In /_\IJK,m/_I=(x+6)^(@),m/_J=(2x+16)^(@), and m/_K=(6x+14)^(@). What is the value of x ? Answer:

Set up equation: In a triangle, the sum of the angles is always 180 degrees. We can set up an equation with the given angle measures to find x. m/_I + m/_J + m/_K = 180^(@) (x+6)^(@) + (2x+16)^(@) + (6x+14)^(@) = 180^(@)Combine like terms: Combine like terms to simplify the equation. x + 6 + 2x + 16 + 6x + 14 = 180 9x + 36 = 180Isolate terms with x: Subtract 36 from both sides to isolate the terms with x. 9x + 36 - 36 = 180 - 36 9x = 144Solve for x: Divide both sides by 9 to solve for x. 9x / 9 = 144 / 9 x = 16

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Math Problems

Grade 8

Solve equations with variables on both sides: fractional coefficients

Full solution

Q. In

\triangle \mathrm{IJK}, \mathrm{m} \angle I=(x+6)^{\circ}, \mathrm{m} \angle J=(2 x+16)^{\circ}

, and

\mathrm{m} \angle K=(6 x+14)^{\circ}

. What is the value of

x

\newline

Answer:

Set up equation: In a triangle, the sum of the angles is always $180$ degrees. We can set up an equation with the given angle measures to find $x$ .
$m/_{I} + m/_{J} + m/_{K} = 180^{\circ}$
$(x+6)^{\circ} + (2x+16)^{\circ} + (6x+14)^{\circ} = 180^{\circ}$
Combine like terms: Combine like terms to simplify the equation. $\newline$ $x + 6 + 2x + 16 + 6x + 14 = 180$ $\newline$ $9x + 36 = 180$
Isolate terms with x: Subtract $36$ from both sides to isolate the terms with $x$ . $\newline$ $9x + 36 - 36 = 180 - 36$ $\newline$ $9x = 144$
Solve for x: Divide both sides by $9$ to solve for x. $\newline$ $\frac{9x}{9} = \frac{144}{9}$ $\newline$ $x = 16$