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mote1985 [20]
1 year ago
10

Select all equations that can result from adding these two equations or subtracting one from the other.

Mathematics
1 answer:
seraphim [82]1 year ago
6 0

The two equations that can be formed by adding or subtracting is

4x−4y = 16 and  2x − 6y = -8 .

the two equations are : x + y = 12 and 3x − 5y = 4

Adding we get :

x + y = 12

<u>3x − 5y = 4</u>

4x -4y = 16

Subtracting we get:

x + y = 12

<u>3x − 5y = 4</u>

-2x + 6y = 8 or 2x - 6y = -8

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side. It demonstrates the equality of the relationship between the expressions printed on the left and right sides.

We have LHS = RHS (left hand side = right hand side) in every mathematical equation. To determine the unknown parameters variable that represents an unknown quantity, equations can be solved. A statement is not an equation if it has no "equal to" sign.

It will be regarded as a phrase. In the following section of this article, you will discover the distinction between an equation and an expression.

to learn more about equations visit:

brainly.com/question/14686792

#SPJ1

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4 years ago
when a number is decreased by 83%, the result is 4. what is the original number to the nearest tenth?
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4 years ago
Use the completing the square method to find the roots of 2x^2+3x-15=13 include the steps, it would be greatly appreciated thank
ValentinkaMS [17]

Answer:

The roots are

x1= \frac{-3+\sqrt{233}}{4}

x2= \frac{-3-\sqrt{233}}{4}

Step-by-step explanation:

we have

2x^{2}+3x-15=13

so

Group terms that contain the same variable, and move the constant to the opposite side of the equation

2x^{2}+3x=13+15

2x^{2}+3x=28

Factor the leading coefficient  

2(x^{2}+(3/2)x)=28

Complete the square. Remember to balance the equation by adding the same constants to each side

2(x^{2}+(3/2)x+(9/16))=28+(9/8)

2(x^{2}+(3/2)x+(9/16))=233/8

(x^{2}+(3/2)x+(9/16))=233/16

Rewrite as perfect squares

(x+(3/4))^{2}=233/16

square root both sides

(x+\frac{3}{4})=(+/-)\sqrt{\frac{233}{16}}\\ \\(x+\frac{3}{4})=(+/-)\frac{\sqrt{233}}{4}\\ \\x= -\frac{3}{4}(+/-)\frac{\sqrt{233}}{4}

x1= -\frac{3}{4}(+)\frac{\sqrt{233}}{4}=\frac{-3+\sqrt{233}}{4}

x2= -\frac{3}{4}(-)\frac{\sqrt{233}}{4}=\frac{-3-\sqrt{233}}{4}

6 0
3 years ago
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