Question

1. AC = 95, AB = 15x - 10, BC = 5x + 5

2. AC = 8x - 16, AB = 3x - 8, BC = 4x
3. AC = x - 0.4, AB = x - 4.9, BC = 0.5x
4. AC = 383, AB = 6x, BC = 8x + 1

Answer 1

Part 1) x=5, AB=65 units, BC=30 units

Part 2) x=8, AB=16 units, BC=32 units

Part 3) x=9, AB=4.1 units, BC=4.5 units

Part 4) x=2 3/4, AB=16 1/2 units, BC=22 1/4 units

Explanation:

The complete question in the attached figure

we have that

Point B is between A and C on segment AC

we know that

`AC=AB+BC` -----> equation A (by addition segment postulate)

Part 1) we have

AC = 95, AB = 15x - 10, BC = 5x + 5

substitute the given values in equation A and solve for x

`95=(15x-10)+(5x+5)`

Combine like terms in the right side

`95=(20x-5)`

Adds 5 both sides

`95+5=20x-5+5`

`100=20x`

Divide by 20 both sides

`100/20=20x/20`

`5=x`

Rewrite

`x=5`

Find the value of AB

`AB=(15x-10)`

substitute the value of x

`AB=(15(5)-10)=65\ units`

Find the value of BC

`BC=(5x+5)`

substitute the value of x

`BC=(5(5)+5)=30\ units`

Part 2) we have

AC = 8x - 16, AB = 3x - 8, BC = 4x

substitute the given values in equation A and solve for x

`8x-16=(3x-8)+(4x)`

Combine like terms in the right side

`8x-16=(7x-8)`

Adds 16 both sides

`8x-16+16=7x-8+16`

`8x=7x+8`

Subtract 7x both sides

`8x-7x=7x+8-7x`

`x=8`

Find the value of AB

`AB=(3x-8)`

substitute the value of x

`AB=(3(8)-8)=16\ units`

Find the value of BC

`BC=(4x)`

substitute the value of x

`BC=4(8)=32\ units`

Part 3) we have

AC = x - 0.4, AB = x - 4.9, BC = 0.5x

substitute the given values in equation A and solve for x

`x-0.4=(x-4.9)+(0.5x)`

Combine like terms in the right side

`x-0.4=(1.5x-4.9)`

Adds 4.9 both sides

`x-0.4+4.9=1.5x-4.9+4.9`

`x+4.5=1.5x`

Subtract x both sides

`x+4.5-x=1.5x-x`

`4.5=0.5x`

Divide by 0.5 both sides

`4.5/0.5=0.5x/0.5`

`9=x`

Rewrite

`x=9`

Find the value of AB

`AB=(x-4.9)`

substitute the value of x

`AB=(9-4.9)=4.1\ units`

Find the value of BC

`BC=(0.5x)`

substitute the value of x

`BC=0.5(9)=4.5\ units`

Part 4) we have

AC = 38 3/4, AB = 6x, BC = 8x + 1/4

substitute the given values in equation A and solve for x

`38(3)/(4)=(6x)+(8x+(1)/(4))`

Convert mixed number to an improper fraction

`38(3)/(4)=38+(3)/(4)=(38*4+3)/(4)=(155)/(4)`

substitute

`(155)/(4)=(6x)+(8x+(1)/(4))`

Multiply by 4 both sides to remove the fraction

`155=(24x)+(32x+1)`

Combine like terms in the right side

`155=56x+1`

Subtract 1 both sides

`155-1=56x+1-1`

`154=56x`

Divide by 56 both sides

`154/56=56x/56`

`154/56=x`

Rewrite

`x=(154)/(56)`

Simplify

`x=(11)/(4)`

Rewrite x as mixed number

`x=(11)/(4)=(8)/(4) +(3)/(4)=2(3)/(4)`

Find the value of AB

`AB=(6x)`

substitute the value of x

`AB=6((11)/(4))=16.5\ units`

Rewrite AB as mixed number

`AB=16.5\ units=16+0.5=16+(1)/(2)=16(1)/(2)\ units`

Find the value of BC

`BC=(8x+(1)/(4))`

substitute the value of x

`BC=8((11)/(4))+(1)/(4)`

`BC=22+(1)/(4)=22(1)/(4)\ units`