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  1.   y = 12

    Step-by-step explanation:

    Ordinarily, in math, when we say "the difference between ..." we mean the second listed item is subtracted from the first. (In English, we often mean "the positive difference between ..." so the order doesn't matter.)

    We'll assume we are only interested in the case where the difference is in the order listed.

      [tex]\dfrac{y+12}{y-4}-\dfrac{y}{y+4}=\dfrac{y+12}{y-4}\cdot\dfrac{y}{y+4}\\\\\dfrac{(y+12)(y+4)-(y-4)(y)}{y^2-16}=\dfrac{(y+12)(y)}{y^2-16}\\\\0=\dfrac{y^2-8y-48}{y^2-16}=\dfrac{(y-12)(y+4)}{(y-4)(y+4)}=\dfrac{y-12}{y-4}\qquad\text{$y\ne -4$}\\\\\boxed{y=12}[/tex]

    There is one solution.

    _____

    Comment on the solution

    If you were to "clear fractions" by multiplying the equation by (y^2-16), you would also obtain the extraneous solution y=-4. That is why we did not solve it that way.

  2.  y = -11

    Step-by-step explanation:

    [tex]\text{Sum:}\\\\.\quad \dfrac{y+1}{y-5}+\dfrac{10}{y+5}\\\\\\=\bigg(\dfrac{y+5}{y+5}\bigg)\dfrac{y+1}{y-5}+\dfrac{10}{y+5}\bigg(\dfrac{y-5}{y-5}\bigg)\\\\\\=\dfrac{y^2+6y+5}{(y+5)(y-5)}+\dfrac{10y-50}{(y+5)(y-5)}\\\\\\=\dfrac{y^2+6y+5+10y-50}{(y+5)(y-5)}\\\\\\=\dfrac{y^2+16y-45}{(y+5)(y-5)}[/tex]

    [tex]\text{Product:}\\\\.\quad \dfrac{(y+1)(10)}{(y-5)(y+5)}\\\\\\=\dfrac{10y+10}{(y-5)(y+5)}\\\\\\\text{Sum = Product:}\\\\\dfrac{y^2+16y-45}{(y+5)(y-5)}=\dfrac{10y+10}{(y+5)(y-5)}\qquad Restriction: y\neq -5, 5\\\\\\\rightarrow y^2+16y-45=10y+10\\\\\rightarrow y^2+6y-55=0\\\\\rightarrow (y+11)(y-5)=0\\\\\rightarrow y+11=0\quad and\quad y-5=0\\\\\rightarrow y=-11\qquad and\qquad y=5\\\\\\\text{Since y = 5 is a restricted value, it is not a valid solution}[/tex]

  3.  6

    Step-by-step explanation:

    [tex]\text{Difference:}\\\\.\quad \dfrac{6}{y-4}-\dfrac{y}{y+2}\\\\\\=\bigg(\dfrac{y+2}{y+2}\bigg)\dfrac{6}{y-4}-\dfrac{y}{y+2}\bigg(\dfrac{y-4}{y-4}\bigg)\\\\\\=\dfrac{6y+12}{(y+2)(y-4)}-\dfrac{y^2-4y}{(y+2)(y-4)}\\\\\\=\dfrac{6y+12-(y^2-4y)}{(y+2)(y-4)}\\\\\\=\dfrac{-y^2+10y+12}{(y+2)(y-4)}[/tex]

    [tex]\text{Product:}\\\\.\quad \dfrac{(6)(y)}{(y-4)(y+2)}\\\\\\=\dfrac{6y}{(y-4)(y+2)}\\\\\\\text{Difference = Product:}\\\\\dfrac{-y^2+10y+12}{(y+2)(y-4)}=\dfrac{6y}{(y+2)(y-4)}\qquad Restriction: y\neq -2, 4\\\\\\\rightarrow -y^2+10y+12=6y\\\\\rightarrow 0=y^2-4y-12\\\\\rightarrow 0=(y-6)(y+2)\\\\\rightarrow 0=y-6\quad and\quad 0=y+2\\\\\rightarrow 6=y\qquad and\quad -2=y\\\\\text{Since y = -2 is a restricted value, it is not a valid solution}\\\text{So, y = 6}[/tex]

  4. Lewis–Randall rule or Lewis fugacity rule states that fugacity of a component in an ideal solution is directly proportional to the mole

    fraction of the component in the solution.

    39 A solution in which the partial molar volumes of the components are the same as their molar volumes in the pure state is called an

    40 Henry’s law and Raoult’s law are identical when

  5. The value of the variable y is [tex]y=6[/tex]

    Explanation:

    The given equation is [tex]\frac{6}{y-4}-\frac{y}{y+2}=\left(\frac{6}{y-4}\right)\left(\frac{y}{y+2}\right)[/tex]

    Taking LCM on LHS of the equation, we get,

    [tex]\frac{6(y+2)-y(y-4)}{(y-4)(y+2)}=\left(\frac{6}{y-4}\right)\left(\frac{y}{y+2}\right)[/tex]

    Simplifying the term on RHS of the equation, we get,

    [tex]\frac{6(y+2)-y(y-4)}{(y-4)(y+2)}=\frac{6 y}{(y-4)(y+2)}[/tex]

    Since, both the sides of the equation have the same denominator, we can cancel them.

    Thus, we have,

    [tex]6(y+2)-y(y-4)=6 y[/tex]

    Multiplying the terms within the bracket, we get,

    [tex]6y+12-y^2-4y=6 y[/tex]

    Adding the like terms, we have,

    [tex]2y+12-y^2=6 y[/tex]

    Subtracting both sides by [tex]6 y[/tex], we have,

    [tex]-4y+12-y^2=0[/tex]

    Factoring the equation, we get,

    [tex](y+2)(y-6)=0[/tex]

    [tex]y=-2, y=6[/tex]

    The values of the variable y are [tex]y=-2, y=6[/tex]

    At the point [tex]y=-2[/tex], the equation [tex]\frac{6}{y-4}-\frac{y}{y+2}=\left(\frac{6}{y-4}\right)\left(\frac{y}{y+2}\right)[/tex] becomes undefined.

    Thus, the value of the variable y is [tex]y=6[/tex]

  6. y = 13.9

    Step-by-step explanation:

    It is given that  

    –5.7 + y = 8.2.    (1)

    Equation (1) is a linear equation in one variable.  

    LHS consists of a constant and a variable term

    RHS consists of only constant term

    Find the value of y

    Solving the equation

    -5.7 + y = 8.2

    moving -5.7 to the RHS

    y = 8.2 + 5.7

    y = 13.9

    Therefore the value of y = 13.9

    -5.7 + y = 8.2

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