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Simultaneous Equation
x-2=[tex]\frac{1}{y}[/tex]
[tex]\frac{-12}{x}[/tex]+[tex]\frac{3}{y}[/tex]=-15

Sagot :

temiajare

Answer:

x = 1, -4

y = -1, -1/6

Step-by-step 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

semsee45

Answer:

[tex]x=1, \quad y=-1[/tex]

[tex]x=-4, \quad y=-\dfrac{1}{6}[/tex]

Step-by-step explanation:

Given equations:

[tex]\begin{cases}x-2=\dfrac{1}{y}\\\\-\dfrac{12}{x}+\dfrac{3}{y}=-15\end{cases}[/tex]

Rearrange the first equation to isolate x:

[tex]\implies x=\dfrac{1}{y}+2[/tex]

[tex]\implies x=\dfrac{1}{y}+\dfrac{2y}{y}[/tex]

[tex]\implies x=\dfrac{1+2y}{y}[/tex]

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

[tex]\implies -\dfrac{12}{x}+\dfrac{3}{y}=-15[/tex]

[tex]\implies -\dfrac{12}{\left(\dfrac{1+2y}{y}\right)}+\dfrac{3}{y}=-15[/tex]

[tex]\implies -\dfrac{12y}{1+2y}+\dfrac{3}{y}=-15[/tex]

[tex]\implies \dfrac{3}{y}-\dfrac{12y}{1+2y}=-15[/tex]

[tex]\implies \dfrac{3(1+2y)}{y(1+2y)}-\dfrac{12y^2}{y(1+2y)}=-15[/tex]

[tex]\implies \dfrac{3(1+2y)-12y^2}{y(1+2y)}=-15[/tex]

[tex]\implies 3(1+2y)-12y^2=-15(y(1+2y))[/tex]

[tex]\implies 3+6y-12y^2=-15(y+2y^2)[/tex]

[tex]\implies 3+6y-12y^2=-15y-30y^2[/tex]

[tex]\implies 3+6y-12y^2+15y+30y^2=0[/tex]

[tex]\implies 18y^2+21y+3=0[/tex]

[tex]\implies 3(6y^2+7y+1)=0[/tex]

[tex]\implies 6y^2+7y+1=0[/tex]

[tex]\implies 6y^2+6y+y+1=0[/tex]

[tex]\implies 6y(y+1)+1(y+1)=0[/tex]

[tex]\implies (6y+1)(y+1)=0[/tex]

[tex]\implies y=-\dfrac{1}{6}, -1[/tex]

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

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

[tex]\begin{aligned}y=-1 \implies x -2& = \dfrac{1}{-1}\\x-2& = -1\\x-2+2&=-1+2\\\implies x&=1\end{aligned}[/tex]

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

[tex]x=1, \quad y=-1[/tex]

[tex]x=-4, \quad y=-\dfrac{1}{6}[/tex]

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