Question

g) \\(\frac{3}{x} + \frac{4}{x+1} = 2\\)

Explanation:

Step1: Eliminate denominators

Multiply both sides by \(x(x + 1)\) (the least common denominator) to get rid of the fractions:
\(3(x + 1)+4x = 2x(x + 1)\)

Step2: Expand and simplify

Expand the left - hand side: \(3x+3 + 4x=2x^{2}+2x\)
Combine like terms on the left: \(7x + 3=2x^{2}+2x\)
Rearrange to form a quadratic equation: \(2x^{2}+2x-7x - 3 = 0\), which simplifies to \(2x^{2}-5x - 3=0\)

Step3: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 2\), \(b=-5\), \(c = - 3\)), we can use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) or factor the equation. Let's factor it:
We need two numbers that multiply to \(2\times(-3)=-6\) and add up to \(-5\). The numbers are \(-6\) and \(1\).
Rewrite the middle term: \(2x^{2}-6x+x - 3 = 0\)
Factor by grouping: \(2x(x - 3)+1(x - 3)=0\), so \((2x + 1)(x - 3)=0\)
Set each factor equal to zero:

• \(2x+1 = 0\) gives \(x=-\frac{1}{2}\)
• \(x - 3=0\) gives \(x = 3\)

Step4: Check for extraneous solutions

We need to check if these solutions make the original denominators zero.
For \(x=-\frac{1}{2}\), \(x=-\frac{1}{2}
eq0\) and \(x + 1=-\frac{1}{2}+1=\frac{1}{2}
eq0\)
For \(x = 3\), \(x = 3
eq0\) and \(x + 1=3 + 1=4
eq0\)

Answer:

\(x = 3\) or \(x=-\frac{1}{2}\)