Question

solve for v. \\(\frac{v - 8}{v - 2} = \frac{v + 7}{v + 8}\\) there may be 1 or 2 solutions. \\(v = \square\\) or \\(v = \square\\)

Explanation:

Step1: Cross - multiply the fractions

To solve the equation \(\frac{v - 8}{v - 2}=\frac{v + 7}{v + 8}\), we use the cross - multiplication property of fractions. If \(\frac{a}{b}=\frac{c}{d}\) (where \(b
eq0\) and \(d
eq0\)), then \(a\times d=c\times b\).
So, \((v - 8)(v + 8)=(v + 7)(v - 2)\)

Step2: Expand both sides

We use the difference of squares formula \((a - b)(a + b)=a^{2}-b^{2}\) on the left - hand side and the distributive property \((a + b)(c + d)=ac+ad+bc+bd\) on the right - hand side.

• Left - hand side: \((v - 8)(v + 8)=v^{2}-8^{2}=v^{2}-64\)
• Right - hand side: \((v + 7)(v - 2)=v\times v-2\times v+7\times v - 2\times7=v^{2}-2v + 7v-14=v^{2}+5v - 14\)

Step3: Simplify the equation

Set the expanded forms equal to each other:
\(v^{2}-64=v^{2}+5v - 14\)
Subtract \(v^{2}\) from both sides of the equation:
\(v^{2}-v^{2}-64=v^{2}-v^{2}+5v - 14\)
\(-64 = 5v-14\)

Step4: Solve for \(v\)

Add 14 to both sides of the equation:
\(-64 + 14=5v-14 + 14\)
\(-50 = 5v\)
Divide both sides by 5:
\(v=\frac{-50}{5}=-10\)

We need to check for extraneous solutions. The original equation has denominators \(v - 2\) and \(v + 8\). When \(v=-10\), \(v - 2=-10 - 2=-12
eq0\) and \(v + 8=-10 + 8=-2
eq0\). So \(v = - 10\) is a valid solution. Wait, let's re - check our expansion:

Wait, \((v + 7)(v - 2)=v^{2}-2v+7v - 14=v^{2}+5v - 14\) (correct). \((v - 8)(v + 8)=v^{2}-64\) (correct). Then \(v^{2}-64=v^{2}+5v - 14\), subtract \(v^{2}\): \(-64 = 5v-14\), add 14: \(5v=-64 + 14=-50\), \(v=-10\). But let's substitute \(v = - 10\) back into the original equation:

Left - hand side: \(\frac{-10 - 8}{-10 - 2}=\frac{-18}{-12}=\frac{3}{2}\)

Right - hand side: \(\frac{-10 + 7}{-10 + 8}=\frac{-3}{-2}=\frac{3}{2}\)

So it works. Wait, but the problem said there may be 1 or 2 solutions. Did we make a mistake? Let's re - do the cross - multiplication.

\(\frac{v - 8}{v - 2}=\frac{v + 7}{v + 8}\)

Cross - multiply: \((v - 8)(v + 8)=(v + 7)(v - 2)\)

\(v^{2}-64=v^{2}+5v - 14\)

Subtract \(v^{2}\): \(-64=5v - 14\)

\(5v=-64 + 14=-50\)

\(v=-10\)

Wait, maybe I missed a solution? Let's check the algebra again.

Starting from \(\frac{v - 8}{v - 2}=\frac{v + 7}{v + 8}\)

Cross - multiply: \((v - 8)(v + 8)=(v + 7)(v - 2)\)

\(v^{2}-64=v^{2}+5v - 14\)

\(0 = 5v+50\)

\(5v=-50\)

\(v=-10\)

Yes, only one solution.

Answer:

\(v=-10\) (Since the problem said there may be 1 or 2 solutions, but in this case, there is 1 solution. If we made a mistake in the initial assumption, but according to the algebra, \(v=-10\) is the solution.)