How do you solve #\frac { 6y } { 2y + 6} + \frac { y + 18} { 3y + 9} = \frac { 7y + 17} { y + 3}#?

Answer 1

No real solution. Read the explanation carefully.

#(6y)/(2y+6)+(y+18)/(3y+9)=(7y+17)/(y+3)#

We can rewrite it as:-

#(6y)/[2(y+3)]+(y+18)/[3(y+3)]=(7y+17)/(y+3)#

#=>(9y+y+18)/(3(y+3))=(7y+17)/(y+3)#

Cross multiply;

#=>(10y+18)(y+3) = 3(7y+17) (y+3)#

#=>(10y+18)(y+3)-3(7y+17) (y+3)=0#

Taking #(y+3)# common;

#=>(y+3)(10y+18-21y-51)=0#

#=>(y+3)(-11y-33)=0#

#=>-11*(y+3)*(y+3)=0#

#=> (y+3)^2=0#

#=> y+3 = 0# #=> color(red)(y=-3)#

BUT there is a catch.

We might think that #y=-3# is the solution but try to substitute this value back into the original equation i.e.

#(6y)/(2y+6)+(y+18)/(3y+9)=(7y+17)/(y+3)#

We find that terms in the denominator become zero.
#2*(-3)+6 = -6+6=0#,
#3*(-3)+9=-9+9=0#,
#-3+3 = 0#.

That means the expressions #(6y)/(2y+6), (y+18)/(3y+9), (7y+17)/(y+3)# become undefined at #y=-3# since division by zero is not defined. Hence we cannot say for sure whether the left-hand-side equals the right-hand-side or not. So #y=-3# is not a solution to this equation.

Where did we go wrong?

Remember the step where we cross multiplied #y+3#. Cross multiplication, or in fact any kind of multiplication on both sides of the equation is allowed only when the number(s) being multiplied is/are non-zero. Therefore cross multiplication in that step is only valid if #y+3!=0# or #y!=-3#. I.e. we are cross multiplying on the assumption that #y=-3 # is not a solution and even if it turns out to be a solution after solving the subsequent equations, we are going to neglect it.

#therefore# we neglect #y=-3# and hence, the given equation has no real solution.

Just to be sure, I checked on wolfram.

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Answer 2

To solve the equation (\frac{6y}{2y+6} + \frac{y+18}{3y+9} = \frac{7y+17}{y+3}), we can follow these steps:

  1. Simplify the fractions on both sides of the equation.
  2. Find a common denominator for the fractions.
  3. Combine the fractions on both sides of the equation.
  4. Solve for the variable, y.

Let's go through each step in detail:

  1. Simplify the fractions:

    • The fraction (\frac{6y}{2y+6}) cannot be simplified further.
    • The fraction (\frac{y+18}{3y+9}) can be simplified by factoring out a common factor of 3 from the numerator and denominator, resulting in (\frac{y+18}{3(y+3)}).
    • The fraction (\frac{7y+17}{y+3}) cannot be simplified further.
  2. Find a common denominator:

    • The denominators in the equation are (2y+6), (3(y+3)), and (y+3).
    • The common denominator for these fractions is (3(y+3)).
  3. Combine the fractions:

    • Rewrite the equation with the common denominator: (\frac{6y}{2y+6} + \frac{y+18}{3(y+3)} = \frac{7y+17}{y+3}).
    • Multiply each fraction by the necessary factors to obtain the common denominator: (\frac{6y}{2y+6} \cdot \frac{3(y+3)}{3(y+3)} + \frac{y+18}{3(y+3)} \cdot \frac{2y+6}{2y+6} = \frac{7y+17}{y+3}).
    • Simplify the numerators: (\frac{18y}{6y+18} + \frac{(y+18)(2y+6)}{3(y+3)(2y+6)} = \frac{7y+17}{y+3}).
  4. Solve for the variable, y:

    • Multiply both sides of the equation by the common denominator to eliminate the fractions: (18y + (y+18)(2y+6) = (7y+17)(3(y+3)(2y+6))).
    • Expand and simplify both sides of the equation: (18y + 2y^2 + 42y + 126 = 42y^3 + 189y^2 + 441y + 306).
    • Rearrange the equation to form a cubic equation: (42y^3 + 189y^2 + 441y + 306 - 18y - 2y^2 - 42y - 126 = 0).
    • Combine like terms: (42y^3 + 187y^2 + 381y + 180 = 0).
    • Solve the cubic equation using numerical methods or factoring techniques.

Please note that solving the cubic equation may require advanced mathematical techniques and may not have a simple algebraic solution.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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