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First of all there is only an expression, there is no equation. Limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "equals", "squared", "cubed" etc.
Second, in order to solve a system with two unknown variables, you need two equations, not just one.
First of all there is only an expression, there is no equation. Limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "equals", "squared", "cubed" etc.
Second, in order to solve a system with two unknown variables, you need two equations, not just one.
First of all there is only an expression, there is no equation. Limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "equals", "squared", "cubed" etc.
Second, in order to solve a system with two unknown variables, you need two equations, not just one.
First of all there is only an expression, there is no equation. Limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "equals", "squared", "cubed" etc.
Second, in order to solve a system with two unknown variables, you need two equations, not just one.
Wiki User
∙ 11y ago
Wiki User
∙ 11y agoFirst of all there is only an expression, there is no equation. Limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "equals", "squared", "cubed" etc.
Second, in order to solve a system with two unknown variables, you need two equations, not just one.
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How would you know if a linear system has a solution?
One way is to look at the graphs of these equations. If they intersect, the point of intersection (x, y) is the only solution of the system. In this case we say that the system is consistent. If their graphs do not intersect, then the system has no solution. In this case we say that the system is inconsistent. If the graph of the equations is the same line, the system has infinitely simultaneous solutions. We can use several methods in order to solve the system algebraically. In the case where the equations of the system are dependent (the coefficients of the same variable are multiple of each other), the system has infinite number of solutions solution. For example, 2x + 3y = 6 4y + 6y = 12 These equations are dependent. Since they represent the same line, all points that satisfy either of the equations are solutions of the system. Try to solve this system of equations, 2x + 3y = 6 4x + 6y = 7 If you use addition or subtraction method, and you obtain a peculiar result such that 0 = 5, actually you have shown that the system has no solution (there is no point that satisfying both equations). When you use the substitution method and you obtain a result such that 5 = 5, this result indicates no solution for the system.
How do you solve linear equations by using elimination?
You multiply one or both equations by some constant (especially chosen for the next step), and add the two resulting equations together. Here is an example: (1) 5x + 2y = 7 (2) 2x + y = 3 Multiply equation (2) by -2; this factor was chosen to eliminate "y" from the resulting equations: (1) 5x + 2y = 7 (2) -2x -2y = -6 Add the two equations together: 3x = 1 Solve this for "x", then replace the result in any of the two original equations to solve for "y".
What is called Substitution method?
When you have a system of equations and you can solve for one in terms of th others and then replace it in the other equations that is substitution. x+y=15 x-y=1 well the second one can be x=1+y x+y=15 then can be (1+y)+y=15 by substitution. 1+2y=15 2y=14 y=7 Then x+7=15 x=8
What is the answer for -4x 4y-8 and x-4y-7 using matrices?
The question contains expressions, not equations. It is not possible to solve linear expressions - whether you use matrices or not.
Is (-1 1) a solution to this system of equations y 3x 4 4y - 3 x -7?
Without any equality signs the given terms can't be considered to be equations
How do you solve 9- 2 equations?
7
What does Xy3 3x2y - 7 equal?
12
Solve by substitution y equals -4x - 7 and y equals 3x?
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).
How do you solve two step equations with Integers?
3*-7=29
How do you solve 3w plus 7 equals x?
You can't solve it - you only have one equation and two unknowns. You need 2 equations to solve this.
What is a system of equations that equals -3 and 7?
Systems of equations don't equal numbers.
Type this equation in standard form 3x2y 7?
Without an equality sign it is not an equation
How do you solve -3y equals -x -7?
4
How do you solve linear equations by using elimination?
You multiply one or both equations by some constant (especially chosen for the next step), and add the two resulting equations together. Here is an example: (1) 5x + 2y = 7 (2) 2x + y = 3 Multiply equation (2) by -2; this factor was chosen to eliminate "y" from the resulting equations: (1) 5x + 2y = 7 (2) -2x -2y = -6 Add the two equations together: 3x = 1 Solve this for "x", then replace the result in any of the two original equations to solve for "y".
How would you know if a linear system has a solution?
One way is to look at the graphs of these equations. If they intersect, the point of intersection (x, y) is the only solution of the system. In this case we say that the system is consistent. If their graphs do not intersect, then the system has no solution. In this case we say that the system is inconsistent. If the graph of the equations is the same line, the system has infinitely simultaneous solutions. We can use several methods in order to solve the system algebraically. In the case where the equations of the system are dependent (the coefficients of the same variable are multiple of each other), the system has infinite number of solutions solution. For example, 2x + 3y = 6 4y + 6y = 12 These equations are dependent. Since they represent the same line, all points that satisfy either of the equations are solutions of the system. Try to solve this system of equations, 2x + 3y = 6 4x + 6y = 7 If you use addition or subtraction method, and you obtain a peculiar result such that 0 = 5, actually you have shown that the system has no solution (there is no point that satisfying both equations). When you use the substitution method and you obtain a result such that 5 = 5, this result indicates no solution for the system.
How do you solve system of equation by substitution?
You do the following: 1) Solve one of the equations for one of the variables 2) Substitute this variable in the other equation or equations 3) Simplify This should normally give you one less equation than the original set, with one less variables. For example:
Solve the system of equations -x minus 3y equals -4 and 3x plus 5y equals 16?
If: -x-3y = -4 and 3x+5y = 16 Then: x = 7 and y = -1
