A "solution" to an equation with 'x' and 'y' in it is a pair of numbers for 'x' and
'y' that make the equation a true statement.
There are an infinite number of solutions to a single equation with two
variables in it.
If you take a piece of graph paper and mark a dot on it for each solution of
this equation ... or even for only a hundred of them ... you'll see a shape on
the graph called a "parabola" ... like a big, wide cup or scoop. It crosses the
x-axis at two places ... one where 'x' is -2, the other where 'x' is 3 ... and
the lowest point of it is where it crosses the y-axis, where 'y' is -6 .
Every point on that big wide curved line is a 'solution' to your equation.
If you mean: x squared + y squared = 26 and x + y = 6 Then it works out that: y = 1 or 5 and therefore x = 5 or 1
If you mean: y-6-2(x+2) then it can be simplified to y-2x-10
y=b+x+x^2 This is a quadratic equation. The graph is a parabola. The quadratic equation formula or factoring can be used to solve this.
x2 - y2 + 9x - 9y =(x2 + 9x) - (y2 + 9y) =x(x + 9) - y(y + 9)================================Another way to go after it:x2 - y2 + 9x - 9y =(x2 - y2) + 9x - 9y =(x + y) (x - y) + 9 (x - y) =(x + y + 9) (x - y)
It is x*x + y*y*y*y
You cannot, in general, solve one equation with two unknown variables. x - y = x - x2 Subtract x from both sides: - y = - x2 Change signs: y = x2 And that is as far as you can go.
X2 - Y2 = (X + Y)(X - Y)
if: y = x2 y = x + 6 then x2 = x + 6 => x2 - x - 6 =0 => (x - 3)(x + 2) = 0 => x = 3 or -2 => y = 9 or 4 respectively, ie the points (3, 9) and (-2, 4).
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There are two numbers that meet those conditions, but they are not integers. You can work this out by taking the two statements and expressing them as equations, then using substitution to solve. We're told that we have two numbers that add up to six and that multiply to make -24. If we call those numbers x and y, we can say: x + y = 6 xy = -24 Now we'll take the first of them, and solve for y: y = 6 - x And substitute it into the latter: x(6 - x) = -24 6x - x2 = -24 x2 - 6x - 24 = 0 We now have a quadratic equation and can solve it this way: x2 - 6x - 24 = 0 x2 - 6x + 9 = 33 (x - 3)2 = 33 x - 3 = ±√33 x = 3 ± √33
If: y = x2-7x+8 and y = -x2+9x-6 Then: x2-7x+8 = -x2+9x-6 So: 2x2-16x+14 = 0 => x2-8x+7 = 0 Therefore: x = 1 and x = 7 By substitution: x =1, y = 2 and x = 7, y = 8 Points of intersection: (1, 2) and (7, 8)
x2 + x2 = 2x2. Don't be thrown by squares (or other powers). y + y = 2y. Just treat x2 like you treat y. But remember, you can only add together the same powers of x, you cannot add together x and x2 (or other powers of x).
Let's call them x and y. x+y=6,xy=7. So, y=6-x and xy=7. We subsitute 6-x for y and get x(6-x)=7. Distribute x. 6x-x2=7. Subtract 7 from both sides. -x2+6x-7=0. Use the Quadratic Formula: x=(-b+or-sqrt(b2-4ac))/2a. Plug in and solve. Your 2 answers are -6+or-sqrt(8)/-2.
equation1: y+1 = x => y = x-1 equation 2: y = x2-6x++5 If, x2-6x+5 = x-1 then, x2-7x+6 = 0, solving this quadratic equation, x = 1 or x = 6 Substituting into equation 1: when x = 1, y = 0 and when x = 6, y = 5 Therefore the solutions are: x = 1, y = 0 and x = 6, y = 5
What are we solving here for, Z, X, or Y?If solve for Z then the answer would be z = x + yIf solve for X then the answer would be z = z - yIf solve for Y than the answer would y = z - xHope this help.
What do you want to convert it to? x2 + y2 = 2x If you want to solve for y: x2 + y2 = 2x ∴ y2 = 2x - x2 ∴ y = (2x - x2)1/2 If you want to solve for x: x2 + y2 = 2x ∴ x2 - 2x = -y2 ∴ x2 - 2x + 1 = 1 - y2 ∴ (x - 1)2 = 1 - y2 ∴ x - 1 = ±(1 - y2)1/2 ∴ x = 1 ± (1 - y2)1/2
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