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omeli [17]
3 years ago
9

3-2(x-1)=2+4x How do you solve

Mathematics
2 answers:
aivan3 [116]3 years ago
8 0

Answer:

x = 1/2

Step-by-step explanation:

3 - 2(x - 1) = 2 + 4x

3 - 2x + 2 = 2 + 4x

-2x + 5 = 2 + 4x

-2x - 4x = 2 - 5

-6x = -3

x = -3/-6

x = 1/2

Darina [25.2K]3 years ago
6 0
You would multiply the negative two by x then the negative to by negative one then you would get 3-2x+2=2+4x then you would ad 2 x to 4x and get 6x the. Subtract 3-2 and you would get 1 the. You would subtract the 2 from the 1 and get -1 and you would get -1=6x so you would divide 6x and 1 and you would get x=-1/6 ( hope this helps)
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2 thirds of the product of 3 octavos and 16
Mazyrski [523]

Answer:4

Step-by-step explanation:

finding the product of 2 numbers means that you have to multiply those numbers

when multiplying fractions , numerators are multiplied by numerators and denominators by denominators

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3 years ago
JUST ANSWER PLEASE!!! QUICK
jeka57 [31]

Answer:

<u>Options 1 and 3</u>

Step-by-step explanation:

We should know that, the system of linear equations can be treated as matrices, i.e: we can modify or make any operations provide that we must apply the same operation for all terms of each equation.

Given:  the solution for the following system is (2,9)

Px + Qy = R  ⇒(1)

Tx + Uy = V  ⇒(2)

We will check which system of equation has the same solution.

<u>System A)</u>  Px + Qy = R

                  (P+T)x + (Q+U)y = R+V  ⇒(3)

So, By summing (1) and (2) we will get the equation (3)

So, system A has the same solution (2,9)

<u>System B)</u> Px + Qy = R

                 (P+2T)x + (Q+2U)y = R-2V  ⇒(4)

By multiplying equation (2) by 2 and add with equation (1), we will get:

 (P+2T)x + (Q+2U)y = R+2V

Which is not the same as equation (4)

So, system B has not the same solution (2,9)

<u>System C)</u> (T-P)x + (U-Q)y = V-R  ⇒(5)

                  Tx + Uy = V  

By multiplying equation (1) by -1 and add with equation (2), we will get the equation (5)

So, system C has the same solution of (2,9)

<u>System D)</u> (T-P)x + (Q+U)y = V-R  ⇒(6)

                  Tx + Uy = V  

We cannot get equation (6) by the same operation over equation (1)

Note the coefficient of x and y⇒ (T-P) and (Q+U)

They must be (T+P) and (Q+U) <u>OR </u>(T-P) and (Q-U)

So, system D has not the same solution of (2,9)

<u>System E)</u> (5T-P)x + (5U-Q)y = V-5R ⇒ (6)

                  Tx + Uy = V  

By subtract equation (1) from 5 times equation (2), we will get:

(5T-P)x + (5U-Q)y = 5V-R

Which is not the same as equation (6)

So, system E has not the same solution (2,9)

As a conclusion, the systems which have the same solution are:

<u>Options 1 and 3</u>

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