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Helen [10]
2 years ago
12

Solve:3m-2(m-3)=1 please help

Mathematics
2 answers:
alina1380 [7]2 years ago
6 0

Answer:

m = -5

Step-by-step explanation:

3m-2(m-3)=1

Distribute

3m -2m +6 = 1

Combine like terms

m+6 =1

Subtract 6 from each side

m+6-6 =1-6

m = -5

mixer [17]2 years ago
5 0

Answer:

3m - 2(m - 3) = 1 \\ 3m - 2m - 6 = 1 \\ m - 6 = 1 \\ m = 1 + 6 \\ m = 7

<h3><u>PLEASE</u><u> MARK</u><u> ME</u><u> BRAINLIEST</u></h3>
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35x+10. How do I solve this?
kicyunya [14]

Hey there Bfarrar07 !

This is pretty simple when you think about it.

With algebra problems like this, you want to set it equal to zero.

So your first step is to take 35x + 10 and set it equal to zero.

35x + 10 = 0

Then bring the 10 to the other side by subtracting it, so you have something like this:

35x + 10 (-10) = -10 (Remember, what you do to one side you have to do to the other!) So when you take 10 from the left, you have to take ten from the right.

You now have 35x = -10

In order to get x by itself, you need to divide the problem by 35.

35x/35 = -10/35

On the left hand side, the 35's cancel out, so you're left with x.

And on the right hand side you have -10/35

So x = -10/35 (Don't forget to simplify!)

-10/35 can be reduced to -5/7

X = - 5/7

Hope this helps!

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3 years ago
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3 years ago
Read 2 more answers
If someone could work out the first half of the problem and explain that would be great!
andrew-mc [135]

Answer:

x = -5

y = -5

Step-by-step explanation:

If you multiply the top equation by negative 2, you can add it to the bottom equation and cancel out the x:

-2(-x + 5y = -20) --> 2x - 10y = 40

Add this to the bottom equation top to bottom:

2x - 10y = 40

-2x + 4y = -10

You get:

-6y = 30

Then solve for y:

y = -5

Plug this back in for either equation and solve for x:

-x + 5(-5) = -20

-x -25 = -20

-x = 5

x = -5

Your answers are:

x = -5

y = -5

7 0
3 years ago
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weeeeeb [17]

Answer:

x = -1, y = 1.5

Step-by-step explanation:

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Can someone tell me whats1+1​
Andreyy89

Answer:

2

Step-by-step explanation:

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