What are the equations to these problems? Adv. Algebra 2?

Question

1. The lengths of the sides of a quadrilateral are consecutive even integers. Twice the length of the shortest plus the longest is 120. Find the lengths of the sides.

2. One angle is 30 degrees more than three times the other. The angles are complementary (add to 90 degrees!) Find each angle.

3.The measure of one angle in a triangle is twice the second. The third is 12 less than the sum of the other two. Find the measures of each angle.

4. Twice the greater of the two consecutive odd integers is 13 less than three times the lesser. Find the integer.

Answer 1

1) where the side lengths are x, x+2, x+4, and x+6
2x+(x+6)=120

2) where the angles are x and 3x+30
x+(3x+30)=90

3) where the angles are x, 2x, and x+2x-12
x+2x+(x+2x-12)=180

4) im not sure

Answer 2

Problem 1: Setting up consecutive even (or odd) integer problems is always the same. Do this:

Let n = 1st even integer (shortest side)
Then n + 2 = 2nd even integer. Remember, even (odd) integers are 2 units apart.
And n + 4 = 3rd even integer
n + 6 = 4th even integer (longest side)
Now you can write the equation.

2n + (n + 6) = 120 Now you solve the equation.
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Problem 2: You can always set up this kind of problem as follows:

Let x = measure of 1st angle
Then 90 - x = measure of the complement. Now write the equation
90 - x = 30 + 3x You solve the equation
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Problem 3. You can always set up these types of problems as follows:

Let x = measure of second angle
Then 2x = measure of 1st angle
and 2x + x - 12 = 3x - 12 = measure of 3rd angle

Remember the sum of the measures of a triangle is always 180.

So, x + 2x + (3x - 12) = 180 You solve the equation.
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Problem 4: The setup for this problem is the same as Prob. 1

Let n = smaller odd integer
Then n + 2 = next odd integer
Now the equation.

2(n + 2) = 3n - 13 You solve the equation.