Question

solve for v.
\\(\frac{v}{v + 50} = \frac{1}{v + 6}\\)
there may be 1 or 2 solutions.
v = \boxed{} or v = \boxed{}

Explanation:

Step1: Cross - multiply the equation

To solve the equation \(\frac{v}{v + 50}=\frac{1}{v + 6}\), we use the cross - multiplication property of proportions. If \(\frac{a}{b}=\frac{c}{d}\), then \(a\times d=b\times c\). So, we have \(v(v + 6)=(v + 50)\times1\).

Step2: Expand both sides

Expand the left - hand side: \(v(v + 6)=v^{2}+6v\).
Expand the right - hand side: \((v + 50)\times1=v + 50\).
So, the equation becomes \(v^{2}+6v=v + 50\).

Step3: Rearrange to form a quadratic equation

Subtract \(v\) and \(50\) from both sides of the equation to get a quadratic equation in standard form \(ax^{2}+bx + c = 0\).
\(v^{2}+6v-v - 50=0\)
Simplify the left - hand side: \(v^{2}+5v - 50=0\).

Step4: Factor the quadratic equation

We need to factor the quadratic equation \(v^{2}+5v - 50=0\). We look for two numbers that multiply to \(- 50\) and add up to \(5\). The numbers are \(10\) and \(-5\) because \(10\times(-5)=-50\) and \(10+( - 5)=5\).
So, \(v^{2}+5v - 50=(v + 10)(v - 5)=0\).

Step5: Solve for \(v\)

Set each factor equal to zero and solve for \(v\):

• If \(v + 10=0\), then \(v=-10\).
• If \(v - 5=0\), then \(v = 5\).

We need to check for extraneous solutions by plugging these values back into the original equation.

• For \(v=-10\):

Left - hand side: \(\frac{-10}{-10 + 50}=\frac{-10}{40}=-\frac{1}{4}\)
Right - hand side: \(\frac{1}{-10+6}=\frac{1}{-4}=-\frac{1}{4}\)

• For \(v = 5\):

Left - hand side: \(\frac{5}{5 + 50}=\frac{5}{55}=\frac{1}{11}\)
Right - hand side: \(\frac{1}{5+6}=\frac{1}{11}\)

Both solutions are valid.

Answer:

\(v=-10\) or \(v = 5\)