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Point U is on line segment TV. Given TV =3x, UV= x + 1, and TU = x determine the numerical length of TU

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We will have to make a sketch of the of the question

From the image above we can deduce that


\begin{gathered} TV=TU+UV \\ \text{Where TU=x} \\ TV=3x \\ UV=x+1 \end{gathered}

By substitution,


\begin{gathered} 3x=x+x+1_{} \\ 3x=2x+1 \\ \text{collect like terms we will have} \\ 3x-2x=1 \\ x=1 \end{gathered}

Therefore,

TU=1

Point U is on line segment TV. Given TV =3x, UV= x + 1, and TU = x determine the numerical-example-1
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