(x-1)/3=(y+1)/4 equation
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The solution
Detail solution
Given the linear equation:
(x-1)/3 = (y+1)/4
Expand brackets in the left part
x/3-1/3 = (y+1)/4
Expand brackets in the right part
x/3-1/3 = y/4+1/4
Move free summands (without x)
from left part to right part, we given:
$$\frac{x}{3} = \frac{y}{4} + \frac{7}{12}$$
Divide both parts of the equation by 1/3
x = 7/12 + y/4 / (1/3)
We get the answer: x = 7/4 + 3*y/4
7 3*re(y) 3*I*im(y)
x1 = - + ------- + ---------
4 4 4
$$x_{1} = \frac{3 \operatorname{re}{\left(y\right)}}{4} + \frac{3 i \operatorname{im}{\left(y\right)}}{4} + \frac{7}{4}$$
x1 = 3*re(y)/4 + 3*i*im(y)/4 + 7/4
Sum and product of roots
[src]
7 3*re(y) 3*I*im(y)
- + ------- + ---------
4 4 4
$$\frac{3 \operatorname{re}{\left(y\right)}}{4} + \frac{3 i \operatorname{im}{\left(y\right)}}{4} + \frac{7}{4}$$
7 3*re(y) 3*I*im(y)
- + ------- + ---------
4 4 4
$$\frac{3 \operatorname{re}{\left(y\right)}}{4} + \frac{3 i \operatorname{im}{\left(y\right)}}{4} + \frac{7}{4}$$
7 3*re(y) 3*I*im(y)
- + ------- + ---------
4 4 4
$$\frac{3 \operatorname{re}{\left(y\right)}}{4} + \frac{3 i \operatorname{im}{\left(y\right)}}{4} + \frac{7}{4}$$
7 3*re(y) 3*I*im(y)
- + ------- + ---------
4 4 4
$$\frac{3 \operatorname{re}{\left(y\right)}}{4} + \frac{3 i \operatorname{im}{\left(y\right)}}{4} + \frac{7}{4}$$
7/4 + 3*re(y)/4 + 3*i*im(y)/4