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3 years ago

Point K is on line JL. given JL =4x, JK= 2x+3, and KL=x, determine the numerical length of kl

Mathematics

1 answer:

balandron [24]3 years ago

5 0

Hi! I'm happy to help!

Our total line is JL (4x), and it is split into two parts: JK, and KL. We have our values, and we know that JK+KL=JL, so we can substitute our values and solve for x:

4x=(2x+3)+(x)

4x=3x+3

To solve for x, we have to isolate it on one side of the equation.

First, let's subtract 3x from both sides so that we can isolate x:

4x=3x+3

-3x -3x

x=3

<u>So, our x=3, which means that KL=3.</u>

I hope this was helpful, keep learning! :D

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$
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(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
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(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
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x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

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Read 2 more answers

Point K is on line JL. given JL =4x, JK= 2x+3, and KL=x, determine the numerical length of kl​

Point K is on line JL. given JL =4x, JK= 2x+3, and KL=x, determine the numerical length of kl