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Vera_Pavlovna [14]
3 years ago
8

2x+y=3, 5x-2y=12 how to solve this

Mathematics
2 answers:
Leviafan [203]3 years ago
7 0
Start by getting one of the equations in terms of one variable, in this case lets put the first equation in terms of y

2x+y=3 ---> y=3-2x

now we plug that into the second equation

5x-2y=12 ---> 5x-2(3-2x)=12

we can then multiply and simplify

5x-2(3-2x)=12 ---> 5x-6+4x=12 ---> 9x-6=12 ---> 9x=18 ---> x=2

we can now plug x=2 into an equation to solve for y

2(2)+y=3 ---> 4+y=3 ---> y=-1
Brums [2.3K]3 years ago
7 0
2x+y=3 =4x+2y=6
5x-2y=12. =5x-2y=12
--------------------------------
9x=18: x=2
4+y=3:y=-1
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Answer:

He should have moved to the right.

Step-by-step explanation:

Ken moved 7 to the right which is correct because he needs to get 7º higher, but when he moved 2º to the left that would be a temperature decrease which is is incorrect.

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3 years ago
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Help is appreciated. Easy I just am always confused
saveliy_v [14]

Answer:

BA=BC

Step-by-step explanation:

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4 years ago
Write the given phrase as an algebraic expression. Use the variable x to represent the unknown number.
Alchen [17]
Hello! I would love to help!

Let's start with this part of the equation: "the sum of a number and seven"

Alright. We know that x represents an unknown number. Do you see a part of the equation that could translate to "an unknown number?"

I see "a number." So let's fill X in for "a number.

Alright. So now we have "the sum of x and 7."

Next, let's remember that sum means adding. So we just need to 7 to x

X+7


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Alright. Now we just have to do the "twice." When it is asking for "twice", it is asking us to multiply our answer by two. But we need to multiply both x and 7. The best way to do that is to put our "x+7" in parenthesis and put a two outside.

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5 0
4 years ago
I the equation 3x^2+6x=12, the value of c is?
ozzi

Answer:

c = -12

Step-by-step explanation:

Quadratic Standard Form: ax² + bx + c

Step 1: Write equation

3x² + 6x = 12

Step 2: Subtract 12 on both sides

3x² + 6x - 12 = 0

Here, we have the standard form of the quadratic. We see that our c = -12

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Which of the following is an extraneous solution of startroot 4 x 41 endroot = x 5? x = –8 x = –2 x = 2 x = 8
Gnom [1K]

The extraneous solution of startroot 4 x 41 endroot = x 5 will be A. -8.

<h3>What is an extraneous solution?</h3>

It should be noted that an extraneous solution simply means a root of a transformed equation which isn't part of the original equation.

✓4x + 41 = x + 5

Square both sides

4x + 41 = x² + 10x + 25

x² + 6x - 16

x(x + 8) - 2(x + 8) = 0

x + 8 = 0

x= 0 + 8 = 8

Learn more about extraneous solution on:

brainly.com/question/295656

#SPJ4

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