You have to do some manipulations to the equation, until the variable remains alone on one side. The manipulations involve doing the same operation to both sides of the equation. Here is an example:
Solve "2x + 3 = 15" for x.
To get the "x" alone on one side, you have to get rid of the "2" and of the "3". Getting rid of the "3" first is simpler in this case. To get rid of the 3, you subtract 3. Note that you always have to do the same operation on both sides.
2x + 3 - 3 = 15 - 3
2x = 12
Next you get rid of the 2. This is done by dividing both sides by 2.
(2/2) x = 12/2
x = 6
This is the solution in this case.
multiply
You plug in what the variable is equal to for that variable then you will be able to finish the problem
To solve a system of equations by substitution, first solve one of the equations for one variable in terms of the other. Then, substitute this expression into the other equation. This will give you an equation with only one variable, which you can solve. Finally, substitute back to find the value of the other variable.
If you don't know something in a math equation you can replace it with a variable and then solve it algebraically.
You use variable in math problems to help you solve the answer. For example, you could do 50+50=a. The (a) would be the variable. Or for that same problem you could do 50+y=100. Then the (y) would be the variable.
Solve the problem using the + sign for the variable. Then solve the problem using the - sign for the variable. Report your answer as the answer that you got using + or the answer that you got using -.
multiply
You plug in what the variable is equal to for that variable then you will be able to finish the problem
There are four steps in an algebraic elimination problem. These steps are: to find a variable with equal or opposite coefficients, if equal then subtract the equations but if opposite then add, solve one variable equation left, and then substitute known variable into other equation and solve. hi
Get the variable by itselfWell my name is george, and thanks to intergers you can easily solve this problem correctly.
To solve a system of equations by substitution, first solve one of the equations for one variable in terms of the other. Then, substitute this expression into the other equation. This will give you an equation with only one variable, which you can solve. Finally, substitute back to find the value of the other variable.
it is the number you would use to solve the problem
It often doesn't matter which one you solve for first. But if you can easily solve one of the equations for one of the variables, that's the one you should solve for.
If you don't know something in a math equation you can replace it with a variable and then solve it algebraically.
Assuming the simplest case of two equations in two variable: solve one of the equations for one of the variables. Substitute the value found for the variable in all places in which the variable appears in the second equation. Solve the resulting equation. This will give you the value of one of the variables. Finally, replace this value in one of the original equations, and solve, to find the other variable.
You use variable in math problems to help you solve the answer. For example, you could do 50+50=a. The (a) would be the variable. Or for that same problem you could do 50+y=100. Then the (y) would be the variable.
If there is one variable. Then put each variable equal to zero and then solve for the other variable.