# How do I solve for a variable when it is on both sides like: 5y + 3 = 4y -1?

### 13 Answers | Add Yours

We want the variable on just one side of the equal sign, so we have to get rid of one of them. In this case 5y + 3 = 4y - 1, we want to get rid of the 5y or the 4y, your choice. Say we choose to get rid of the 4y. Since it is a positive 4y we would have to subtract 4y. 5y + 3 = 4y **- 4y** -1. But let's not forget, if we do it to one side of the equal sign, we must do it to the other, so; 5y **- 4y** + 3 = 4y **- 4y** - 1. This gives us y + 3 = -1, eliminating the variable from one of the sides of the equal sign.

You can add, subtract, multiply, divide, or perform any other mathematical opperation to both sides of the equation. So this includes adding or subtracting by the variable itself.

So to solve your problem, move the 4y onto the other side by subtracting it from both sides, like this:

5y + 3 = 4y - 1

5y - 4y + 3 = 4y - 1 - 4y

y + 3 = -1

**y = -4**

Plug it in and check:

5(-4) + 3 = 4(-4) - 1

-20 + 3 = -16 - 1

-17 = -17 check.

5y + 3 = 4y - 1

You first subtract 3 on both sides

5y + 3 = 4y - 1

-3 -3

5y = 4y - 4

Then you subtract 4y from both of the sides

5y = 4y- 4

-4y -4y

The final answer is y = -4

5y + 3 = 4y -1

To solve problems like these you have to move the " y " on one side. In this equation in order to have the " y " all on the same you will have to subtract 4y on both sides.

By subtracting 4y on both sides, your equation should look like

**y + 3 = -1 **now subtract 3 from both sides

By subtracting 3 from both sides, your equation should look like

**y = -4** which is your answer

5y + 3 = 4y -1

You move the variable on one side

5y-4y=-1-3

1y=-4

therefore y=-4

5y + 3 = 4y -1

bring all the like terms on the same side

5y-4y=-1-3

simplify

y=-4

5y + 3 = 4y -1?

You can move variables about in a similar fashion to how you would manipulate numbers:

5y - 4y = -1 - 3

y = -4

5y+3=4y-1

5y+3+1=4y-1+1

5y+4=4y

5y+4-5y=4y-5y

4=-1y

4/-1=-1y/-1

y=-4

You have to get them both on the same side. If they are positive variables, take them away and if they are negative variables, make them positive to cancel them out.

5y-4y+3=4y-4y-1=

y+3=-1

The rest would be to solve for y:

y+3+1=-1+1

y+4=0 where y=-4

you can solve like this:

5y-4y=-1-3

y=-4

in these kind of equations we always join the similar variables and then subtract or add

This is an equation in one unknown or variable y.We solve the equation in one unknown by making the variable to one side and the known or the number to the other side. We do this by simple operartions like adding the equals to both sides, or subtracting the equals to both sides , or multiplying by the equals both sides , or dividing by the equals(but not by zero) both sides of the equation:

5y+3=4y-1. Here 5y on the left and 4y on the right are the unknowns. 3 on left and -1 on the right are the numbers and they are known. To make the unknowns on the left and the knowns on the right with a purpose of making the unknown to be determined through the knowns, we do the following operations , one by one:

Subtract 4y from both sides:

5y+3-4y= 4y-1-4y.

Simplify by collecting the like terms together:

5y-4y+3=4y-4y-1

y+3=-1

Subtract 3 from both sides:

y+3-3=-1-3

Simplify.

y= -4.

Verification:

Substitute the obtained solution, y=-4 in the original equation and see whether you get the same value on both sides:

Left:5(-4)+3=-20+3=-17

Righr:4(-4)-1=-16-1=-17

Hope this helps.

The problem of a variable being on both the side of the equation can simply be solved by transferring all the terms containing the variable on one side of the equation. There is one simple rule for transferring a term from one side of the equation to the other. Just reverse the sign of the term - that is plus (+) to minus (-) and minus to plus, when transferring the terms from one side to the other.

The equation given is a first order equation - that is, there are no terms that contain variable (y) in form of other than raised to the power of 1. This means there are no terms containing y^2, Y^3 , or 1/y which are exponents of y raised to the power of 2, 3, and -1 respectively.

After understanding these basic rules we can now solve the equations as follow:

The equation is 5y + 3 = 4y - 1

Rewrite the equation transferring all the terms containing 'y' to the left hand side, and all plain numbers to the right hand side.

Therefore: 5y -4y = - 1 - 3

Therefore: y = - 4