Question

Someone please help me with this.

Answer 1

Answer:

The solution is (-2,-11)

Step-by-step explanation:

The given system is ;

$\frac{1+y}{2y} -\frac{3x+1}{y} =0, y\ne0$

Multiply through by 2y.

$1+y-2(3x+1)=0$

Expand;

$1+y-6x-2=0$

$y-6x=1$

$y=6x+1...(1)$

Also we have;

$\frac{2y+8}{12x+10}+\frac{4x}{y+7}=3...(2)$

Put equation (1) into (2)

$\frac{2(6x+1)+8}{12x+10}+\frac{4x}{6x+1+7}=3$

$\frac{12x+2+8}{12x+10}+\frac{4x}{6x+1+7}=3$

$\frac{12x+10}{12x+10}+\frac{4x}{6x+8}=3$

$\frac{12x+10}{12x+10}+\frac{4x}{6x+8}=3$ for $x\ne -\frac{5}{6},x\ne -\frac{4}{3}$

$1+\frac{4x}{6x+8}=3$

$\frac{4x}{6x+8}=3-1$

$\frac{4x}{6x+8}=2$

Cross multiply;

$4x=2(6x+8)$

$4x=12x+16$

$4x-12x=16$

$-8x=16$

Divide through by -8.

x=-2

Put x=-2 into equation (1).

y=6(-2)+1

y=-11

The solution is (-2,-11)