Unfortunately, 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" etc. Also use ^ to indicate powers (eg x-squared = x^2).
However, the question looks as if it might be a linear equation: an equation of a straight line. If that is the case then there are infinitely many solutions. The coordinates of each point on the line represents a solution.
First get y in terms of x: 5x-2y=-12 ----- -2y=-12-5x ------ y=6+(5/2)x Sub into other equation: 2x+3y=-1 ---- 2x+3(6+(5/2)x)=-1 Solving for x gets: x= -2 Sub into other equation: 5x-2y= -12 --- 5(-2)-2y= -12 --- y=1 So x= -2, and y= 1
A straight line through the points (0, 1) and (-0.4, 0).
Equations: 2x-3y = 4 and 5x+2y = 1 Multiply all terms in 1st equation by 5 and all terms in 2nd equation by 2:- So: 10x-15y = 20 and 10x+4y = 2 Subtract the 2nd equation from the 1st equation: -19y = 18 => y = -18/19 By substitution the solutions are: x = 11/19 and y = -18/19
The equation "x plus 2y equals 5" is: x + 2y = 5 2y = 5 - x y = 5/2 - 1/2x y = -1/2x + 5/2
Zero is an acceptable value for a variable. As a general rule, if you solve an equation and get x equals zero, that's just what it is: zero. There is one solution. Example: 10x +1 = 5x +1 5x = 1 - 1 5x = 0 x = 0 There is one solution, zero. If you solve for x and get two solutions, such as 2 and 0, there are two answers. Later on, if you've studied domain and range (the numbers you can use for x so that the equation is still defined, (i.e. not 1/0, or the square root of negative 1), you may find that x = 0 does not always work. You need to check first, to see if zero is in the domain, before you say it's an answer.
Without an equality sign it cannot be classed as an equation therefore it follows that there is no solution.
Both of them are equations!The solution is (x, y) = (2, 1).
2y - x = 3 y = -5x + 7 Rewrite the second y + 5x = 7 Mutiply it by 2 2y + 10x = 14 Now subtract the first equation from this: 2y + 10x = 14 2y - x = 3 11x = 11 Divide both sides by 11: x = 1 Substitute for x in the first equation: 2y - 1 = 3 Add 1 to both sides: 2y = 4 Divide both sides by 2: y = 2 Answer: (x, y) = (1, 2)
Multiply the bottom equation by -1
The equation given, 5x = 2y + 1, does not have a unique solution since it contains two variables, x and y. To find a specific solution, you would need additional information to uniquely determine the values of x and y.
If those are two different equations, such that 2y=5x-1 and x=y+2, then you subsitute y+2 in for x in the first equation, and when you solve the whole thing, you get that x=-1 and y=-3, or the point (-1,-3)
First get y in terms of x: 5x-2y=-12 ----- -2y=-12-5x ------ y=6+(5/2)x Sub into other equation: 2x+3y=-1 ---- 2x+3(6+(5/2)x)=-1 Solving for x gets: x= -2 Sub into other equation: 5x-2y= -12 --- 5(-2)-2y= -12 --- y=1 So x= -2, and y= 1
It is not possible to do so because the question contains one equation (5x + y = 1) and one expression (3x + 2y + 2). An expression cannot be solved.
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.
Double first equation: 10x + 2y = 10Subtract second equation: 7x = 7x = 1, y = 0
3x+2y=20 5x-2y=1 (3x+2y) + (5x-2y)= 20+1 3x+5x= 21....... 8x=21.........x=2 5/8 Subsitute x=3(2 5/8)+ 2y=20........... 7 7/8 +2y=20.........2y= 20- 7 7/8....... y=6 1/16
x-2y+bz=3 ax+2z=2 5x+2y=1 find values of a and b for which the system of linear equation has infinitely many solutions