Question

solve for x.

1. d • x - 3 e x - 1 • f

20

1. t • x + 20 u 3 • v

2x + 35

1. r • 8 q 3x - 1 • p

5x - 1

1. t • 2x - 12 u 3 • v

x + 1
points a, b, and c are collinear. point b is between a and c. solve for x.

1. find x if ab = 2x - 5, bc = 2x - 9,

and ac = 14.

1. ac = 12, bc = 3x, and ab = 8x + 1.

find x.

1. find x if ab = 2x - 2, bc = 2x - 10,

and ac = 16.

1. find x if ac = x + 23, ab = 27 + 2x,

and bc = 6.

Explanation:

Step1: Set up the equation for problem 9

Since \(DE + EF=DF\), we have \((x - 3)+(x - 1)=20\).

Step2: Simplify the left - hand side of the equation

Combine like terms: \(x-3+x - 1=2x-4\). So the equation becomes \(2x-4 = 20\).

Step3: Solve for \(x\)

Add 4 to both sides: \(2x=20 + 4=24\). Then divide both sides by 2: \(x=\frac{24}{2}=12\).

Step1: Set up the equation for problem 10

Since \(TU+UV = TV\), we have \((x + 20)+3=2x+35\).

Step2: Simplify the left - hand side of the equation

\(x+20 + 3=x + 23\). So the equation is \(x + 23=2x+35\).

Step3: Solve for \(x\)

Subtract \(x\) from both sides: \(23=x + 35\). Then subtract 35 from both sides: \(x=23-35=-12\).

Step1: Set up the equation for problem 11

Since \(RQ+QP=RP\), we have \(8+(3x - 1)=5x-1\).

Step2: Simplify the left - hand side of the equation

\(8+3x-1=3x + 7\). So the equation is \(3x+7=5x-1\).

Step3: Solve for \(x\)

Subtract \(3x\) from both sides: \(7=2x-1\). Add 1 to both sides: \(8 = 2x\). Divide by 2: \(x = 4\).

Step1: Set up the equation for problem 12

Since \(TU+UV=TV\), we have \((2x-12)+3=x + 1\).

Step2: Simplify the left - hand side of the equation

\(2x-12 + 3=2x-9\). So the equation is \(2x-9=x + 1\).

Step3: Solve for \(x\)

Subtract \(x\) from both sides: \(x-9=1\). Add 9 to both sides: \(x=10\).

Step1: Set up the equation for problem 13

Since \(AB + BC=AC\), we have \((2x-5)+(2x-9)=14\).

Step2: Simplify the left - hand side of the equation

\(2x-5+2x-9=4x-14\). So the equation is \(4x-14 = 14\).

Step3: Solve for \(x\)

Add 14 to both sides: \(4x=28\). Divide by 4: \(x = 7\).

Step1: Set up the equation for problem 14

Since \(AB+BC=AC\), we have \((8x + 1)+3x=12\).

Step2: Simplify the left - hand side of the equation

\(8x+1+3x=11x + 1\). So the equation is \(11x+1=12\).

Step3: Solve for \(x\)

Subtract 1 from both sides: \(11x=11\). Divide by 11: \(x = 1\).

Step1: Set up the equation for problem 15

Since \(AB+BC=AC\), we have \((2x-2)+(2x-10)=16\).

Step2: Simplify the left - hand side of the equation

\(2x-2+2x-10=4x-12\). So the equation is \(4x-12 = 16\).

Step3: Solve for \(x\)

Add 12 to both sides: \(4x=28\). Divide by 4: \(x = 7\).

Step1: Set up the equation for problem 16

Since \(AB+BC=AC\), we have \((27 + 2x)+6=x + 23\).

Step2: Simplify the left - hand side of the equation

\(27+2x+6=2x+33\). So the equation is \(2x+33=x + 23\).

Step3: Solve for \(x\)

Subtract \(x\) from both sides: \(x+33=23\). Subtract 33 from both sides: \(x=-10\).

Answer:

1. \(x = 12\)
2. \(x=-12\)
3. \(x = 4\)
4. \(x = 10\)
5. \(x = 7\)
6. \(x = 1\)
7. \(x = 7\)
8. \(x=-10\)