How do you determine if an ordered pair is a solution to a given equation?

How do you determine if an ordered pair is a solution to a given equation?

To figure out if an ordered pair is a solution to an equation, you could perform a test. Identify the x-value in the ordered pair and plug it into the equation. When you simplify, if the y-value you get is the same as the y-value in the ordered pair, then that ordered pair is indeed a solution to the equation.

How do you find the number of solutions in an equation?

If solving an equation yields a statement that is true for a single value for the variable, like x = 3, then the equation has one solution. If solving an equation yields a statement that is always true, like 3 = 3, then the equation has infinitely many solutions.

How did you know if the given ordered pair is not a solution of the system?

To see if an ordered pair is a solution to an inequality, plug it into the inequality and simplify. If you get a true statement, then the ordered pair is a solution to the inequality. If you get a false statement, then the ordered pair is not a solution.

How do you know how many solutions a quadratic equation has?

The discriminant can be positive, zero, or negative, and this determines how many solutions there are to the given quadratic equation.

1. A positive discriminant indicates that the quadratic has two distinct real number solutions.
2. A discriminant of zero indicates that the quadratic has a repeated real number solution.

How many solutions exist for the given equation 3x 13?

Answer: There is no solution to the equation 3x + 13 = 3(x + 6) + 1.

How do you know if there is one solution?

A system of linear equations has one solution when the graphs intersect at a point. No solution. A system of linear equations has no solution when the graphs are parallel.

How do you tell if a system of equations has no solution or infinitely many without graphing?

3 Answers By Expert Tutors Two equations have parallel lines (no solution to the system) if the slopes are equal and and y-intercepts are not. Adding the equations gives an obviously false statement. This system of equations has no solution.

How do you find solutions?

Determine whether a number is a solution to an equation.

1. Substitute the number for the variable in the equation.
2. Simplify the expressions on both sides of the equation.
3. Determine whether the resulting equation is true. If it is true, the number is a solution. If it is not true, the number is not a solution.

How do you find the number and type of solution?

The discriminant of the Quadratic Formula is the quantity under the radical, b2−4ac b 2 − 4 a c . It determines the number and the type of solutions that a quadratic equation has. If the discriminant is positive, there are 2 real solutions. If it is 0 , there is 1 real repeated solution.

Is it possible for an equation to have a solution?

It is false as 3 will never equal 5. So, it is impossible for the equation to have a solution. He shows you a 2nd example that starts out with variables and then becomes 3=2. Any value you pick for “x” and insert into this equation will result in the 2 sides not being equal. So, no value of “x” can make the equation be true. Hope this helps.

Why are there infinite number of solutions to equations?

You now have the same thing on both sides. At this point you can tell that no matter what value you put in for x, the equation will always be true. That is why it has an infinite number of solutions. What Sal did is went ahead and solved it by eliminating the x terms. So what Sal is saying is that 2=2 no matter what x you input.

Is there a solution to the equation 3 = 5?

Direct link to Kim Seidel’s post “Sal wrote: `3=5` which is…” Notice – his version has no “x”. It is false as 3 will never equal 5. So, it is impossible for the equation to have a solution. He shows you a 2nd example that starts out with variables and then becomes 3=2.

How to calculate the sum of two positive numbers?

Find two positive numbers whose sum is 300 and whose product is a maximum. Solution Find two positive numbers whose product is 750 and for which the sum of one and 10 times the other is a minimum. Solution Let x x and y y be two positive numbers such that x +2y =50 x + 2 y = 50 and (x+1)(y +2) ( x + 1) ( y + 2) is a maximum.