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Q: How would solving a literal equation differ from a one or two step equation?

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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.

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.

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

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)

It is called solving the equation. * * * * * I would suggest that the answer is "evaluating it".

It really depends on the type of equation. Sometimes you can know, from experience with similar equations. But in many cases, you have to actually do the work of trying to solve the equation.

None. Properties of inequalities are not that relevant when solving equalities.

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