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$In /_\BCD, bar(BD) is extended through point D to point E,m/_CDE=(6x+7)^(@), m/_BCD=(3x+9)^(@), and m/_DBC=(x+20)^(@). What is the value of x? Answer:$

In $\triangle \mathrm{BCD}, \overline{B D}$ is extended through point $\mathrm{D}$ to point $\mathrm{E}, \mathrm{m} \angle C D E=(6 x+7)^{\circ}$ , $\mathrm{m} \angle B C D=(3 x+9)^{\circ}$ , and $\mathrm{m} \angle D B C=(x+20)^{\circ}$ . What is the value of $x ?$ $\newline$ Answer:

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Math Problems

Grade 8

Solve equations with variables on both sides: fractional coefficients

Full solution

Q. In

\triangle \mathrm{BCD}, \overline{B D}

is extended through point

\mathrm{D}

to point

\mathrm{E}, \mathrm{m} \angle C D E=(6 x+7)^{\circ}

\mathrm{m} \angle B C D=(3 x+9)^{\circ}

, and

\mathrm{m} \angle D B C=(x+20)^{\circ}

. What is the value of

x ?

\newline

Answer:

Triangle Interior Angles: We know that the sum of the interior angles in any triangle is $180$ degrees. Since triangle BCD is extended to form an exterior angle at point D, the exterior angle CDE is equal to the sum of the two non-adjacent interior angles, which are angle BCD and angle DBC. $\newline$ So, we can write the equation: $\newline$ $m/\_CDE = m/\_BCD + m/\_DBC$ $\newline$ Substitute the given angle measures into the equation: $\newline$ $(6x + 7) = (3x + 9) + (x + 20)$
Exterior Angle Calculation: Now, we will combine like terms on the right side of the equation: $\newline$ $(6x + 7) = (3x + 9) + (x + 20)$ $\newline$ $(6x + 7) = 3x + 9 + x + 20$ $\newline$ $(6x + 7) = 4x + 29$
Combine Like Terms: Next, we will isolate the variable $x$ by subtracting $4x$ from both sides of the equation: $\newline$ $6x + 7 - 4x = 4x + 29 - 4x$ $\newline$ $2x + 7 = 29$
Isolate Variable $x$ : Now, we will subtract $7$ from both sides to solve for $x$ : $2x + 7 - 7 = 29 - 7$ $2x = 22$
Solve for x: Finally, we will divide both sides by $2$ to find the value of x: $\newline$ $\frac{2x}{2} = \frac{22}{2}$ $\newline$ $x = 11$