The difference is that first you have to understand the problem and translate it into an equation (or equations).
The 1st step would be to give an example of the equation to be solved.
if you solve by plugging in the known values ahead of time you won't have a general formula for the variable in the literal equation. Therefor if the known values change, you would have to start all over again, making each problem more individualized. Once the literal equation is solved for some variable, if the known values change all you have to do is plug in those new numbers to your literal equation, and out pops your answer
When solving equations remember that whatever operations are performed on the LHS of the equation must be performed on its RHS to keep the equation in balance.
Eradicate the fractions.
The 1st step would have been to show a particular quadratic equation in question.
The 1st step would be to give an example of the equation to be solved.
The first step not possible in solving an equation algebraically is not to provide an equation in the first place in which it appears to be so in this case.
if you solve by plugging in the known values ahead of time you won't have a general formula for the variable in the literal equation. Therefor if the known values change, you would have to start all over again, making each problem more individualized. Once the literal equation is solved for some variable, if the known values change all you have to do is plug in those new numbers to your literal equation, and out pops your answer
When solving equations remember that whatever operations are performed on the LHS of the equation must be performed on its RHS to keep the equation in balance.
The first step would be to find the equation that you are trying to solve!
Eradicate the fractions.
The 1st step would have been to show a particular quadratic equation in question.
You would have a reasonable shot at finding the correct solution.
Write the quadratic equation in the form ax2 + bx + c = 0 then the roots (solutions) of the equation are: [-b ± √(b2 - 4*a*c)]/(2*a)
Solving for a variable involves isolating that variable in an equation to determine its value. This process typically includes using algebraic operations such as addition, subtraction, multiplication, or division to manipulate the equation. The goal is to express the variable in terms of known quantities or constants. For example, in the equation (2x + 3 = 11), solving for (x) would yield (x = 4).
Solving a literal equation for a variable allows for a general solution that can be applied to various scenarios, rather than just a single instance. This approach provides flexibility and insight into the relationships between variables, making it easier to understand how changes in one variable affect others. Additionally, it can simplify complex problems by allowing you to rearrange equations for different contexts without needing to repeatedly substitute values.
It is called solving the equation. * * * * * I would suggest that the answer is "evaluating it".