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How do you find the point of intersection using these to equations?

x=-2y-3

4y-x=9

 Nov 25, 2014

Best Answer 

 #2
avatar+130511 
+11

x=-2y-3

4y-x=9

Notice that we can substitute the first thing into the second....this gives

4x - (-2y - 3) = 9    simplify

6y + 3 = 9      subtract 3 from both sides

6y = 6            divide both sides by 6

y = 1       and subtituting this back into x = -2y - 3 ,  we find that x = -2(1) - 3   = -5

So our answer is x = -5 and y =1

 

 

 Nov 25, 2014
 #1
avatar+7188 
+11

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{x, y}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{x}}={\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{y}}{\mathtt{\,-\,}}{\mathtt{3}}\\
{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{y}}{\mathtt{\,-\,}}{\mathtt{x}}={\mathtt{9}}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{5}}\\
{\mathtt{y}} = {\mathtt{1}}\\
\end{array} \right\}$$

.
 Nov 25, 2014
 #2
avatar+130511 
+11
Best Answer

x=-2y-3

4y-x=9

Notice that we can substitute the first thing into the second....this gives

4x - (-2y - 3) = 9    simplify

6y + 3 = 9      subtract 3 from both sides

6y = 6            divide both sides by 6

y = 1       and subtituting this back into x = -2y - 3 ,  we find that x = -2(1) - 3   = -5

So our answer is x = -5 and y =1

 

 

CPhill Nov 25, 2014
 #3
avatar+130511 
0

Wow.....thanks h7....I didn't even know you could do that on this calculator!!....did you figure that out just fooling around with it ????

 

 Nov 25, 2014
 #4
avatar+7188 
+3

No....I just answered.....I don't know what this calc can do......now you know!

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{x, y}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{x}}={\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{y}}{\mathtt{\,-\,}}{\mathtt{3}}\\
{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{y}}{\mathtt{\,-\,}}{\mathtt{x}}={\mathtt{9}}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{5}}\\
{\mathtt{y}} = {\mathtt{1}}\\
\end{array} \right\}$$
!

 Dec 1, 2014

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