How do you solve an equation for a given variable? For example, how would you do: 5(2m+3) -(1-2m) = 2[3(3+2m) - (3-m)]

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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5(2m+3) - (1-2m) = 2[3(3+2m) - (3-m)]

First we will expand brackets starting from the inside out:

5*2m + 5*3 -1 +2m = 2[3*3+3*2m -3 +m]

10m + 15 -1 +2m = 2[9 +6m -3 +m]

Group similar terms:

12m + 14 = 2[7m +6]

12m + 14 = 14m + 12

Group similar"

2m = 2

==> m= 1

 

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givingiswinning | Student, Grade 10 | (Level 1) Valedictorian

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5 x 2m+5 x 3 - 1 + 2m = 2[3 x 3 + 3 x 2m - 3 +m]

10m + 15 - 1 + 2m =  2x [ 9 + 6m - 3 + m].

12m + 14 = 2 [ 7m + 6 ]

12m + 14 = 14m + 12

14 - 12 = 14m - 12m

2 = 2m

1=m

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fashionableb1 | Student, Grade 10 | (Level 1) Salutatorian

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on, a dad and a grandpa.

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fashionableb1 | Student, Grade 10 | (Level 1) Salutatorian

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5 x 2m+5 x 3 - 1 + 2m = 2[3 x 3 + 3 x 2m - 3 +m]

10m + 15 - 1 + 2m =  2x [ 9 + 6m - 3 + m].

12m + 14 = 2 [ 7m + 6 ]

12m + 14 = 14m + 12

14 - 12 = 14m - 12m

2 = 2m

1=m

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atyourservice | Student, Grade 11 | (Level 3) Valedictorian

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5(2m+3) -(1-2m) = 2[3(3+2m) - (3-m)]

follow PEMDAS

first distribute the number in front of the parenthesis (multiplication)

10m + 15 - 1 +2m = 2 [ 9 + 6m - 3 + m ]

combine like terms

10m + 2m + 15 - 1 = 2 [ 9 - 3 + 6m + m]

12m + 14 = 2 [ 6 + 7m]

distribute the 2 to the numbers inside of the bracket:

12 m + 14 = 12 + 14m

move like terms to the same side

12 m - 14m = 12 - 14

-2m = -2

divide by -2

m = 1

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manager25 | Student, Grade 9 | (Level 2) eNoter

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5(2m+3) -(1-2m) = 2[3(3+2m) - (3-m)]

expanding will result into:

10m +15  -1 + 2m = 6(3+2m) - 2(3-m)

12m +14 = 18+12m -6 + 2m

12m +14 = 14m +12

all m on one side will result into :

12m-14m= 12-14

-2m=-2

m=-2/-2

m=1

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll start by removing the brackets both sides:

5(2m+3) -(1-2m) = 2[3(3+2m) - (3-m)]

10m + 15 - 1 + 2m = 6(3+2m) - 2(3-m)

Now, we'll remove the brackets form the right side:

10m + 15 - 1 + 2m = 18 + 12m - 6 + 2m

We'll subtract 18 both side:

10m + 15 - 1 + 2m - 18 = 18 - 18 + 12m - 6 + 2m

10m + 15 - 1 + 2m - 18 = 12m - 6 + 2m

We'll add 6 both sides:

10m + 15 - 1 + 2m - 18 + 6 = 12m - 6 + 6 + 2m

10m + 15 - 1 + 2m - 18 + 6 = 12m + 2m

We'll combine like terms from the right side:

10m + 15 - 1 + 2m - 18 + 6 = 14 m

We'll subtract 14m both sides:

10m + 15 - 1 + 2m - 18 + 6 - 14m = 0

We'll combine like terms again:

-2m + 2 = 0

WE'll subtract 2 both sides:

-2m = -2

We'll divide by -2:

m = 1

krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

Given:

5(2m + 3) - ( 1 - 2m) = 2[3(3 + 2m) - (3 - m)]

To solve this equation first thing we need to do is to simplify the equation. We can do this as follows by removing the bracket and then by bringing together similar terms on each side of the equation.

==> 5(2m + 3) - ( 1 - 2m) = 6(3 + 2m) - 2(3 - m)

==> 10m + 15 - 1 + 2m = 18 + 2m - 6 + 2m

==> 12m + 14 = 4m + 12

Next we transfer all the terms containing m on left hand side and other terms on right hand side:

==> 12m - 4m = 12 - 14

==> 8m = - 2

Therefore:

m = -2/8 = -1/4 = - 0.25

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neela | High School Teacher | (Level 3) Valedictorian

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The given equation 5(2m+3)-(1-2m)=2[3(3+2m)-(3-m)]

Solution:

To solve the equation, we open the brackets fillowing the rules of signs, We add equals or subtract equals, multiply or divide by equals both sides (but do not multiply by or divide by zeros) in such a way that all the variables go to one side of the equation and numbers or known goes to another side.

Here m is the variable.

5*2m+5*3-1+2m = 2[3*3+3*2m-3+m]

10m+15-1+2m =  2*[9+6m-3+m].

12m+14 = 2[7m+6]

12m+14 =14m+12

14-12 =14m-12m

2=2m

2/2=2m/2

1=m.

So m =1.

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