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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
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]