Question

1. \\(\frac{2x + 5}{6} = \frac{7}{x - 6}\\)

Explanation:

Step1: Cross - multiply the fractions

To solve the equation \(\frac{2x + 5}{6}=\frac{7}{x - 6}\), we use the cross - multiplication property of proportions. If \(\frac{a}{b}=\frac{c}{d}\), then \(a\times d=b\times c\).
So, \((2x + 5)(x - 6)=6\times7\)

Step2: Expand the left - hand side

Using the FOIL method to expand \((2x + 5)(x - 6)\):
\(2x\times x-2x\times6 + 5\times x-5\times6=42\)
\(2x^{2}-12x + 5x-30 = 42\)
Combine like terms:
\(2x^{2}-7x-30 = 42\)

Step3: Rearrange the equation to standard quadratic form

Subtract 42 from both sides of the equation:
\(2x^{2}-7x-30 - 42=0\)
\(2x^{2}-7x-72 = 0\)

Step4: Solve the quadratic equation

We can solve the quadratic equation \(ax^{2}+bx + c = 0\) (here \(a = 2\), \(b=-7\), \(c = - 72\)) using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) or by factoring. Let's try factoring.
We need two numbers that multiply to \(2\times(-72)=-144\) and add up to \(-7\). The numbers are \(-16\) and \(9\) since \(-16\times9=-144\) and \(-16 + 9=-7\).
Rewrite the middle term:
\(2x^{2}-16x+9x - 72 = 0\)
Factor by grouping:
\(2x(x - 8)+9(x - 8)=0\)
\((2x + 9)(x - 8)=0\)
Set each factor equal to zero:
\(2x+9 = 0\) or \(x - 8=0\)
For \(2x+9 = 0\), we have \(2x=-9\), so \(x=-\frac{9}{2}\)
For \(x - 8=0\), we have \(x = 8\)
We also need to check for extraneous solutions. The original equation has denominators \(6\) and \(x - 6\). So \(x
eq6\). Both \(x = 8\) and \(x=-\frac{9}{2}\) do not make the denominator \(x - 6\) equal to zero.

Answer:

\(x = 8\) or \(x=-\frac{9}{2}\)