We have to solve the equation: 5(u-5)+10=12u+3(u+5)
First open the brackets
=> 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?
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.