How do you solve the simultaneous equation $4x + 5y = 0$ and $8x - 15y = 5$?

Hint: In this question, a linear equation of two variables is given. Here we will use the substitution method to solve these two linear equations. To solve the equations using the substitution method, we should follow the below steps:
1.Select one equation and solve it for one of its variables.
2.On the other equation, substitute for the variable that we get from the first step.
3.Solve the new equation.
4.Substitute the value that we found into any equation involving both variables and solve for the other variable.
5.Check the solution in both original equations.

Complete step-by-step answer:
Here, we want to solve the equations using the substitution method.
$ \Rightarrow 4x + 5y = 0$ ...(1)
$ \Rightarrow 8x - 15y = 5$ ...(2)
The first step is to select one equation and solve it for one of its variables.
Let us take equation (1), and convert it into the form of x variable.
So, subtract 5y on both sides.
$ \Rightarrow 4x + 5y - 5y = 0 - 5y$
That is equal to,
$ \Rightarrow 4x = - 5y$
Now, divide both sides by 4.
$ \Rightarrow \dfrac{{4x}}{4} = - \dfrac{{5y}}{4}$
So, the value of x will be,
$ \Rightarrow x = - \dfrac{5}{4}y$
In the second step, on the other equation, substitute for the variable that we get from the first step.
 Substitute $x = - \dfrac{5}{4}y$ in equation (2).
$ \Rightarrow 8x - 15y = 5$
$ \Rightarrow 8\left( { - \dfrac{5}{4}y} \right) - 15y = 5$
$ \Rightarrow 2\left( { - 5y} \right) - 15y = 5$
That is equal to,
$ \Rightarrow - 10y - 15y = 5$
Therefore,
$ \Rightarrow - 25y = 5$
Now, divide by -25 into both sides.
$ \Rightarrow \dfrac{{ - 25y}}{{ - 25}} = \dfrac{5}{{ - 25}}$
So,
$ \Rightarrow y = - \dfrac{1}{5}$
Now, put the value of y in equation (1).
$ \Rightarrow 4x + 5y = 0$
Put the value of y.
$ \Rightarrow 4x + 5\left( { - \dfrac{1}{5}} \right) = 0$
That is equal to
$ \Rightarrow 4x - 1 = 0$
Let us add 1 to both sides.
 $ \Rightarrow 4x - 1 + 1 = 0 + 1$
That is equal to,
$ \Rightarrow 4x = 1$
Now, divide both sides by 4.
$ \Rightarrow \dfrac{{4x}}{4} = \dfrac{1}{4}$
So,
$ \Rightarrow x = \dfrac{1}{4}$
Hence, we find the value of x is $\dfrac{1}{4}$ and the value of y is $ - \dfrac{1}{5}$ using substitution methods.

Note:
To check whether our answer is correct or not, feed the x and y values in each equation.
Let us take equation (1) and put the values.
$ \Rightarrow 4x + 5y = 0$
Put $x = \dfrac{1}{4}$ and $y = - \dfrac{1}{5}$.
$ \Rightarrow 4\left( {\dfrac{1}{4}} \right) + 5\left( { - \dfrac{1}{5}} \right) = 0$
$ \Rightarrow 1 - 1 = 0$
That is equal to,
$ \Rightarrow 0 = 0$
Now, let us take equation (2) and put the values.
$ \Rightarrow 8x - 15y = 5$
Put $x = \dfrac{1}{4}$ and $y = - \dfrac{1}{5}$.
$ \Rightarrow 8\left( {\dfrac{1}{4}} \right) - 15\left( { - \dfrac{1}{5}} \right) = 5$
$ \Rightarrow 2 + 3 = 5$
That is equal to,
$ \Rightarrow 5 = 5$