How do I solve 1/x+1/y=7, 2/x+3/y=17, where x and y is not equal to 0 by cross-multiplication method?
Equations can be written as1/x+1/y=7(y+x)/xy=7x+y=7xyAnd 2/x+3/y=17(2y+3x)/xy=173x+2y=17xyEquation 2 - 2(equation 1)3x+2y - 2x - 2y = 17xy-14xyx=3xy1=3yy=1/3From equation 11/x+1/y =71/x=4x =1/4
How do I solve for x and y in x + y =5 and xy=6?
Let’s first solve this problem by inspection. Notice that x and y can’t both be negative because then their addition will not produce the positive result required by the equation x + y = 5. And one of the unknowns cannot be negative (and the other positive) because then the product will not produce a positive result as required by the equation x y = 6. So x and y must both be positive and add up to 5. So the candidate solutions for the equation x + y = 5 are (0, 5), (1, 4) and (2, 3). Of these, only the pair (2, 3) will satisfy the equation x y = 6. So by inspection, we have that the solution is x = 2 and y = 3 as well as x = 3 and y = 2.Now let’s solve the problem analytically. From the first equation, we have that x = 5 – y. Plug this into the second equation, collect terms, and simplify to get y^2 – 5y + 6 = 0. This is a quadratic equation of the form ay^2 + by + c = 0, whose two solutions can be written as y = {-b ± (b^2 – 4ac)^1/2 }/2a. Substitute a =1, b = -5 and c = 6 into this result to find that our quadratic equation for y has two solutions: y = 3 and y = 2. Substitute these solutions for y back into the equation x = 5 – y to find that x = 2 and x = 3. This is the same result we found above by inspection.
How do I solve [math]y (x + y + 1)dx + x (x + 3y + 2)dy = 0?[/math]
We have differential equation of the form: M(x,y) dx + N(x,y) dy = 0Equation is exact if ∂M/∂y = ∂N/∂xM(x,y) = xy+y²+y —-> ∂M/∂y = x+2y+1N(x,y) = x²+3xy+2x —→ ∂N/∂x = 2x+3y+2Equation is not exact. However we can find integrating factor if (∂M/∂y − ∂N/∂x)/N is a function off x only or (∂N/∂x − ∂M/∂y)/M is a function of y only.(∂N/∂x − ∂M/∂y)/M = ((2x+3y+2)−(x+2y+1))/(y(x+y+1)) = (x+y+1)/(y(x+y+1)) = 1/yIntegrating factor = e^∫ 1/y dy = e^ln y = yMultiply through by yy²(x+y+1) dx + xy(x+3y+2) dy = 0(xy²+y³+y²) dx + (x²y+3xy²+2xy) dy = OEquation is now exact.Solution is F(x,y) = C, where∂F/∂x = M(x,y) = (xy²+y³+y²)F(x,y) = ∫ (xy²+y³+y²) dx = ½ x²y² + xy³ + xy² + G(y)∂F/∂y = N(x,y) = (x²y+3xy²+2xy)F(x,y) = (x²y+3xy²+2xy) dy = ½ x²y² + xy³ + xy² + H(x)Comparing both values of F(x,y), we see that G(y) = 0, H(x) = 0Solution:½ x²y² + xy³ + xy² = cx²y² + 2xy³ + 2xy² = Cxy² (x + 2y + 2) = C
What does it mean to solve for y in terms of x?
Solve for ‘y' means, solve the equation to get the value of y.And in terms of x , means, value of y not necessarily in pure constant form, but in the form of x.Example: Solve the following for y in terms of x3y +x = 7=> y = (7-x)/3Here, i solved the equation for y, in terms of x.
3x-y=7 ; x+4y=11 solve this equation by cramer's rule?
3x - y = 7 … (1)x + 4y = 11 … (2)Multiply (1) by 4 to get12x - 4y = 28 … (3)x + 4y = 11 … (2)Add (3) and (2), to get13x = 39, or x = 3Put this value x = 3 in (1) and you get9 - y = 7or 9–7 = y or y = 2.Hence x =3, and y =2.
Using the two equations above, solve for x. x + y = 8 and 3x = y + 10?
x + y + 3x = 8 + y + 10 4x = 18 x = 18/4 or 9/2 Answer: x = 9/2 or 4.5 Proof—1st & 2nd equation, substitute x with 4.5: 4.5 + y = 8 y = 3.5 3(4.5) = y + 10 13.5 = y + 10 y = 3.5
What is the solution of (X^3+3xy^2) dx+ (y^3+3x^2y) dy=0?
3xy^2 and 3x^2y provokes use of exact differential.
What is the solution to this ODE: (3x^2 + 4xy + y^2) dx + (2x^2 + 2xy + 9) dy = 0?
[math](3x^2 + 4xy + y^2) dx + (2x^2 + 2xy + 9) dy = 0[/math]Equation of the form [math]\quad M(x,y) dx + N(x,y) dy = 0[/math] is exact when [math]\quad\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}[/math]In that case, solution is of the form [math]F(x,y) = C[/math], where[math]\quad M(x,y) = \frac{\partial F}{\partial x}[/math] [math]\quad N(x,y) = \frac{\partial F}{\partial y}[/math][math]M(x,y) = 3x^2 + 4xy + y^2 \Longrightarrow \frac{\partial M}{\partial y} = 4x + 2y[/math] [math]N(x,y) = 2x^2 + 2xy + 9 \Longrightarrow \frac{\partial N}{\partial x} = 4x + 2y[/math]Equation is exact, so we find [math]F(x,y)[/math] by integrating:[math]F(x,y) = \int M(x,y) dx = \int (3x^2 + 4xy + y^2) dx = x^3 + 2x^2y + xy^2 + g(y)[/math] [math]F(x,y) = \int N(x,y) dy = \int (2x^2 + 2xy + 9) dy = 2x^2y + xy^2 + 9y + h(x)[/math]Comparing both results, we see that [math]g(y) = 9y, h(x) = x^3,[/math] and therfore:[math]F(x) = x^3 + 2x^2y + xy^2 + 9y[/math]Solution:[math]\boxed{\boldsymbol{x^3 + 2x^2y + xy^2 + 9y = C}}[/math]
Solve the simultaneous question x-y=3; xy+10x+y=150?
SUBSTITUION! :) how exciting. x-y=3 can be reworked: x=y+3 x-3=y now we can substitute (x-3) for y in the second equation: x(x-3)+10x+(x-3)=150 x^2-3x+10x+x-3=150 <