(ax - ab) + (bx + ab)}{(x+a)(x-b)} = \frac{a+b}{x+c}.
  3. Simplify numerator: Simplify the numerator on the left side: (ax+bx)/(x+a)(xb)=(a+b)/(x+c)(ax + bx)/(x+a)(x-b) = (a+b)/(x+c).
  4. Combine like terms: Combine like terms in the numerator: (a+b)x/(x+a)(xb)=(a+b)/(x+c)(a+b)x/(x+a)(x-b) = (a+b)/(x+c).
  5. Cancel out common factor: Since (a+b)(a+b) appears in both numerators, we can cancel it out, assuming a+b0a+b \neq 0:x(x+a)(xb)=1(x+c).\frac{x}{(x+a)(x-b)} = \frac{1}{(x+c)}.
  6. Cross-multiply: Cross-multiply to get rid of the fractions: x(x+c)=(x+a)(xb)x(x+c) = (x+a)(x-b).
  7. Expand both sides: Expand both sides: x2+cx=x2+axbxabx^2 + cx = x^2 + ax - bx - ab.
  8. Subtract x2x^2: Subtract x2x^2 from both sides to simplify: cx=axbxabcx = ax - bx - ab.
  9. Combine like terms: Combine like terms: cx=(ab)xabcx = (a - b)x - ab.
  10. Subtract (ab)x(a - b)x: Subtract (ab)x(a - b)x from both sides:\newlinecx(ab)x=abcx - (a - b)x = -ab.
  11. Factor out xx: Factor out xx on the left side:\newlinex(ca+b)=abx(c - a + b) = -ab.
  12. Divide to solve for xx: Divide both sides by (ca+b)(c - a + b) to solve for xx:x=ab(ca+b).x = \frac{-ab}{(c - a + b)}.

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