Question

point m lies between points l and n on (overline{ln}). if (ln = 12x + 16), what is the length of (overline{ln}) in units?

diagram: line segment with endpoints l (left) and n (right), point m between them. segment (lm) is labeled (10x + 8), segment (mn) is labeled (5x - 4).

options:

• 16 units
• 40 units
• 48 units
• 64 units

Explanation:

Step1: Set up the equation

Since M is between L and N, \( LN = LM + MN \). So, \( 12x + 16=(10x + 8)+(5x - 4) \).

Step2: Simplify the right - hand side

Simplify \( (10x + 8)+(5x - 4) \): \( 10x+5x + 8 - 4=15x + 4 \). Now our equation is \( 12x + 16 = 15x+4 \).

Step3: Solve for x

Subtract \( 12x \) from both sides: \( 16=15x - 12x+4 \), which simplifies to \( 16 = 3x+4 \). Then subtract 4 from both sides: \( 16 - 4=3x \), so \( 12 = 3x \). Divide both sides by 3: \( x = 4 \).

Step4: Find the length of LN

Substitute \( x = 4 \) into the expression for \( LN \): \( LN=12x + 16 \). Plugging in \( x = 4 \), we get \( 12\times4+16=48 + 16=64 \). Wait, no, wait. Wait, let's check again. Wait, \( LM = 10x + 8 \), \( MN=5x - 4 \), \( LN=12x + 16 \). So \( 10x + 8+5x - 4=12x + 16 \). Combine like terms: \( 15x + 4=12x + 16 \). Subtract \( 12x \): \( 3x+4 = 16 \). Subtract 4: \( 3x=12 \), so \( x = 4 \). Then \( LN=12\times4 + 16=48 + 16 = 64 \)? Wait, but let's check \( LM=10\times4 + 8=48 \), \( MN=5\times4-4 = 16 \), \( 48 + 16=64 \), and \( 12\times4+16 = 64 \). So that's correct. Wait, but the options are 16,40,48,64. So the length of \( \overline{LN} \) is 64 units? Wait, no, wait, maybe I made a mistake. Wait, let's re - do the equation. \( LM + MN=LN \), so \( (10x + 8)+(5x - 4)=12x + 16 \). \( 10x+5x+8 - 4=12x + 16 \), \( 15x + 4=12x + 16 \), \( 15x-12x=16 - 4 \), \( 3x = 12 \), \( x = 4 \). Then \( LN=12\times4+16=48 + 16 = 64 \). So the length of \( \overline{LN} \) is 64 units.

Answer:

64 units (corresponding to the option "64 units")