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

+ z = 3
z = 3 +

then substitute the result (3 - 0a) into the second equation:

3a - 3(3 + ) = 7
3a + (-3 x 3) + (-3 x ) = 7
3a - 9 + = 7
3a + = 7 + 9
3a = 16
a = \( \frac{16}{3} \)
a = 5\(\frac{1}{3}\)

To solve this equation, repeatedly do the same thing to both sides of the equation until the variable is isolated on one side of the equal sign and the answer on the other.

c - 5 = \( \frac{c}{9} \)
9 x (c - 5) = c
(9 x c) + (9 x -5) = c
9c - 45 = c
9c - 45 - c = 0
9c - c = 45
8c = 45
c = \( \frac{45}{8} \)
c = 5\(\frac{5}{8}\)

To solve this equation, repeatedly do the same thing to both sides of the equation until the variable is isolated on one side of the > sign and the answer on the other.

-4x - 9 > -6 - 7x
-4x > -6 - 7x + 9
-4x + 7x > -6 + 9
3x > 3
x > \( \frac{3}{3} \)
x > 1

The first step to solve a quadratic equation that's set to zero is to factor the quadratic equation:

y² - 7y + 10 = 0
(y - 2)(y - 5) = 0

For this expression to be true, the left side of the expression must equal zero. Therefore, either (y - 2) or (y - 5) must equal zero:

If (y - 2) = 0, y must equal 2
If (y - 5) = 0, y must equal 5

So the solution is that y = 2 or 5

To solve this equation, isolate the variable for which you are solving (c) on one side of the equation and put everything else on the other side.

8c + 3y = -6c - 8y + 6
8c = -6c - 8y + 6 - 3y
8c + 6c = -8y + 6 - 3y
14c = -11y + 6
c = \( \frac{-11y + 6}{14} \)
c = \( \frac{-11y}{14} \) + \( \frac{6}{14} \)
c = -\(\frac{11}{14}\)y + \(\frac{3}{7}\)