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3 years ago

7

9x−9y=0 3x−4y=10 solve by elimination

1 answer:

NikAS [45]3 years ago

3 0

Answer:

(-10,-10)

Step-by-step explanation:

9x-9y=0

3x-4y=10

In elimination, we want both equations to have the same form and like terms to be lined up. We have that. We also need one of the columns with variables to contain opposites or same terms. Neither one of our columns with the variables contain this.

We can do a multiplication to the second equation so that the first terms of each are either opposites or sames. It doesn't matter which. I like opposites because you just add the equations together. So I'm going to multiply the second equation by -3.

I will rewrite the system with that manipulation:

9x-9y=0

-9x+12y=-30

----------------------Add them up!

0+3y=-30

3y=-30

y=-10

So now once you find a variable, plug into either equation to find the other one.

I'm going to use 9x-9y=0 where y=-10.

So we are going to solve for x now.

9x-9y=0 where y=-10.

9x-9(-10)=0 where I plugged in -10 for y.

9x+90=0 where I simplified -9(-10) as +90.

9x =-90 where I subtracted 90 on both sides.

x= -10 where I divided both sides by 9.

The solution is (x,y)=(-10,-10)

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Step-by-step explanation:

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3 years ago

Determine the measure of each segment then indicate whether the statements are true or false

kupik [55]

Answer:

$d_{AB}\ne d_{JK}$

$d_{AB}\ne \:d_{GH}$

$d_{GH}\ne \:d_{JK}$

Therefore,

Option (A) is false

Option (B) is false

Option (C) is false

Step-by-step explanation:

Considering the graph

Given the vertices of the segment AB

A(-4, 4)

B(2, 5)

Finding the length of AB using the formula

$d_{AB}\:=\:\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$=\sqrt{\left(2-\left(-4\right)\right)^2+\left(5-4\right)^2}$

$=\sqrt{\left(2+4\right)^2+\left(5-4\right)^2}$

$=\sqrt{6^2+1}$

$=\sqrt{36+1}$

$=\sqrt{37}$

$d_{AB}\:=\sqrt{37}$

$d_{AB}=6.08$ units

Given the vertices of the segment JK

J(2, 2)

K(7, 2)

From the graph, it is clear that the length of JK = 5 units

so

$d_{JK}=5$ units

Given the vertices of the segment GH

G(-5, -2)

H(-2, -2)

Finding the length of GH using the formula

$d_{GH}\:=\:\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$=\sqrt{\left(-2-\left(-5\right)\right)^2+\left(-2-\left(-2\right)\right)^2}$

$=\sqrt{\left(5-2\right)^2+\left(2-2\right)^2}$

$=\sqrt{3^2+0}$

$=\sqrt{3^2}$

$\mathrm{Apply\:radical\:rule\:}\sqrt[n]{a^n}=a,\:\quad \mathrm{\:assuming\:}a\ge 0$

$d_{GH}\:=\:3$ units

Thus, from the calculations, it is clear that:

$d_{AB}=6.08$

$d_{JK}=5$

$d_{GH}\:=\:3$

Thus,

$d_{AB}\ne d_{JK}$

$d_{AB}\ne \:d_{GH}$

$d_{GH}\ne \:d_{JK}$

Therefore,

Option (A) is false

Option (B) is false

Option (C) is false

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Answer:

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Step-by-step explanation:

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