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solve for v. ( v + \frac{28}{v} = 12 - \frac{7}{v} ) if there is more t…

Question

solve for v.

( v + \frac{28}{v} = 12 - \frac{7}{v} )

if there is more than one solution, separate them with commas
if there is no solution, click on
o solution\.

Explanation:

Step1: Eliminate the denominator

Multiply both sides of the equation \( v+\frac{28}{v}=12 - \frac{7}{v} \) by \( v \) (assuming \( v
eq0 \)) to get rid of the fractions. We have \( v\times v + v\times\frac{28}{v}=v\times12 - v\times\frac{7}{v} \), which simplifies to \( v^{2}+28 = 12v-7 \).

Step2: Rearrange into standard quadratic form

Move all terms to one side of the equation to form a quadratic equation. Subtract \( 12v \) and add \( 7 \) to both sides: \( v^{2}-12v + 28 + 7=0 \), so \( v^{2}-12v + 35 = 0 \).

Step3: Factor the quadratic equation

Factor the quadratic \( v^{2}-12v + 35 \). We need two numbers that multiply to \( 35 \) and add up to \( - 12 \). The numbers are \( - 5 \) and \( - 7 \). So, \( (v - 5)(v - 7)=0 \).

Step4: Solve for \( v \)

Set each factor equal to zero:

  • For \( v - 5=0 \), we get \( v = 5 \).
  • For \( v - 7=0 \), we get \( v = 7 \).

We need to check if these solutions are valid (since we multiplied by \( v \) earlier, we must ensure \( v
eq0 \), and both \( 5 \) and \( 7 \) are not zero, so they are valid.

Answer:

5, 7