Math Problem Statement

Solve and explains the answers to both A and B

Solve and explain the answers to both A and B based on the image of two geometric segments with algebraic expressions.

Solution

Let's solve the problems step by step for both parts A and B based on the image.

Part (a):

You have a segment ACAC with point BB in between. From the image, it seems:

  • Segment AB=x+10AB = x + 10
  • Segment BC=2x+6BC = 2x + 6
  • The whole segment AC=5AC = 5

Since BB lies on the segment ACAC, we can use the segment addition postulate:

AB+BC=ACAB + BC = AC

Substitute the given expressions into this equation:

(x+10)+(2x+6)=5(x + 10) + (2x + 6) = 5

Now, simplify:

x+10+2x+6=5x + 10 + 2x + 6 = 5

3x+16=53x + 16 = 5

Subtract 16 from both sides:

3x=5163x = 5 - 16

3x=113x = -11

Now, divide by 3:

x=113x = \frac{-11}{3}

This is the value of xx. To find ABAB and BCBC, substitute x=113x = \frac{-11}{3} into the expressions for ABAB and BCBC:

AB=(113)+10=113+303=193AB = \left( \frac{-11}{3} \right) + 10 = \frac{-11}{3} + \frac{30}{3} = \frac{19}{3}

BC=2(113)+6=223+183=43BC = 2\left( \frac{-11}{3} \right) + 6 = \frac{-22}{3} + \frac{18}{3} = \frac{-4}{3}

Thus:

  • AB=193AB = \frac{19}{3}
  • BC=43BC = \frac{-4}{3}

Part (b):

Again, you have a segment ACAC with point BB in between. The details are:

  • Segment AB=9x+7AB = 9x + 7
  • Segment BC=3x+20BC = -3x + 20
  • The whole segment AC=53AC = 53

By the segment addition postulate:

AB+BC=ACAB + BC = AC

Substitute the given expressions:

(9x+7)+(3x+20)=53(9x + 7) + (-3x + 20) = 53

Simplify:

9x+73x+20=539x + 7 - 3x + 20 = 53

6x+27=536x + 27 = 53

Subtract 27 from both sides:

6x=53276x = 53 - 27

6x=266x = 26

Now divide by 6:

x=266=133x = \frac{26}{6} = \frac{13}{3}

This is the value of xx. To find ABAB and BCBC, substitute x=133x = \frac{13}{3} into the expressions for ABAB and BCBC:

AB=9(133)+7=1173+7=39+7=46AB = 9\left( \frac{13}{3} \right) + 7 = \frac{117}{3} + 7 = 39 + 7 = 46

BC=3(133)+20=13+20=7BC = -3\left( \frac{13}{3} \right) + 20 = -13 + 20 = 7

Thus:

  • AB=46AB = 46
  • BC=7BC = 7

Summary of Results:

  • (a) x=113x = \frac{-11}{3}, AB=193AB = \frac{19}{3}, and BC=43BC = \frac{-4}{3}
  • (b) x=133x = \frac{13}{3}, AB=46AB = 46, and BC=7BC = 7

Would you like further details on any part of this solution, or do you have any questions?

Related Questions:

  1. How do we calculate segment lengths when multiple variables are involved?
  2. What does the segment addition postulate imply in geometry?
  3. How do we solve equations with fractions and integers combined?
  4. Can we always determine the position of points on a line given algebraic expressions?
  5. What would happen if the total length was less than the sum of the segments in part (a)?

Tip: Always simplify both sides of an equation before isolating the variable to avoid errors.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Algebra
Geometry
Segment Addition Postulate

Formulas

Segment Addition Postulate: AB + BC = AC
Linear Equation Solution: Combine like terms, isolate variable

Theorems

Segment Addition Postulate

Suitable Grade Level

Grades 8-10