Question

Simultaneous Equation

x-2=
`(1)/(y)`

`(-12)/(x)`+
`(3)/(y)`=-15

Answer 1

x = 1, -4

y = -1, -1/6

Explanation:

x-2=1/y ---------------------------- equ 1

-12/x + 3/y = -15 ---------------- equ 2 isolate y in equ 1

y = 1/x-2 x ≠ 2 for equ 2, sub y with 1/x-2

-12/x + 3/(1/x-2) = -15

-12/x + 3x - 6 = -15

-12/x + 3x = -15 + 6

-12+3x^2/x = 9

-12 + 3x^2 = 9x

3x^2 - 9x = 12

3(x^2 - 3x)/3 = 12/3

x^2 - 3x = -4

x(x-3) = -4

x = -4, x = -4-3

x = -4, x = 1 plug x = 1, x = -4 into equ 1

1-2 = 1/y

-1 = 1/y

y = -1

-4-2 = 1/y

-6 = 1/y

y = -1/6

Answer 2

`x=1, \quad y=-1`

`x=-4, \quad y=-(1)/(6)`

Explanation:

Given equations:

`\begin{cases}x-2=(1)/(y)\\\\-(12)/(x)+(3)/(y)=-15\end{cases}`

Rearrange the first equation to isolate x:

`\implies x=(1)/(y)+2`

`\implies x=(1)/(y)+(2y)/(y)`

`\implies x=(1+2y)/(y)`

Substitute the expression for x into the second equation and solve for y:

`\implies -(12)/(x)+(3)/(y)=-15`

`\implies -(12)/(\left((1+2y)/(y)\right))+(3)/(y)=-15`

`\implies -(12y)/(1+2y)+(3)/(y)=-15`

`\implies (3)/(y)-(12y)/(1+2y)=-15`

`\implies (3(1+2y))/(y(1+2y))-(12y^2)/(y(1+2y))=-15`

`\implies (3(1+2y)-12y^2)/(y(1+2y))=-15`

`\implies 3(1+2y)-12y^2=-15(y(1+2y))`

`\implies 3+6y-12y^2=-15(y+2y^2)`

`\implies 3+6y-12y^2=-15y-30y^2`

`\implies 3+6y-12y^2+15y+30y^2=0`

`\implies 18y^2+21y+3=0`

`\implies 3(6y^2+7y+1)=0`

`\implies 6y^2+7y+1=0`

`\implies 6y^2+6y+y+1=0`

`\implies 6y(y+1)+1(y+1)=0`

`\implies (6y+1)(y+1)=0`

`\implies y=-(1)/(6), -1`

Substitute the found values of y into the first equation and solve for x:

`\begin{aligned}y=-(1)/(6) \implies x -2& = (1)/(\left(-(1)/(6)\right))\\x-2& = -6\\x-2+2&=-6+2\\\implies x&=-4\end{aligned}`

`\begin{aligned}y=-1 \implies x -2& = (1)/(-1)\\x-2& = -1\\x-2+2&=-1+2\\\implies x&=1\end{aligned}`

Therefore, the solutions of the given system of equations are:

`x=1, \quad y=-1`

`x=-4, \quad y=-(1)/(6)`