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What are the solutions to the simultaneuos equations of x equals 2y -2 and x square equals y square plus 7?
Wiki User
∙ 13y ago
multiplying the first equation by the power of route 7, you will work out a formula to give you and equal equation to the one above, thus allowing you to work out the solutions to this eqation as being y= 14 and x= 21
Improved Answer:-
x = 2y-2 and x2 = y2+7
(2y-2)2 = y2+7
(2y-2)(2y-2) = y2+7
4y2-8y+4 = y2+7
3y2-8y-3 = 0
(3y+1)(y-3) = 0
Therefore y = -1/3 or y = 3
So when x = 4 then y = 3 and when x = -8/3 then y = -1/3
Wiki User
∙ 13y ago
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What are the solutions to the simultaneous equations of x square plus y square plus 4x plus 6y minus 40 equals 0 and x minus y equals 10?
Rearrange the second equation as x = 10+y and then substitute it into the first equation which will create a quadratic equation in the form of: 2y2+30y+100 = 0 and when solved y = -10 or y = -5 Therefore the solutions are: x = 0, y = -10 and x = 5, y = -5
What are the roots of the equation x²-10x plus 25 equals 0?
Equations have 'solutions', not roots.Both of the solutions to this quadratic equation are x=5 .(The expression on the left side is a perfect square. So the graph of theexpression is a parabola whose nose rests on the x-axis at x=5, andboth x-intercepts are at the same point.)
What is a discriminant and how does help to solve equations?
In the quadratic formula, the discriminant is b2-4ac. If the discriminant is positive, the equation has two real solutions. If it equals zero, the equation has one real solution. If the discriminant is negative, it has two imaginary solutions. This is because you find the square root of the discriminant and add or subtract it from -b and divide the sum or difference by 2a. If the square root is of a positive number, then you get two different solutions, one from adding the discriminant to -b and one from subtracting the discriminant from -b. If the square root is of zero, then it equals zero, and the solution is -b/2a. If the square root is of a negative number, then you have two imaginary solutions because you can't take the square root of a negative number and get a real number. One solution is from subtracting the discriminant from -b and dividing by 2a, and the other is from adding it to -b and dividing by 2a. The parabola on the left has a positive discriminant. The parabola in the middle has a discriminant of zero. The parabola on the right has a negative discriminant.
How many solutions to x2-10 equals 0?
x2-10 = 0 x2 = 10 x = the square root of 10
Which values are solutions to the inequality x2 equals 16?
x2 = 16take the root square for both sides the result will be :X = +4 or -4
How can you find out how many solutions an equation has?
By solving it. There is no single easy way to solve all equations; different types of equations required different methods. You have to learn separately how to solve equations with integer polynomials, rational equations (where polynomials can also appear in the denominator), equations with square roots and other roots, trigonometric equations, and others.Sometimes, the knowledge of a type of equations can help you quickly guess the number of solutions. Here are a few examples. An equation like:sin(x) = 0.5has an infinite number of solutions, because the sine function is periodic. An equation with a polynomial - well, in theory, you can factor a polynomial of degree "n" into "n" linear factors, meaning the polynomial can have "n" solutions. However, it may have multiple solutions, that is, some of the factors may be equal. Also, some of the solutions may be complex. A real polynomial of odd degree has at least one real solution.By solving it. There is no single easy way to solve all equations; different types of equations required different methods. You have to learn separately how to solve equations with integer polynomials, rational equations (where polynomials can also appear in the denominator), equations with square roots and other roots, trigonometric equations, and others.Sometimes, the knowledge of a type of equations can help you quickly guess the number of solutions. Here are a few examples. An equation like:sin(x) = 0.5has an infinite number of solutions, because the sine function is periodic. An equation with a polynomial - well, in theory, you can factor a polynomial of degree "n" into "n" linear factors, meaning the polynomial can have "n" solutions. However, it may have multiple solutions, that is, some of the factors may be equal. Also, some of the solutions may be complex. A real polynomial of odd degree has at least one real solution.By solving it. There is no single easy way to solve all equations; different types of equations required different methods. You have to learn separately how to solve equations with integer polynomials, rational equations (where polynomials can also appear in the denominator), equations with square roots and other roots, trigonometric equations, and others.Sometimes, the knowledge of a type of equations can help you quickly guess the number of solutions. Here are a few examples. An equation like:sin(x) = 0.5has an infinite number of solutions, because the sine function is periodic. An equation with a polynomial - well, in theory, you can factor a polynomial of degree "n" into "n" linear factors, meaning the polynomial can have "n" solutions. However, it may have multiple solutions, that is, some of the factors may be equal. Also, some of the solutions may be complex. A real polynomial of odd degree has at least one real solution.By solving it. There is no single easy way to solve all equations; different types of equations required different methods. You have to learn separately how to solve equations with integer polynomials, rational equations (where polynomials can also appear in the denominator), equations with square roots and other roots, trigonometric equations, and others.Sometimes, the knowledge of a type of equations can help you quickly guess the number of solutions. Here are a few examples. An equation like:sin(x) = 0.5has an infinite number of solutions, because the sine function is periodic. An equation with a polynomial - well, in theory, you can factor a polynomial of degree "n" into "n" linear factors, meaning the polynomial can have "n" solutions. However, it may have multiple solutions, that is, some of the factors may be equal. Also, some of the solutions may be complex. A real polynomial of odd degree has at least one real solution.
What are the solutions to the simultaneous equations of x square plus y square plus 4x plus 6y minus 40 equals 0 and x minus y equals 10?
Rearrange the second equation as x = 10+y and then substitute it into the first equation which will create a quadratic equation in the form of: 2y2+30y+100 = 0 and when solved y = -10 or y = -5 Therefore the solutions are: x = 0, y = -10 and x = 5, y = -5
What are the roots of the equation x²-10x plus 25 equals 0?
Equations have 'solutions', not roots.Both of the solutions to this quadratic equation are x=5 .(The expression on the left side is a perfect square. So the graph of theexpression is a parabola whose nose rests on the x-axis at x=5, andboth x-intercepts are at the same point.)
Which of the numbers are solutions to the equation x2 equals 3?
If: x2 = 3 Then: x = square root of 3
What are the solutions to the equation y2 equals 169?
y2 = 169 Square root both sides: y = 13
What are the solutions to the equation x2 equals 81?
x2 = 81 Square root both sides:- x = +/- 9
What is x2 -2x plus 2 equals 0?
It is a quadratic equation with no real roots or real solutions. In the complex domain, the solutions are 1 +/- i where i is the imaginary square root of -1.
Why do quadratic equations have to solutions?
Because all positive numbers have 2 square roots, eg 9 = 3 x 3 and also -3 x -3
What two numbers multiply to give you 20 but add to give you positive one?
Let the two numbers be x and y. Our two equations can then be written as: xy = 20 and x + y = 1 These equations have no real solutions. However, they do have complex solutions. x= 0.5 + 4.44i and y = (0.5 - 4.44i) were i is the square root of negative 1.
What is the nature of the solutions of the equation x2 plus 11 equals 0?
The two solutions are the conjugate complex numbers, +i*sqrt(11) and -i*sqrt(11), where i is the imaginary square root of -1.
What are the solutions to the simultaneous equations of y equals -2x and x2 plus y2 equals 80?
If: y = -2x then y ^2 = 4x^2 If: x^2 + y^2 = 80 then x^2 +4x^2 = 80 So: 5x^2 = 80 Divide all terms by 5: x^2 = 16 Square root both sides: x = -4 or +4 By substitution into the original equation solutions are: (-4, 8) and (4, -8)
What is a discriminant and how does help to solve equations?
In the quadratic formula, the discriminant is b2-4ac. If the discriminant is positive, the equation has two real solutions. If it equals zero, the equation has one real solution. If the discriminant is negative, it has two imaginary solutions. This is because you find the square root of the discriminant and add or subtract it from -b and divide the sum or difference by 2a. If the square root is of a positive number, then you get two different solutions, one from adding the discriminant to -b and one from subtracting the discriminant from -b. If the square root is of zero, then it equals zero, and the solution is -b/2a. If the square root is of a negative number, then you have two imaginary solutions because you can't take the square root of a negative number and get a real number. One solution is from subtracting the discriminant from -b and dividing by 2a, and the other is from adding it to -b and dividing by 2a. The parabola on the left has a positive discriminant. The parabola in the middle has a discriminant of zero. The parabola on the right has a negative discriminant.
