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How do you solve for v in #u(v+2) + w(v-3) = z(v-1)#?

Answer 1

#v=(3w-z-2u)/(w-z+u)#

Technique:

Multiply out the brackets. All terms with #v# in on the left. All other terms on the right Factor out #v# Rearrange so that #v# is on its own. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #uv+2u+wv-3w=zv-z#

#uv+wv-zv=3w-z-2u#

#v(u+w-z)=3w-z-2u#

#v=(3w-z-2u)/(w-z+u)#

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Answer 2

Julia Adams

To solve for ( v ), first distribute ( u ) and ( w ) into their respective parentheses, then combine like terms and isolate ( v ). Here are the steps:

Distribute ( u ) and ( w ) into their respective parentheses: [ uv + 2u + wv - 3w = zv - z ]
Combine like terms: [ uv + wv + 2u - 3w = zv - z ]
Rearrange the equation to isolate ( v ) terms on one side: [ uv + wv - zv = -2u + 3w - z ]
Factor out ( v ) on the left side: [ v(u + w - z) = -2u + 3w - z ]
Divide both sides by ( u + w - z ) to solve for ( v ): [ v = \frac{-2u + 3w - z}{u + w - z} ]

So, ( v = \frac{-2u + 3w - z}{u + w - z} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.