Which is the solution of the equation 2x-10=-3x+5?

giorgiana1976 | Student

In order to solve the linear equation, we'll have to find a value for x so that the equation to hold, even after the substitution of the unknown x with the found value was made.

A proper manner to solve this type of equation would be to move to the left side, the terms which contain the unknown x, changing the sign of the terms moved. The terms which are not containing the unknown, will be moved to the right side, of course, with the opposite sign.


2x+3x = 10+5

5x = 15

Now, to move 5 across the equal sign, we need to divide both sides of the equation with 5.

5x/5 = 15/5

x = 3

udonbutterfly | Student

2x-10=-3x+5 In order to solve for x you are going to add 10 to both sides

2x -10 +10 =-3x +5 +10 The 10 on the left side of the equation will be cancelled out.

2x = -3x + 15 Now add 3x from each side.

2x +3x = -3x + 15 +3x The 3x on the right side will the be cancelled out.

5x = 15 Now divide 5 on both sides/

5x/ 5 = 15/ 5 The negative from -x will be cancelled out. 

x= 3 Here is your answer!

jess1999 | Student

2x - 10 = -3x + 5

First, add 3x on both sides that way " x " can be on the same side.

By adding 3x on the same side, your equation should look like

5x - 10 = 5 now, add 10 on both sides

By adding, your equation should look like

5x = 15 now divide 5 on both sides

By diving, your equation should look like

x = 3 which is your answer

jscorpio43 | Student

in order to solve this equation, we must variables on one side and numbers on the other side

so we need to get rid of the -3x to do this we must adda positive 3x to both sides-3x+3x   2x+3x

2x+3x =5x

5x-10=5 add 10 to both sides 5x=15 dividing both sides by 5 now leaving us with x alone, x=15/5=3 our answer is x=3

neela | Student

To 2x-10 = -3x+5


This is a linear equation with one variable x.We solve equations  by simple operarations like (i) adding suitable equals to both sides of the equation or (ii) subtracting suitable equals to both sides or (iii) multiplying bu suitable equals both sides or (iv) dividing by suitable equals(but not dividing by zeros) both sides until we arrive at a n equation of the type x = a numerical (or known) value which is the solution.

2x-10 = -3x+5. Add 10. This makes 10 vanish on left.

2x=-3x+5+10 . Or

2x=-3x+15. Add 3x. This makes x's vanish on the right.

2x+3x = 15. Or

5x = 15. Divid by 5. This makes  left reduce from 5x to x and rnumber  on the right which is the solution.

5x/5 = 15/5 . Or

x = 3. So the entire procedure consists the steps enumerated from (i) to (iv).