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$In /_\KLM, bar(KM) is extended through point M to point N,m/_MKL=(3x+1)^(@), m/_LMN=(7x-8)^(@), and m/_KLM=(2x+11)^(@). What is the value of x ? Answer:$

In $\triangle \mathrm{KLM}, \overline{K M}$ is extended through point $\mathrm{M}$ to point $\mathrm{N}, \mathrm{m} \angle M K L=(3 x+1)^{\circ}$ , $\mathrm{m} \angle L M N=(7 x-8)^{\circ}$ , and $\mathrm{m} \angle K L M=(2 x+11)^{\circ}$ . What is the value of $x$ ? $\newline$ Answer:

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Algebra 1

Transformations of quadratic functions

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Q. In

\triangle \mathrm{KLM}, \overline{K M}

is extended through point

\mathrm{M}

to point

\mathrm{N}, \mathrm{m} \angle M K L=(3 x+1)^{\circ}

\mathrm{m} \angle L M N=(7 x-8)^{\circ}

, and

\mathrm{m} \angle K L M=(2 x+11)^{\circ}

. What is the value of

x

\newline

Answer:

Understand angles relationship: Understand the relationship between the angles in the problem. $\newline$ The sum of the angles in triangle $KLM$ and the exterior angle at $M$ (angle $LMN$ ) must be equal to the straight line angle at $M$ , which is $180$ degrees.
Set up equation: Set up the equation based on the angle sum property and the exterior angle theorem. $\newline$ The exterior angle theorem states that the measure of an exterior angle (angle LMN) is equal to the sum of the measures of the two non-adjacent interior angles (angles MKL and KLM). $\newline$ So, we have: $\newline$ $m/_{\text{MKL}} + m/_{\text{KLM}} = m/_{\text{LMN}}$ $\newline$ Substitute the given expressions: $\newline$ $(3x + 1) + (2x + 11) = (7x - 8)$
Combine terms and solve: Combine like terms and solve for $x$ .
$(3x + 1) + (2x + 11) = (7x - 8)$
$3x + 2x + 1 + 11 = 7x - 8$
$5x + 12 = 7x - 8$
Move $x$ and constant terms: Move the $x$ terms to one side and the constant terms to the other side. $5x + 12 - 5x = 7x - 8 - 5x$ $12 = 2x - 8$
Isolate x value: Isolate x by adding $8$ to both sides of the equation. $\newline$ $12 + 8 = 2x - 8 + 8$ $\newline$ $20 = 2x$
Divide to find x: Divide both sides by $2$ to find the value of $x$ . $\frac{20}{2} = \frac{2x}{2}$ $10 = x$