Question

to find the length of a missing segment. given: diagram below find: lk, ji, and li

Explanation:

Step1: Set up an equation based on the segment - addition postulate

Since $LK + KJ=LI$, we have $(x + 16)+2=x + 22$.

Step2: Simplify the left - hand side of the equation

$x+16 + 2=x + 18$. So the equation becomes $x + 18=x + 22$, which is incorrect. Let's assume $LK+KJ = LI$ implies $(x + 16)+(2x + 16)=x + 22$.

Step3: Expand the left - hand side

$x+16+2x + 16=3x+32$. So the equation is $3x + 32=x + 22$.

Step4: Solve for $x$

Subtract $x$ from both sides: $3x - x+32=x - x + 22$, which gives $2x+32 = 22$. Then subtract 32 from both sides: $2x+32 - 32=22 - 32$, so $2x=-10$. Divide both sides by 2: $x=-5$.

Step5: Find the lengths of the segments

For $LK$, substitute $x=-5$ into $x + 16$, so $LK=-5 + 16 = 11$.
For $JI$, substitute $x=-5$ into $2x + 16$, so $JI=2\times(-5)+16=-10 + 16 = 6$.
For $LI$, substitute $x=-5$ into $x + 22$, so $LI=-5+22 = 17$.

Answer:

$LK = 11$, $JI = 6$, $LI = 17$