Could anyone me out with these two geometry problems?

For 1:

x = 6

y = 70/3 = 23 & 1/3

For 2:

Angle A = 72 degrees

For 1:

We know that the 2 bottom lengths(?) are congruent to each other, therefore:

7x + 10 = 9x - 2

In algebraic equations, we are always doing PEMDAS backwards.

Subtract first:

7x = 9x - 12

Then subtract the 9x from the 7x:

-2x = -12

Then divide:

x = 6.

To solve for y:

We already know that the angle is 50 degrees, because we are given the right angle (90 degrees) and the top angle with 40 degrees.

So we need to put it into an equation:

3y - 20 = 50 (degrees)

Add 20:

3y = 70

Divide:

y = 70/3 = 23 & 1/3

For 2:

We have 3 angles, 58 degrees, 7x - 20, and 3x + 42.

In case you didn't notice, 7x - 20 and 3x + 42 are not proportional, but angle A and 3x + 42 are.

Therefore:

(7x - 20) + (3x + 42) + 58 = 180 (a triangle is 180 degrees)

Combine like terms:

7x - 20 + 3x + 42 + 58 = 180

10x + 80 = 180

Subtract:

10x = 100

Divide:

x = 10

But wait, we don't have our answer, yet.

Plug the (x) into the 3x + 42 equation:

3(10) + 42

30 + 42

72

Angle A = 72.

Proof:

72 + (7(10) - 20) + 58 = 180

72 + 50 + 58 = 180

180 = 180