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nignag [31]
3 years ago
10

Which ordered pair (p,r) is the solution to the given system of linear equations?

Mathematics
1 answer:
Misha Larkins [42]3 years ago
8 0

Answer:

(p,r) = (1/3, 2/9)

Step-by-step explanation:

Here, we want to solve a system of equations

We can rewrite the second equation by dividing through by 2

So we have;

4p + 3r = 2

and

5p - 3r = 1

Add both equations:

9p = 3

p = 3/9

p = 1/3

Recall ;

5p - 3r = 1

3r = 5p - 1

Substitute the value or p here

3r = 5(1/3)-1

3r = 5/3 - 1

3r = 2/3

r = 2/9

So we have the solution set as;

(p,r) = (1/3 , 2/9)

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<h2>Answer</h2>

After the dilation \frac{5}{3} around the center of dilation (2, -2), our triangle will have coordinates:

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<h2>Explanation</h2>

First, we are going to translate the center of dilation to the origin. Since the center of dilation is (2, -2) we need to move two units to the left (-2) and two units up (2) to get to the origin. Therefore, our first partial rule will be:

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Next, we are going to perform our dilation, so we are going to multiply our resulting point by the dilation factor \frac{5}{3}. Therefore our second partial rule will be:

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(x,y)→(\frac{5}{3} x-\frac{10}{3} ,\frac{5}{3} y+\frac{10}{3} )

Now, the only thing left to create our actual rule is going back from the origin to the original center of dilation, so we need to move two units to the right (2) and two units down (-2)

(x,y)→(\frac{5}{3} x-\frac{10}{3}+2,\frac{5}{3} y+\frac{10}{3}-2)

(x,y)→(\frac{5}{3} x-\frac{4}{3} ,\frac{5}{3}y+ \frac{4}{3})

Now that we have our rule, we just need to apply it to each point of our triangle to perform the required dilation:

R=(2,1)

R'=(\frac{5}{3} x-\frac{4}{3} ,\frac{5}{3}y+ \frac{4}{3})

R'=(\frac{5}{3} (2)-\frac{4}{3} ,\frac{5}{3}(1)+ \frac{4}{3})

R'=(\frac{10}{3} -\frac{4}{3} ,\frac{5}{3}+ \frac{4}{3})

R'=(2,3)

S=(2,-2)

S'=(\frac{5}{3} (2)-\frac{4}{3} ,\frac{5}{3}(-2)+ \frac{4}{3})

S'=(\frac{10}{3} -\frac{4}{3} ,-\frac{10}{3}+ \frac{4}{3})

S'=(2,-2)

T=(-1,-2)

T'=(\frac{5}{3} (-1)-\frac{4}{3} ,\frac{5}{3}(-2)+ \frac{4}{3})

T'=(-\frac{5}{3} -\frac{4}{3} ,-\frac{10}{3}+ \frac{4}{3})

T'=(-3,-2)

Now we can finally draw our triangle:

8 0
3 years ago
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