Question

parallel lines cut by a transversal coloring activity! 1 find the value of x. light blue alt interior 14x - 13 = 8x + 5 8x + 13 - 8x + 13 6x = 18 x = 3 2 find the value of x. (9x + 1)° (11x - 23)° x = 12 yellow 3 find the value of x. (3x + 5)° (7x - 15)° x = 5 4 find the value of y. (3y + 1)° (6x - 13)° (8x - 61)° 5 find the value of y. (21y - 1)° (4x + 23)° (9x - 38)° 6 find the value of y. (9x + 42)° (15x)° (4y - 13)°

Explanation:

Step1: For problem 1

Set up equation using alternate - interior angles property. Alternate - interior angles are equal. So, \(14x - 13=8x + 5\).

Step2: Solve for \(x\) in problem 1

Subtract \(8x\) from both sides: \(14x-8x - 13=8x-8x + 5\), which gives \(6x-13 = 5\). Then add 13 to both sides: \(6x-13 + 13=5 + 13\), so \(6x=18\). Divide both sides by 6: \(x = 3\).

Step3: For problem 2

Set up equation using corresponding - angles property. \(9x + 1=11x-23\).

Step4: Solve for \(x\) in problem 2

Subtract \(9x\) from both sides: \(9x-9x + 1=11x-9x-23\), so \(1 = 2x-23\). Add 23 to both sides: \(1+23=2x-23 + 23\), then \(24 = 2x\). Divide by 2: \(x = 12\).

Step5: For problem 3

Set up equation using vertical - angles property. \(3x + 5=7x-15\).

Step6: Solve for \(x\) in problem 3

Subtract \(3x\) from both sides: \(3x-3x + 5=7x-3x-15\), so \(5 = 4x-15\). Add 15 to both sides: \(5 + 15=4x-15 + 15\), then \(20 = 4x\). Divide by 4: \(x = 5\).

Step7: For problem 4

First, find \(x\) using vertical - angles. \(6x-13=8x - 61\). Subtract \(6x\) from both sides: \(-13=2x-61\). Add 61 to both sides: \(48 = 2x\), so \(x = 24\). Then use the fact that \(3y + 1=6x-13\). Substitute \(x = 24\) into it: \(3y+1=6\times24-13\), \(3y + 1=144-13=131\). Subtract 1 from both sides: \(3y=130\), \(y=\frac{130}{3}\).

Step8: For problem 5

First, find \(x\) using vertical - angles. \(4x + 23=9x-38\). Subtract \(4x\) from both sides: \(23 = 5x-38\). Add 38 to both sides: \(61 = 5x\), \(x=\frac{61}{5}\). Then use the fact that \(21y-1=9x-38\). Substitute \(x=\frac{61}{5}\) into it: \(21y-1=9\times\frac{61}{5}-38\), \(21y-1=\frac{549}{5}-38=\frac{549 - 190}{5}=\frac{359}{5}\). Add 1 to both sides: \(21y=\frac{359}{5}+1=\frac{359 + 5}{5}=\frac{364}{5}\), \(y=\frac{364}{105}\).

Step9: For problem 6

Set up equation using corresponding - angles. \(9x + 42=15x\). Subtract \(9x\) from both sides: \(42 = 6x\). Divide by 6: \(x = 7\). Then use the fact that \(4y-13=9x + 42\). Substitute \(x = 7\) into it: \(4y-13=9\times7+42\), \(4y-13=63 + 42=105\). Add 13 to both sides: \(4y=118\), \(y=\frac{59}{2}\).

Answer:

1. \(x = 3\)
2. \(x = 12\)
3. \(x = 5\)
4. \(y=\frac{130}{3}\)
5. \(y=\frac{364}{105}\)
6. \(y=\frac{59}{2}\)