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4 years ago

3x-2y=2, 5x-5y=-18 Solve by elimination

2 answers:

8 0

5(3x-2y=2)

15x -10y = 10

2(5x-5y=-18)

10x - 10y = -36
- 15x - 10y = 10

-5x = 46
/-5 /-5

x = 9.2

3(9.2) - 2y = 2

27.6 - 2y = 2
-27.6 -27.6

-2y = 25.6
/-2 /-2

y = 12.8

x = 9.2; y = 12.8

navik [9.2K]4 years ago

7 0

We have $\begin{bmatrix}3x-2y=2\\ 5x-5y=-18\end{bmatrix}$

$\mathrm{Multiply\:}3x-2y=2\mathrm{\:by\:}5:\quad 15x-10y=10$
$\mathrm{Multiply\:}5x-5y=-18\mathrm{\:by\:}3:\quad 15x-15y=-54$

$\begin{bmatrix}15x-10y=10\\ 15x-15y=-54\end{bmatrix}$

15x - 15y = -54
-
15x - 10y = 10
/
-5y = -64

$\begin{bmatrix}15x-10y=10\\ -5y=-64\end{bmatrix}$

$-5y=-64 \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}-5 \ \textgreater \ \frac{-5y}{-5}=\frac{-64}{-5} \ \textgreater \ y=\frac{64}{5}$

$\mathrm{For\:}15x-10y=10\mathrm{\:plug\:in\:}\ \:y=\frac{64}{5}$

$15x-10\cdot \frac{64}{5}=10 \ \textgreater \ 10\cdot \frac{64}{5} \ \textgreater \ \mathrm{Multiply\:fractions}:\ \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c}$

$\frac{64\cdot \:10}{5} \ \textgreater \ \mathrm{Multiply\:the\:numbers:}\:64\cdot \:10=640 \ \textgreater \ \frac{640}{5}$

$\mathrm{Divide\:the\:numbers:}\:\frac{640}{5}=128 \ \textgreater \ 15x-128=10$

$\mathrm{Add\:}128\mathrm{\:to\:both\:sides} \ \textgreater \ 15x-128+128=10+128 \ \textgreater \ 15x=138$

$\mathrm{Divide\:both\:sides\:by\:}15 \ \textgreater \ \frac{15x}{15}=\frac{138}{15} \ \textgreater \ x=\frac{46}{5}$

$Therefore\:the\:solutions\:are \ \textgreater \ y=\frac{64}{5},\:x=\frac{46}{5}$

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$m=\dfrac{3}{2}$

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Given points are: ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 )

The average of x-coordinate will be:

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1) Finding $(\overline{x},\overline{y})$

Average of the x coordinates:

$\overline{x} = \dfrac{1+2+3}{3}$

$\overline{x} = 2$

Average of the y coordinates:

similarly for y

$\overline{y} = \dfrac{3+3+6}{3}$

$\overline{y} = 4$

2) Finding the line through $(\overline{x},\overline{y})$ with slope m.

Given a point and a slope, the equation of a line can be found using:

$(y-y_1)=m(x-x_1)$

in our case this will be

$(y-\overline{y})=m(x-\overline{x})$

$(y-4)=m(x-2)$

$y=mx-2m+4$

this is our equation of the line!

3) Find the squared vertical distances between this line and the three points.

So what we up till now is a line, and three points. We need to find how much further away (only in the y direction) each point is from the line.

Distance from point (1,3)

We know that when x=1, y=3 for the point. But we need to find what does y equal when x=1 for the line?

we'll go back to our equation of the line and use x=1.

$y=m(1)-2m+4$

$y=-m+4$

now we know the two points at x=1: (1,3) and (1,-m+4)

to find the vertical distance we'll subtract the y-coordinates of each point.

$d_1=3-(-m+4)$

$d_1=m-1$

finally, as asked, we'll square the distance

$(d_1)^2=(m-1)^2$

Distance from point (2,3)

we'll do the same as above here:

$y=m(2)-2m+4$

$y=4$

vertical distance between the two points: (2,3) and (2,4)

$d_2=3-4$

$d_2=-1$

squaring:

$(d_2)^2=1$

Distance from point (3,6)

$y=m(3)-2m+4$

$y=m+4$

vertical distance between the two points: (3,6) and (3,m+4)

$d_3=6-(m+4)$

$d_3=2-m$

squaring:

$(d_3)^2=(2-m)^2$

3) Add up all the squared distances, we'll call this value R.

$R=(d_1)^2+(d_2)^2+(d_3)^2$

$R=(m-1)^2+4+(2-m)^2$

4) Find the value of m that makes R minimum.

Looking at the equation above, we can tell that R is a function of m:

$R(m)=(m-1)^2+4+(2-m)^2$

you can simplify this if you want to. What we're most concerned with is to find the minimum value of R at some value of m. To do that we'll need to derivate R with respect to m. (this is similar to finding the stationary point of a curve)

$\dfrac{d}{dm}\left(R(m)\right)=\dfrac{d}{dm}\left((m-1)^2+4+(2-m)^2\right)$