We have to solve the equation: 5(u-5)+10=12u+3(u+5)

First open the brackets

5(u-5)+10=12u+3(u+5)

=> 5u - 25 + 10 = 12u + 3u + 15

bring the terms with u to one side and the numeric terms to the other.

=> 5u - 12u - 3u = 15 + 25 - 10

=> -10u = 30

=> u = -30/10

=> u = -3

**Therefore u = -3**

What are the right steps to solve?

5(u-5)+10=12u+3(u+5).

Solution:

We open the brackets of each side and add or subtract the like terms:

LHS:5(u-5) -12u = 5u-5*5+10 = 5u-25+10 = 5u-15.

RHS: 12u+3(u+5) = 12u+3u+3*5 = 15u+15.

Now we rewrite the equation with the above simplified expressions of LHS abs RHS:

5u-15 = 15u +15.

We move u's to left and numbers to right as below:

Add 15 to both sides:

5u-15+15 = 15u+15+15.

5u = 15u+30.

We subtract 15u from both sides:

5u-15u = 15u+30-15u = 15u-15u+30 = 30.

-10u = 30.

We divide both sides by -10

-10u/-10 = 30/-10.

u = -3.

Therefore u = -3 is the solution.

The first step is to remove the brackets both sides. For this reason, we'll apply the distributive property of multiplication, over addition.

5u - 25 + 10 = 12u + 3u + 15

Now, we'll isolate the terms in u to the left side and the numbers alone, to the right side. For this rason, we'll subtract 12u + 3u both sides and we'll add 25 and subtract 10 both sides:

5u - 15u = 15 + 15

We'll combine like terms:

-10u = 30

We'll divide by -10 to get u:

u = 30/-10

**u = -3**

**For u = -3, the given identity is true.**