Question

(07.06 LC) Which statement is true about the equation fraction 4 over 5z − fraction 1 over 5z + 4 = fraction 3 over 5z + 6? It has no solution. It has one solution. It has two solutions. It has infinitely many solutions.

Answer 1

`$z=-(48)/(55) $`

It has one solution.

Explanation:

`$(4)/(5z)-(1)/(5z+4)=(3)/(5z+6) $`

Once we have fractions we already know the values when the equation is not true. I'm talking about division when the denominator is 0. It is undefined.

The values for
`z` are:

`$0 ; -(4)/(5) \text{ and } -(6)/(5) $`

Now let's find
`z`. First, we have to find the least common multiplier:

In this case:
`(5z+4)(5z+6)(5z)`

`$(4)/(5z)-(1)/(5z+4)=(3)/(5z+6) $`

`$(4(5z+4)(5z+6))/((5z+4)(5z+6)(5z))-(1(5z)(5z+6))/((5z+4)(5z+6)(5z))=(3(5z)(5z+4))/((5z+4)(5z+6)(5z)) $`

I will not proceed in the way above because it would take some time to type, instead, multiply
`(5z+4)(5z+6)(5z)` both sides.

`4(5z+4)(5z+6)-1(5z)(5z+6)=3(5z)(5z+4)`

`(20z+16)(5z+6)-(5z)(5z+6)=15z(5z+4)`

`100z^2+200z+96-25z^2-30z=75z^2+60z`

`200z+96-30z=60z`

`200z-90z=-96`

`110z=-96`

`$z=-(96)/(110) $`

`$z=-(48)/(55) $`

Answer 2

Answer: no solution

Step-by-step explanation: 4/5z - 1/5z = 3/5z

3/5z + 4 = 3/5z + 6

No solution