How do you know if a value is a solution for an inequality?    how is this different from deyermining if a value is a solution to the equation?



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william1941's profile pic

Posted on (Answer #1)

Assume that you have an inequality of the form  ax^2 + bx + c > 0 . ( just an expression followed by a > or < sign and a constant value) If on substituting what is given as the solution for the inequation the relation is true, you know that it is a solution for the inequalition. The same goes for an equalition, the difference being that you will have an = sign in a equation instead of a < or > sign as in an inequation.

Usually the number of solutions available for an inequation is a lot more than that for an equation. An equation with an expression of order n can have a maximum of n solutions whereas the number of solutions for an inequation may even be infinite.

neela's profile pic

Posted on (Answer #2)

To know whether a particular value is a solution:

When both sides of an equation are algebraic expresions we put the solution value of the variable in place of the variable in the expressions. We arrive at a value in the left and another value in the right. Then the equality does not hold. If the calculared values are equal on both side, then we say the euality hods. Or equality is true.

In case of an inequality like the expression on the left < epression on the right, we  put the solution value for the unknown in left  and right side. The calculated value on the left should be less than the calculated value on the right then the inequality  holds.


7x+1  =  5x+11

We verify if x = 4 is a solution.

Left side value with x =4.: 7*4+1 = 29.

Right side value with x =4 : 5*4 +11 = 31.

  29 is not equal to 31. So  x =4 is not a solution.

Try with any other values for x.


5 x < 3x+for inequality: 10.

5x-3x < 10.

2x < 10

Divide by 2:

x <  5

This means for any  x < 5 is a solution. 1 is a solution.. Because if we put x =1, in 5x < 3x+10, we get:

 5*1 < 3*1+10.  

  5 < 13.

Similarly , we can verify if 6 is a solution:

5*6 <  3*5+10

30 < 15 +10

30 < 25. How this this. This is because we tried whether 6 is solution. So it is not a solution.

giorgiana1976's profile pic

Posted on (Answer #3)

We'll recognize the solution of an equation and the solution of an inequation in this way:

- the solution for an equation is a distinct value

For instance: x = a, where a is a real number or a complex number.

- the solution for an inequation is a set of valuesor an interval of values.

For instance: x is in the set A = {x belongs to R}, or x belongs to an interval:

[a, +infinite) or (-infinite a], if these intervals contain also the value of a.

(a, +infinite) or (-infinite a), if these intervals do not contain the value of a.

(a,b)  if the interval does not contain the values of a and b.


[a,b]  if the interval does contain the values of a and b.

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