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sammy [17]
3 years ago
8

What is the solution to the system of equations represented by the two equations?

Mathematics
2 answers:
SpyIntel [72]3 years ago
6 0
There are 2 placeholder
it's easier if we had one
get rid of one with subsitutuion
y=-2x+7
y=1/3x
therefor
-2x+7=y=1/3x
-2x+7=1/3x
solve for x
multipl yboth sides by 3
-6x+21=x
add 6x to both sides
21=7x
divide both sides by 7
3=x
subsitute
y=1/3x
y=1/3(3)
y=3/3
y=1

the solution is x=3
y=1 or the point
(x,y)
(3,1)
Iteru [2.4K]3 years ago
5 0
If you would like to solve the following system of equations, you can do this like this:

y = - 2 * x + 7
y = 1/3 * x
____________
1/3 * x = - 2 * x + 7
1/3 * x + 2 * x = 7
1/3 * x + 6/3 * x = 7
7/3 * x = 7      /*3/7
x = 7 * 3 / 7
x = 3

y = 1/3 * x = 1/3 * 3 = 1

The correct result would be: x = 3 and y = 1.
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m=\frac{2506.833}{604.833}=4.1447  

Nowe we can find the means for x and y like this:  

\bar x= \frac{\sum x_i}{n}=\frac{77}{6}=12.833  

\bar y= \frac{\sum y_i}{n}=\frac{587}{6}=97.833  

And we can find the intercept using this:  

b=\bar y -m \bar x=97.833-(4.1447*12.833)=44.644  

So the line would be given by:  

y=4.1447 x +44.644  

Step-by-step explanation:

m=\frac{S_{xy}}{S_{xx}}  

Σx = 77, Σy = 587, Σx2 = 1593, Σy2 = 70,533, and Σxy = 10040.

Where:  

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}  

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}  

With these we can find the sums:  

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=1593-\frac{77^2}{6}=604.833  

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=10040-\frac{77*587}{6}=2506.833  

And the slope would be:  

m=\frac{2506.833}{604.833}=4.1447  

Nowe we can find the means for x and y like this:  

\bar x= \frac{\sum x_i}{n}=\frac{77}{6}=12.833  

\bar y= \frac{\sum y_i}{n}=\frac{587}{6}=97.833  

And we can find the intercept using this:  

b=\bar y -m \bar x=97.833-(4.1447*12.833)=44.644  

So the line would be given by:  

y=4.1447 x +44.644  

5 0
3 years ago
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