Question

1. if $lm = 22$ and $mn = 15$, find $ln$.
2. if $ln = 54$ and $lm = 31$, find $mn$.
3. if $df = 9x - 39$, find $ef$.

d 47 e 3x + 10 f

1. if $hj = 7x - 27$, find the value of $x$.

h 3x - 5 i x - 1 j

Explanation:

Step1: Use segment - addition postulate for question 1

Since $LN=LM + MN$, substituting $LM = 22$ and $MN = 15$, we get $LN=22 + 15$.
$LN=37$

Step2: Use segment - addition postulate for question 2

Since $LN=LM + MN$, then $MN=LN - LM$. Substituting $LN = 54$ and $LM = 31$, we get $MN=54 - 31$.
$MN = 23$

Step3: Use segment - addition postulate for question 4

Since $DF=DE + EF$, and $DF = 9x-39$, $DE = 47$, $EF=3x + 10$. Then $9x-39=47+(3x + 10)$.
First, simplify the right - hand side: $9x-39=3x+57$.
Subtract $3x$ from both sides: $9x-3x-39=3x-3x + 57$, which gives $6x-39=57$.
Add 39 to both sides: $6x-39 + 39=57+39$, so $6x=96$.
Divide both sides by 6: $x = 16$. Then $EF=3x + 10=3\times16+10=48 + 10=58$.

Step4: Use segment - addition postulate for question 6

Since $HJ=HI+IJ$, and $HJ = 7x-27$, $HI = 3x-5$, $IJ=x - 1$. Then $7x-27=(3x-5)+(x - 1)$.
Simplify the right - hand side: $7x-27=3x-5+x - 1=4x-6$.
Subtract $4x$ from both sides: $7x-4x-27=4x-4x-6$, so $3x-27=-6$.
Add 27 to both sides: $3x-27 + 27=-6 + 27$, then $3x=21$.
Divide both sides by 3: $x = 7$.

Answer:

1. $LN = 37$
2. $MN = 23$
3. $EF = 58$
4. $x = 7$