Um there are an infinite number of equations but some simple ones are: X + 1 = 6 X + 2 = 7 123553X = 617765
Three different equations that have ( x = 5 ) as a solution are: ( 2x - 10 = 0 ) (Solving gives ( x = 5 )). ( x^2 - 25 = 0 ) (Factoring gives ( (x - 5)(x + 5) = 0 ), so ( x = 5 )). ( 3x + 15 = 30 ) (Simplifying gives ( 3x = 15 ), thus ( x = 5 )).
To find the solution of the system of equations ( y = -2 + 5 ) and ( y = x - 5 ), we first simplify the first equation to ( y = 3 ). Next, we set ( 3 = x - 5 ) from the second equation and solve for ( x ), yielding ( x = 8 ). Therefore, the solution to the system is the point ( (8, 3) ).
There is no such pair. The solution to equation 1 and equation 2 is x = 1, y = 1. The solution to equation 2 and equation 3 is x = 1, y = 1. And the solution to equation 1 and equation 3 is any point on the line 3x + 2y = 5 - an infinite number of solutions. The fact that the determinant for equations 1 and 3 is zero (or that they are not independent) does not mean that there is no solution. It means that there is no UNIQUE solution. In this particular case, the two equations are equivalent and so have an infinite number of solutions.
To determine the number of solutions for the system of equations, we first need to clarify and rewrite the equations correctly. It seems there may be a formatting issue. If the equations are (2x + y = 5) and (2y = 4x + 10), we can analyze them. If the second equation can be simplified to the first, the system has infinitely many solutions (they are the same line). If they yield different lines, there will be no solution. Please confirm the equations for a precise answer.
x = -3/5 and y = -24/5
The answer will depend on statement 3 5 - whatever that may be!
equivalent equations
Three different equations that have ( x = 5 ) as a solution are: ( 2x - 10 = 0 ) (Solving gives ( x = 5 )). ( x^2 - 25 = 0 ) (Factoring gives ( (x - 5)(x + 5) = 0 ), so ( x = 5 )). ( 3x + 15 = 30 ) (Simplifying gives ( 3x = 15 ), thus ( x = 5 )).
Do you mean: 4x+7y = 47 and 5x-4y = -5 Then the solutions to the simultaneous equations are: x = 3 and y = 5
7
To find the solution of the system of equations ( y = -2 + 5 ) and ( y = x - 5 ), we first simplify the first equation to ( y = 3 ). Next, we set ( 3 = x - 5 ) from the second equation and solve for ( x ), yielding ( x = 8 ). Therefore, the solution to the system is the point ( (8, 3) ).
There is no such pair. The solution to equation 1 and equation 2 is x = 1, y = 1. The solution to equation 2 and equation 3 is x = 1, y = 1. And the solution to equation 1 and equation 3 is any point on the line 3x + 2y = 5 - an infinite number of solutions. The fact that the determinant for equations 1 and 3 is zero (or that they are not independent) does not mean that there is no solution. It means that there is no UNIQUE solution. In this particular case, the two equations are equivalent and so have an infinite number of solutions.
To determine the number of solutions for the system of equations, we first need to clarify and rewrite the equations correctly. It seems there may be a formatting issue. If the equations are (2x + y = 5) and (2y = 4x + 10), we can analyze them. If the second equation can be simplified to the first, the system has infinitely many solutions (they are the same line). If they yield different lines, there will be no solution. Please confirm the equations for a precise answer.
Tell whether the ordered pair (5, -5) is a solution of the system
Without any equality signs they can't be considered to be equations. But if you mean: x+2y =12 and y = x+3 then the solution is as follows x+2y = 12 -x+y = 3 Adding the equations together: 3y = 15 and y = 5 By substitution: x = 2 Solution: x = 2 and y = 5
If you mean: 6x-3y = -33 and 2x+y = -1 Then solving the simultaneous equations by substitution: x = -3 and y = 5
x = -3/5 and y = -24/5