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AlekseyPX
1 year ago
7

Suppose that the relation H is defined as follows.

Mathematics
1 answer:
Norma-Jean [14]1 year ago
6 0

Answer:

  • Domain = {-8, 2}
  • Range = {-2, -1, 7}

Step-by-step explanation:

The domain is the set of x values, and the range is the set of y values.

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Step-by-step explanation:

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Range is the set of y-values that can be outputted by function f(x).

We see that our x-values span from 4 to infinity. Since it is an open dot, we cannot include it in our domain:

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The given system of equation that is 2x+y=3 and 6x=9-3y has infinite number of solutions.

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<u>Solution:</u>

Need to determine number of solution given system of equation has.

\begin{array}{l}{2 x+y=3} \\\\ {6 x=9-3 y}\end{array}

Let us first bring the equation in standard form for comparison

\begin{array}{l}{2 x+y-3=0} \\\\ {6 x+3 y-9=0}\end{array}

\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}

To check how many solutions are there for system of equations a_{1} x+b_{1} y+c_{1}=0 \text{ and }a_{2} x+b_{2} y+c_{2}=0, we need to compare ratios of \frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \text { and } \frac{c_{1}}{c_{2}}

In our case,  

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a_{2}  = 6, b_{2} = 3,\text{ and }c_{2} = -9

\begin{array}{l}{\Rightarrow \frac{a_{1}}{a_{2}}=\frac{2}{6}=\frac{1}{3}} \\\\ {\Rightarrow \frac{b_{1}}{b_{2}}=\frac{1}{3}} \\\\ {\Rightarrow \frac{c_{1}}{c_{2}}=\frac{-3}{-9}=\frac{1}{3}} \\\\ {\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{1}{3}}\end{array}

As \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}, so given system of equations have infinite number of solutions.

Hence, we can conclude that system has infinite number of solutions.

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