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Vlada [557]
3 years ago
13

How many Solutions does this system have? (1 point)

Mathematics
1 answer:
mixas84 [53]3 years ago
5 0

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

Option -C.

<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,  

a_{1} = 2, b_{1}= 1\text{ and }c_{1}= -3

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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Step-by-step explanation:

4 0
2 years ago
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Va rog e urgent am nevoie de raspuns
Lady_Fox [76]

Answer: 4

<u>Explanation:</u>

f(x) = 2x - 1

f(√2) = 2√2 - 1

f(1) = 2(1) - 1

     =  2 - 1

     =     1

f(√3)  = 2√3 - 1

*******************************************************

\frac{f(\sqrt{2})-f(1)}{\sqrt{2}-1} +\frac{f(\sqrt{3})-f(\sqrt{2})}{\sqrt{3}-\sqrt{2}}

= \frac{(2\sqrt{2}-1)-1}{\sqrt{2}-1} +\frac{(2\sqrt{3}-1)-(2\sqrt{2}-1)}{\sqrt{3}-\sqrt{2}}

= \frac{2\sqrt{2}-2}{\sqrt{2}-1} +\frac{2\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}

= \frac{2\sqrt{2}-2}{\sqrt{2}-1}(\frac{\sqrt{2}+1}{\sqrt{2}+1})+\frac{2\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}})

= \frac{4+2\sqrt{2}-2\sqrt{2}-2}{2 - 1} + \frac{6 +2\sqrt{6}-2\sqrt{6}-4}{3-2}

= \frac{2}{1} +\frac{2}{1}

= 4

3 0
3 years ago
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