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Isha sharma writes three liner equations in two variables as under 3x plus 2y equals 5 5x-7y equals -2 6x plus 4y equals 10which pairs has no solution?
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.
Is (2-11) a solution for -3x 2y 5?
To determine if (2, -11) is a solution for the equation (-3x + 2y = 5), substitute (x = 2) and (y = -11) into the equation. This gives (-3(2) + 2(-11) = -6 - 22 = -28), which does not equal 5. Therefore, (2, -11) is not a solution to the equation.
Suppose x equals 5 is a solution to the equation 4zx plus 1 equals j where z and j are constants. Find a solution to the equation 8zx plus 5 equals 2j plus 3?
It is also x = 5 8zx + 5 = 2j + 3 → 8zx + 2 = 2j → 4zx + 1 = j Which is the first equation, which has a solution x = 5.
Which equation has a solution of -20?
10 (-2) 5 (-4)
Find two solutions for the equation y 3x - 5?
-2
Solution of equation 2x-5 equals 2-1?
2x -5 = 2-12x -5 +5= 2-1+52x = 62x/2 = 6/2x = 3
Isha sharma writes three liner equations in two variables as under 3x plus 2y equals 5 5x-7y equals -2 6x plus 4y equals 10which pairs has no solution?
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.
Anyone who can simplify 2x2 9x - 5?
Presumably this is a quadratic equation question asking you to find the solution of 2x2+9x-5 = 0 First factorise the expression in the equation in order to find its solution. (2x-1)(x+5) = 0 Solution: x = 1/2 or x = -5
Is (2-11) a solution for -3x 2y 5?
To determine if (2, -11) is a solution for the equation (-3x + 2y = 5), substitute (x = 2) and (y = -11) into the equation. This gives (-3(2) + 2(-11) = -6 - 22 = -28), which does not equal 5. Therefore, (2, -11) is not a solution to the equation.
Suppose x equals 5 is a solution to the equation 4zx plus 1 equals j where z and j are constants. Find a solution to the equation 8zx plus 5 equals 2j plus 3?
It is also x = 5 8zx + 5 = 2j + 3 → 8zx + 2 = 2j → 4zx + 1 = j Which is the first equation, which has a solution x = 5.
Is 2 a solution of the inequality 2x 5 9?
2 is a solution of the equation, but not if it's an inequality.
What does a golden ratio mean in maths?
It is [1 + sqrt(5)]/2 It is the solution to the equation 1/x = x/(1+x)
What is the solution set of x-5-2?
The solution set of the equation x - 5 - 2 is x = 7. Therefore, the solution set is {7}.
Which equation has a solution of -20?
10 (-2) 5 (-4)
Find two solutions for the equation y 3x - 5?
-2
Does an equation with an integer coefficient always have an integer solution?
No, an equation with integer coefficients does not always have an integer solution. For example, the equation (2x + 3 = 5) has the integer solution (x = 1), but the equation (x^2 + 1 = 0) has no real solutions, let alone integer ones. The existence of integer solutions depends on the specific form and constraints of the equation.
Is 5 1 a solution of y -3x - 2?
I understand the equation to be y = -3x - 2 and the point to be (5,1). I substitute 1 for each appearance of y and 5 for each appearance of x: 1 = -3(5) - 2 = -15 - 2 = -17, which is not a true statement. Therefore, that is not a solution. To get a solution, set x=1, and calculate y by substituting this value (1) for x wherever it appears: y = -3(1) - 2 = -3 - 2 = -5. Therefore, (-5,1) is a solution. (I suspect that this is what you meant to put in the question.)
