Question

name: date: pd: solving equations holiday color by number 10. 4x + 7 = 2x +15 11. a d (3x + 38)° b (5x + 20)° c e color: color: color: 3. -3 + 5x = 4x - 10 14. 15. a d (4x + 2)° b (5x - 13)° c e or: color: color: color:

Explanation:

Step1: Solve \(4x + 7 = 2x + 15\)

Subtract \(2x\) from both sides: \(4x - 2x + 7 = 15\) → \(2x + 7 = 15\)
Subtract \(7\): \(2x = 15 - 7\) → \(2x = 8\)
Divide by \(2\): \(x = \frac{8}{2} = 4\)

Step2: Solve \(9 - 2x = x + 3\)

Add \(2x\) to both sides: \(9 = x + 2x + 3\) → \(9 = 3x + 3\)
Subtract \(3\): \(3x = 9 - 3\) → \(3x = 6\)
Divide by \(3\): \(x = \frac{6}{3} = 2\)

Step3: Solve \(-3 + 5x = 4x - 10\)

Subtract \(4x\) from both sides: \(-3 + 5x - 4x = -10\) → \(-3 + x = -10\)
Add \(3\): \(x = -10 + 3 = -7\)

Step4: Solve for \(x\) in vertical angles (\(3x + 38 = 5x + 20\))

Subtract \(3x\): \(38 = 2x + 20\)
Subtract \(20\): \(2x = 38 - 20 = 18\)
Divide by \(2\): \(x = \frac{18}{2} = 9\)

Step5: Solve for \(x\) in vertical angles (\(4x + 2 = 5x - 13\))

Subtract \(4x\): \(2 = x - 13\)
Add \(13\): \(x = 2 + 13 = 15\)

Step6: For problem 14 (subtracting sets of triangles and circles)

First set: 6 triangles, 8 circles; Second set: 2 triangles, 4 circles.
Triangles: \(6 - 2 = 4\); Circles: \(8 - 4 = 4\) (though color depends on problem's key, but solving the count: 4 triangles, 4 circles)

Answer:

s (x-values for equations 10,12,13,11,15; counts for 14):

1. \(x = 4\)
2. \(x = 9\)
3. \(x = 2\)
4. \(x = -7\)
5. Triangles: 4, Circles: 4 (color key dependent)
6. \(x = 15\)