5/y-3=y+7/2y-6+1=
an equation with extraneou solution.
I need help with this problem
Do you mean
5/(y-3) = [(y+7)/(2y-6)] + 1 ?
If not, put the parentheses where they belong. Otherwise, there are too many possible answers.
In the problem there are no parentheses
this how the problem is in the book.
5/y-3=y+7/2y-6+
5 over y-3= yplus seven over 2y-6 +1 there is no parenheses
To solve the equation 5/y - 3 = y + 7/2y - 6 + 1, we need to first simplify the equation and then solve for y.
Step 1: Simplify the equation by finding a common denominator for the fractions. The common denominator for y and 2y is 2y, so we can rewrite the equation as follows:
5/y - 3 = (y(2y) + 7)/(2y) - 6 + 1
Simplifying further:
5/y - 3 = (2y^2 + 7)/(2y) - 6 + 1
Step 2: Simplify the right side of the equation:
= (2y^2 + 7)/(2y) - 5
Step 3: Multiply both sides of the equation by 2y to eliminate the fraction:
2y * (5/y - 3) = 2y * ((2y^2 + 7)/(2y) - 5)
This gives us:
10 - 6y = 2y^2 + 7 - 10y
Step 4: Rearrange the terms and simplify to bring the equation to the form 2y^2 + (10y - 6y) - 3 = 0:
2y^2 + 4y - 3 = 0
Step 5: Now, we can solve this quadratic equation by factoring or using the quadratic formula. However, before doing that, let's remember that we are looking for an equation with an extraneous solution.
An extraneous solution is a root of a transformed equation that does not satisfy the original equation. So, let's say that we will solve the equation and check if the solution(s) satisfy the original equation. If they do not satisfy the original equation, they will be considered extraneous.
Step 6: Solving the quadratic equation 2y^2 + 4y - 3 = 0:
Considering that the quadratic equation does not factor nicely, we can use the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / (2a)
For our equation:
a = 2, b = 4, and c = -3.
Substituting these values into the quadratic formula, we get:
y = (-4 ± √(4^2 - 4 * 2 * -3)) / (2 * 2)
Simplifying further:
y = (-4 ± √(16 + 24)) / 4
= (-4 ± √40) / 4
= (-4 ± 2√10) / 4
= -1 ± 0.5√10
So, the possible solutions for y are:
y = -1 + 0.5√10 or y = -1 - 0.5√10
Step 7: Now, let's check if these solutions satisfy the original equation 5/y - 3 = y + 7/2y - 6 + 1:
Checking y = -1 + 0.5√10:
5/(-1 + 0.5√10) - 3 = (-1 + 0.5√10) + 7/(2(-1 + 0.5√10)) - 6 + 1
Performing the calculations, we see that this solution does not satisfy the original equation.
Checking y = -1 - 0.5√10:
5/(-1 - 0.5√10) - 3 = (-1 - 0.5√10) + 7/(2(-1 - 0.5√10)) - 6 + 1
Again, performing the calculations, we see that this solution does not satisfy the original equation either.
Therefore, there are no valid solutions for y in the original equation. This means that the equation has an extraneous solution.