Question

(\frac{x^{2}-29}{x^{2}-10x + 21}=\frac{6}{x - 7}+\frac{5}{x - 3})

Explanation:

Step1: Factor the denominator

Factor \(x^{2}-10x + 21\) as \((x - 7)(x - 3)\). The equation becomes \(\frac{x^{2}-29}{(x - 7)(x - 3)}=\frac{6}{x - 7}+\frac{5}{x - 3}\).

Step2: Multiply through by \((x - 7)(x - 3)\)

Multiply each term by \((x - 7)(x - 3)\) to eliminate the denominators: \(x^{2}-29=6(x - 3)+5(x - 7)\).

Step3: Expand the right - hand side

Expand \(6(x - 3)+5(x - 7)\): \(6x-18 + 5x-35=11x-53\). So the equation is \(x^{2}-29 = 11x-53\).

Step4: Rearrange into standard quadratic form

Rearrange the equation to \(x^{2}-11x + 24 = 0\).

Step5: Factor the quadratic equation

Factor \(x^{2}-11x + 24\): We need two numbers that multiply to 24 and add to - 11. The numbers are - 3 and - 8. So \((x - 3)(x - 8)=0\).

Step6: Solve for x

Set each factor equal to zero: \(x-3 = 0\) or \(x - 8=0\). But \(x = 3\) makes the original denominators \(x - 3\) and \((x - 7)(x - 3)\) equal to zero, so we discard \(x = 3\). When \(x=8\), we check the original equation:
Left - hand side: \(\frac{8^{2}-29}{8^{2}-10\times8 + 21}=\frac{64 - 29}{64-80 + 21}=\frac{35}{5}=7\)
Right - hand side: \(\frac{6}{8 - 7}+\frac{5}{8 - 3}=6 + 1=7\)

Answer:

\(x = 8\)