How many solutions does the system have? (x+2y=2 2x + 4y = -8

Mathematics

2 answers:

SIZIF [17.4K]2 years ago

6 0

Answer:

<h2>0 - no solution</h2>

Step-by-step explanation:

$\text{Let}\\\\\left\{\begin{array}{ccc}ax+by=c\\dx+ey=f\end{array}\right\\\\\text{in its simplest form, i.e. that a and b, d and e are relatively first}$

$\text{if}\ a=d,\ b=e,\ c=f,\ \text{then a system of equations has infinitely many solutions}\\\\\text{Example:}\\\\\left\{\begin{array}{ccc}x+3y=6\\-2x-6y=-12&\text{divide both sides by (-2)}\end{array}\right\\\\\left\{\begin{array}{ccc}x+3y=6\\x+3y=6\end{array}\right$

$\text{if}\ a=d,\ b=e,\ c\neqf,\ \text{then the system of equations has no solution}\\\\\text{Example:}\\\\\left\{\begin{array}{ccc}3x+9y=12&\text{divide both sides by 3}\\x+3y=1\end{array}\right\\\\\left\{\begin{array}{ccc}x+3y=4\\x+3y=1\end{array}\right$

$\text{In other cases it has one solution.}$

$\text{We have:}\\\\\left\{\begin{array}{ccc}x+2y=2\\2x+4y=-8&\text{divide both sides by 2}\end{array}\right\\\\\left\{\begin{array}{ccc}x+2y=2\\x+2y=-4\end{array}\right\\\\\text{Conclusion:}\\\\\text{The system of equations has no solution.}$

$\text{Now I will show it}$

$\left\{\begin{array}{ccc}x+2y=2\\x+2y=-8&\text{change the sings}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}x+2y=2\\-x-2y=8\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad0=10\qquad\bold{FALSE}$