Question

4/x-5/y=x+y/xy+3/10 and 3xy=10(y-x)​

Answer 1

Answer:

x=20

y=-4

I have solved it on pages

Answer 2

Correct Question: Solve for x and y if 4/x - 5/y = (x + y)/xy + 3/10 and 3xy = 10(y - x).​

Answer:

The values are $x=0$, $y=0$ and $x=2$, $y=5$.

Step-by-step explanation:

Step 1 of 3

Consider the equations as follows:

$\frac{4}{x} -\frac{5}{y}=\frac{x+y}{xy}+\frac{3}{10}$ . . . . . (1)

$3xy=10(y-x)$ . . . . . (2)

Simplify the equation (1) as follows:

⇒ $\frac{4y-5x}{xy} =\frac{x+y}{xy}+\frac{3}{10}$

⇒ $\frac{4y-5x}{xy} -\frac{x+y}{xy}=\frac{3}{10}$

Further simplify the left-hand side as follows:

⇒ $\frac{4y-5x-x-y}{xy}=\frac{3}{10}$

⇒ $\frac{3y-6x}{xy}=\frac{3}{10}$

Cross multiplying the equation, we get

⇒ $10(3y-6x)=3xy$ . . . . . (3)

Step 2 of 3

Equating equation (2) and (3) as follows:

⇒ $10(3y-6x)=10(y-x)$

⇒ $30y-60x=10y-10x$

Now,

⇒ $30y-10y=60x-10x$

⇒ $20y=50x$

Taking 10 common out from both the sides.

⇒ $2y=5x$

⇒ $y=\frac{5}{2} x$ . . . . (4)

Step 3 of 3

Substitute the value $y=\frac{5}{2} x$ in the equation (2) as follows:

⇒ $3x\left(\frac{5x}{2} \right)=10(\frac{5x}{2} -x)$

Now, solve to find the value of $x$.

⇒ $\frac{15x^2}{2}=\frac{50x}{2} -10x$

⇒ $\frac{15x^2}{2}=25x-10x$

⇒ $\frac{15x^2}{2}=15x$

Cross multiplying the equation as follows:

⇒ $15x^2=30x$

⇒ $15x^2-30x=0$

Taking $15x$ common, we get

⇒ $15x(x-2)=0$

⇒ $x=0$ or $x=2$

Case1. When $x=0$. Then,

From (4),

⇒ $y=\frac{5}{2}\times 0$

⇒ $y=0$

Case2. When $x=2$. Then,

From (4),

⇒ $y=\frac{5}{2}\times 2$

⇒ $y=5$

Therefore, the values are $x=0$, $y=0$ and $x=2$, $y=5$.

#SPJ2