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icang [17]
3 years ago
14

Solve the system algebraically. Check your work.

Mathematics
1 answer:
Luden [163]3 years ago
8 0

answer.

Answer:

x=2 and y=0 is the required result.

Step-by-step explanation:

We have been given system of equations:

5x+2y=105x+2y=10     (1)

And 3x+2y=63x+2y=6      (2)

We will use elimination method:

Multiply 1st equation by 3 and 2nd equation by 5 we get:

15x+6y=3015x+6y=30      (3)

15x+10y=3015x+10y=30      (4)

Now subtract  (4) from (3) we get:

-4y=0−4y=0

y=0y=0

Now, put y=0 in (1) equation:

5x+2(0)=105x+2(0)=10

5x=105x=10

x=2x=2

Hence, x=2 and y=0

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3 years ago
Solve the equation for y. Then find the value of Y for each value of X.
Hunter-Best [27]

Answer:

y=9-7x

If x= -1, y=16

If x= 0, y=9

If x=2, y= -5

Step-by-step explanation:

Because you are solving the equation for y, you want to have y by itself on one side. The only way to do that in this problem is by subtracting the 7x on both sides.

y+7x=9

-7x -7x ---> y=9-7x

The next part of the question says to find the value of Y for each value of X. Using the given X values, we plug them into the equation we just got in order to find Y.

y=9-7(-1) ----> y=16

y=9-7(0) ----> y=9

y=9-7(2) ----> y=-5

4 0
3 years ago
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Is the following statement always, never, or sometimes true?
pickupchik [31]
Sometimes...

2^-3 = 1/(2^3) = 1/8

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3 years ago
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I don’t have to have ads. When I wasn’t signed up I had to but I logged in and now I don’t have to watch ads.

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Please help me I need help please​
MakcuM [25]

Answer:

p = 4

Step-by-step explanation:

Given equation:

x^2+(p-3)y^2-4x+6y-16=0

<u>Standard equation of a circle:</u>

(x-a)^2+(y-b)^2=r^2

(where (a,b) is the centre of the circle, and r is the radius)

If you expand this equation, you will see that the coefficient of y^2 is always one.

Therefore, p-3=1

\implies p=1+3=4

<u>Additional information</u>

To rewrite the given equation in the standard form:

\implies x^2+y^2-4x+6y-16=0

\implies x^2-4x+y^2+6y=16

\implies (x-2)^2-4+(y+3)^2-9=16

\implies (x-2)^2+(y+3)^2=16+4+9

\implies (x-2)^2+(y+3)^2=29

So this is a circle with centre (2, -3) and radius √29

3 0
2 years ago
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