Symbols not visible. Please resubmit using words eg "plus", "equals" etc
The easiest way to solve this system of equations is to solve for a variable in one of the equations. In the second equation, y = 3x. This can be substituted into the first equation: y = -4x - 7; 3x = = -4x - 7; 7x = -7; x = -1. Since we have determined that x equals -1, we can then substitute -1 into either equation to find our corresponding y-value. Thus: y = 3x; y = 3(-1) y = -3. Thus, the solution to this system of equations is (-1, -3).
3x + y + z = 63x - y + 2z = 9y + z = 3y + z = 3y = 3 - z (substitute 3 - z for y into the first equation of the system)3x + y + z = 63x + (3 - z) + z = 63x + 3 = 63x = 3x = 1 (substitute 3 - z for y and 1 for x into the second equation of the system)3x - y + 2z = 93(1) - (3 - z) + 2z = 93 - 3 + z + 2z = 93z = 9z = 3 (which yields y = 0)y = 3 - z = 3 - 3 = 0So that solution of the system of the equations is x = 1, y = 0, and z = 3.
Add the two equations and get 4x = 12 so x = 3 and y = -1
The system is inconsistent because there is no solution, i.e., no ordered pair, that satisfies both equations. You can see that this will be the case by seeing that their graphs have the same slope (2) but different y-intercepts (2 and 3/4 respectively). So the lines are parallel and will not intersect.
x = y = 3
-1
(3, 6)-------------------Let's see.(6) = 3(3) - 33(3) - (6) = 36 = 9 - 39 - 6 = 36 = 63 = 3========== (3, 6) is a solution to the system of equations. The only solution? I do not know.
Without any equality signs the given expressions can't be considered to be equations.
If "equations-" is intended to be "equations", the answer is y = -2. If the first equation is meant to start with -3x, the answer is y = 0.2
the system of equations 3x-6y=20 and 2x-4y =3 is?Well its inconsistent.
Solving equations in two unknowns requires two independent equations. Since you have only one equation there is no solution.
Without any equality signs the given terms can't be considered to be equations.
Without any equality signs the given terms can't be considered to be equations
No. Solving equations in two unknowns requires two independent equations. Since you have only one equation there is no solution.
No
The easiest way to solve this system of equations is to solve for a variable in one of the equations. In the second equation, y = 3x. This can be substituted into the first equation: y = -4x - 7; 3x = = -4x - 7; 7x = -7; x = -1. Since we have determined that x equals -1, we can then substitute -1 into either equation to find our corresponding y-value. Thus: y = 3x; y = 3(-1) y = -3. Thus, the solution to this system of equations is (-1, -3).
3x + y + z = 63x - y + 2z = 9y + z = 3y + z = 3y = 3 - z (substitute 3 - z for y into the first equation of the system)3x + y + z = 63x + (3 - z) + z = 63x + 3 = 63x = 3x = 1 (substitute 3 - z for y and 1 for x into the second equation of the system)3x - y + 2z = 93(1) - (3 - z) + 2z = 93 - 3 + z + 2z = 93z = 9z = 3 (which yields y = 0)y = 3 - z = 3 - 3 = 0So that solution of the system of the equations is x = 1, y = 0, and z = 3.