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Komok [63]
3 years ago
5

3(x + 1) - 2x = -6 what is X i need halp

Mathematics
2 answers:
lara [203]3 years ago
5 0
3(x+1)-2x=-6
3x+3-2x=-6
(Like terms) 3x-2x=1x
3+1x=-6
-6-3+-9
Now you’re going to divide -9 by 3 which you will get -3
That will be your answer, hope that help you
Mashutka [201]3 years ago
4 0

Answer:x=−9

Step-by-step explanation:

Apply the distributive property.

3 x + 3 ⋅ 1 − 2 x = − 6

Multiply  3  by  1

.

3 x + 3 − 2 x = − 6

Subtract 2x from 3x.

x + 3 = − 6

Move all terms not containing  x  to the right side of the equation.

Subtract 3 from both sides of the equation.

x = − 6 − 3

Subtract 3 from − 6 .

x = − 9

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Answer:

The solutions of the system of equations are the points

(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})  

(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})  

Step-by-step explanation:

we have

y=-x^{2} +2x+10 ----> equation A

y=x+2 ----> equation B

Solve the system by substitution

substitute equation B in equation A

x+2=-x^{2} +2x+10

solve for x

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The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}

in this problem we have

-x^{2} +x+8=0

so

a=-1\\b=1\\c=8

substitute in the formula

x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}

x=\frac{-1\pm\sqrt{33}} {-2}

x=\frac{-1+\sqrt{33}} {-2}  -----> x=\frac{1-\sqrt{33}} {2}  

x=\frac{-1-\sqrt{33}} {-2}  -----> x=\frac{1+\sqrt{33}} {2}  

<em>Find the values of y</em>

For x=\frac{1-\sqrt{33}} {2}  

y=x+2

y=\frac{1-\sqrt{33}} {2}+2  ---->y=\frac{5-\sqrt{33}} {2}  

For x=\frac{1+\sqrt{33}} {2}  

y=x+2

y=\frac{1+\sqrt{33}} {2}+2  ---->y=\frac{5+\sqrt{33}} {2}  

therefore

The solutions of the system of equations are the points

(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})  

(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})  

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