Solve for x and y: 3x/(y−1) = 1 and (2x+2)(y−5) = 1:2

Tunapewa milinganyo miwili:

1. $\frac{3x}{y-1} = 1$
2. $(2x+2)(y-5) = 12$ Kutoka mlinganyo wa kwanza, zidisha pande zote mbili kwa $(y-1)$:

$$3x = y-1$$

Tenga $y$:

$$y = 3x+1 \quad \text{(Mlinganyo 3)}$$

Badilisha $y = 3x+1$ kwenye mlinganyo wa pili:

$$(2x+2)((3x+1)-5) = 12$$ $$(2x+2)(3x-4) = 12$$

Panua mabano:

$$2x(3x) + 2x(-4) + 2(3x) + 2(-4) = 12$$ $$6x^2 - 8x + 6x - 8 = 12$$ $$6x^2 - 2x - 8 = 12$$

Toa 12 kutoka pande zote mbili:

$$6x^2 - 2x - 20 = 0$$

Gawanya mlinganyo mzima kwa 2:

$$3x^2 - x - 10 = 0$$

Tumia fomula ya quadratic $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Hapa $a=3$, $b=-1$, $c=-10$.

$$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-10)}}{2(3)}$$ $$x = \frac{1 \pm \sqrt{1 + 120}}{6}$$ $$x = \frac{1 \pm \sqrt{121}}{6}$$ $$x = \frac{1 \pm 11}{6}$$

Kuna thamani mbili zinazowezekana kwa $x$:

$$x_1 = \frac{1 + 11}{6} = \frac{12}{6} = 2$$ $$x_2 = \frac{1 - 11}{6} = \frac{-10}{6} = -\frac{5}{3}$$

Tumia Mlinganyo 3 ($y = 3x+1$).

Kwa $x_1 = 2$:

$$y_1 = 3(2) + 1 = 6 + 1 = 7$$

Kwa $x_2 = -\frac{5}{3}$:

$$y_2 = 3\left(-\frac{5}{3}\right) + 1 = -5 + 1 = -4$$

Kutoka mlinganyo wa kwanza, $y-1 \neq 0$, hivyo $y \neq 1$. Thamani zetu za $y$ ni 7 na -4, ambazo zote ni halali.

Suluhisho ni:

$$x = 2, y = 7$$

au

$$x = -\frac{5}{3}, y = -4$$

Majibu ni $x=2, y=7 \text{ au } x=-\frac{5}{3}, y=-4$.