If a linear system has more unknowns than equations, then the system can-not have a unique solution. True or False

Mathematics

1 answer:

Oduvanchick [21]3 years ago

5 0

Answer:

The statement is true.

Step-by-step explanation:

If a linear system has more unknowns than equations, then the system can-not have a unique solution. This is true.

When more unknowns are there than equations, then the system of polynomial equations is said to be under determined.

This means that if any solutions exist, it will not be unique.

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Write the linear equation that gives the rule for this table.

Maksim231197 [3]

to get the equation of any straight line all we need is two points, well, let's just grab them from the table.

hmmm let's use (-49 , -39) and hmmm say (67 , 77)

$(\stackrel{x_1}{-49}~,~\stackrel{y_1}{-39})\qquad (\stackrel{x_2}{67}~,~\stackrel{y_2}{77}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{77}-\stackrel{y1}{(-39)}}}{\underset{run} {\underset{x_2}{67}-\underset{x_1}{(-49)}}}\implies \cfrac{77+39}{67+49}\implies \cfrac{116}{116}\implies 1$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-39)}=\stackrel{m}{1}(x-\stackrel{x_1}{(-49)}) \\\\\\ y+39=1(x+49)\implies y+39=x+49\implies y=x+10$