Answer:
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
Step-by-step explanation:
The complete question in the attached figure
we have that
Point B is between A and C on segment AC
we know that
[tex]AC=AB+BC[/tex] -----> 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
[tex]95=(15x-10)+(5x+5)[/tex]
Combine like terms in the right side
[tex]95=(20x-5)[/tex]
Adds 5 both sides
[tex]95+5=20x-5+5[/tex]
[tex]100=20x[/tex]
Divide by 20 both sides
[tex]100/20=20x/20[/tex]
[tex]5=x[/tex]
Rewrite
[tex]x=5[/tex]
Find the value of AB
[tex]AB=(15x-10)[/tex]
substitute the value of x
[tex]AB=(15(5)-10)=65\ units[/tex]
Find the value of BC
[tex]BC=(5x+5)[/tex]
substitute the value of x
[tex]BC=(5(5)+5)=30\ units[/tex]
Part 2) we have
AC = 8x - 16, AB = 3x - 8, BC = 4x
substitute the given values in equation A and solve for x
[tex]8x-16=(3x-8)+(4x)[/tex]
Combine like terms in the right side
[tex]8x-16=(7x-8)[/tex]
Adds 16 both sides
[tex]8x-16+16=7x-8+16[/tex]
[tex]8x=7x+8[/tex]
Subtract 7x both sides
[tex]8x-7x=7x+8-7x[/tex]
[tex]x=8[/tex]
Find the value of AB
[tex]AB=(3x-8)[/tex]
substitute the value of x
[tex]AB=(3(8)-8)=16\ units[/tex]
Find the value of BC
[tex]BC=(4x)[/tex]
substitute the value of x
[tex]BC=4(8)=32\ units[/tex]
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
[tex]x-0.4=(x-4.9)+(0.5x)[/tex]
Combine like terms in the right side
[tex]x-0.4=(1.5x-4.9)[/tex]
Adds 4.9 both sides
[tex]x-0.4+4.9=1.5x-4.9+4.9[/tex]
[tex]x+4.5=1.5x[/tex]
Subtract x both sides
[tex]x+4.5-x=1.5x-x[/tex]
[tex]4.5=0.5x[/tex]
Divide by 0.5 both sides
[tex]4.5/0.5=0.5x/0.5[/tex]
[tex]9=x[/tex]
Rewrite
[tex]x=9[/tex]
Find the value of AB
[tex]AB=(x-4.9)[/tex]
substitute the value of x
[tex]AB=(9-4.9)=4.1\ units[/tex]
Find the value of BC
[tex]BC=(0.5x)[/tex]
substitute the value of x
[tex]BC=0.5(9)=4.5\ units[/tex]
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
[tex]38\frac{3}{4}=(6x)+(8x+\frac{1}{4})[/tex]
Convert mixed number to an improper fraction
[tex]38\frac{3}{4}=38+\frac{3}{4}=\frac{38*4+3}{4}=\frac{155}{4}[/tex]
substitute
[tex]\frac{155}{4}=(6x)+(8x+\frac{1}{4})[/tex]
Multiply by 4 both sides to remove the fraction
[tex]155=(24x)+(32x+1)[/tex]
Combine like terms in the right side
[tex]155=56x+1[/tex]
Subtract 1 both sides
[tex]155-1=56x+1-1[/tex]
[tex]154=56x[/tex]
Divide by 56 both sides
[tex]154/56=56x/56[/tex]
[tex]154/56=x[/tex]
Rewrite
[tex]x=\frac{154}{56}[/tex]
Simplify
[tex]x=\frac{11}{4}[/tex]
Rewrite x as mixed number
[tex]x=\frac{11}{4}=\frac{8}{4} +\frac{3}{4}=2\frac{3}{4}[/tex]
Find the value of AB
[tex]AB=(6x)[/tex]
substitute the value of x
[tex]AB=6(\frac{11}{4})=16.5\ units[/tex]
Rewrite AB as mixed number
[tex]AB=16.5\ units=16+0.5=16+\frac{1}{2}=16\frac{1}{2}\ units[/tex]
Find the value of BC
[tex]BC=(8x+\frac{1}{4})[/tex]
substitute the value of x
[tex]BC=8(\frac{11}{4})+\frac{1}{4}[/tex]
[tex]BC=22+\frac{1}{4}=22\frac{1}{4}\ units[/tex]
