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

Answer: no solution

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

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

No solution

Answer 2

Answer:

$z=-\frac{48}{55}$

It has one solution.

Step-by-step explanation:

$\frac{4}{5z}-\frac{1}{5z+4}=\frac{3}{5z+6}$

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:

$0 ; -\frac{4}{5} \text{ and } -\frac{6}{5}$

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

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

$\frac{4}{5z}-\frac{1}{5z+4}=\frac{3}{5z+6}$

$\frac{4(5z+4)(5z+6)}{(5z+4)(5z+6)(5z)}-\frac{1(5z)(5z+6)}{(5z+4)(5z+6)(5z)}=\frac{3(5z)(5z+4)}{(5z+4)(5z+6)(5z)}$

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.

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

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

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

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

$200z-90z=-96$

$110z=-96$

$z=-\frac{96}{110}$

$z=-\frac{48}{55}$