The perimeter of a triangle is the sum of the lengths of its sides:

p = x + y + z
p = 7cm + 6cm + 15cm = 28cm

To solve this equation, replace the variables with the values given and then solve the now variable-free equation. (Remember order of operations, PEMDAS, Parentheses, Exponents, Multiplication/Division, Addition/Subtraction.)

-2a(a - x)
-2(-8)(-8 + 8)
-2(-8)(0)
(16)(0)
0

To combine like terms, add or subtract the coefficients (the numbers that come before the variables) of terms that have the same variable raised to the same exponent.

2a - 3a = -1a

You need to find the value of a so solve the first equation in terms of y:

6a + y = -9
y = -9 - 6a

then substitute the result (-9 - 6a) into the second equation:

-9a + 8(-9 - 6a) = -2
-9a + (8 x -9) + (8 x -6a) = -2
-9a - 72 - 48a = -2
-9a - 48a = -2 + 72
-57a = 70
a = \( \frac{70}{-57} \)
a = -1\(\frac{13}{57}\)

For parallel lines with a transversal, the following relationships apply:

• angles in the same position on different parallel lines equal each other (a° = w°, b° = x°, c° = z°, d° = y°)
• alternate interior angles are equal (a° = z°, b° = y°, c° = w°, d° = x°)
• all acute angles (a° = c° = w° = z°) and all obtuse angles (b° = d° = x° = y°) equal each other
• same-side interior angles are supplementary and add up to 180° (e.g. a° + d° = 180°, d° + c° = 180°)

Applying these relationships starting with z° = 40, the value of b° is 140.