GIVEN:HZ=4a-5

RZ=3a=5
FIND:HZ

To find the value of HZ, we first need to simplify the given equations.

Given:
HZ = 4a - 5
RZ = 3a + 5

Looking at the second equation, it seems like there's a typo. It says "3a = 5" which means the equation is not solvable as it is contradictory. So, we'll disregard that equation for now and focus on the first equation.

HZ = 4a - 5

To find the value of HZ, we need to know the value of 'a'. If the value of 'a' is given, you can substitute it into the equation to calculate HZ.

If 'a' is not given, we cannot solve for HZ with the information provided. We need either the value of 'a' or more equations involving 'a' to find the value of HZ.

To find the value of HZ, we need to use the given equations:

HZ = 4a - 5
RZ = 3a + 5

However, there seems to be an error in the second equation. It should be RZ = 3a - 5 instead of RZ = 3a = 5. Let's proceed assuming that it is 3a - 5.

Now, let's solve the equations to find the value of HZ.

First, we'll isolate a in the second equation:

RZ = 3a - 5

Add 5 to both sides to get:

RZ + 5 = 3a

Now, divide both sides by 3 to solve for a:

(RZ + 5) / 3 = a

Once we have the value of a, we can substitute it into the first equation to find HZ.

3 a = 5

so
a = 5/3

4a - 5 = 20/3 - 5

= 20/3 - 15/3 = 5/3