Solve each rational equation. State the restrictions. 2𝑥+3/𝑥−1=3/𝑥 +2
clearly, the user needs to learn how to use parentheses. It should have been
(2𝑥+3)/(𝑥−1) = 3/(𝑥 +2)
which has no real solutions.
Excluded values are x=1 or x=-2
Apologies for the confusion. You are correct. Let's solve the equation (2𝑥+3)/(𝑥−1) = 3/(𝑥 +2) with the proper formatting.
To begin, let's find a common denominator for the fractions. The common denominator is (𝑥−1)(𝑥+2).
Multiplying both sides of the equation by (𝑥−1)(𝑥+2), we get:
(2𝑥+3)(𝑥+2) = 3(𝑥−1)
Expanding both sides, we have:
2𝑥^2 + 7𝑥 + 6 = 3𝑥 - 3
Combining like terms, we get:
2𝑥^2 + 4𝑥 + 9 = 0
However, this quadratic equation doesn't factor nicely and does not have any real solutions. Therefore, there are no real solutions to the equation (2𝑥+3)/(𝑥−1) = 3/(𝑥 +2).
The excluded values, as you correctly pointed out, are 𝑥 = 1 and 𝑥 = -2.
To solve the rational equation 2𝑥+3/𝑥−1=3/𝑥 +2, we will first simplify each numerator and then find a common denominator to combine the fractions.
Starting with the left side of the equation: 2𝑥+3/𝑥−1
We can find a common denominator by multiplying the first term's denominator (𝑥−1) with the second term's denominator (𝑥), which gives us (𝑥−1)(𝑥).
Now, we need to multiply the numerators of each term by their respective denominator, so we get 2𝑥(𝑥)+3(𝑥).
Our left side becomes : (2𝑥^2 + 3𝑥) / (𝑥−1)(𝑥)
Now, moving to the right side of the equation : 3/𝑥 + 2
Here, we can rewrite 2 as 2𝑥/𝑥 to make the denominators match.
This gives us: (3+ 2𝑥) / 𝑥
Now that both sides have the same denominator, we can combine the fractions:
(2𝑥^2 + 3𝑥) / (𝑥−1)(𝑥) = (3+ 2𝑥) / 𝑥
To eliminate the fractions, we can cross multiply:
(2𝑥^2 + 3𝑥) ⋅ 𝑥 = (𝑥−1)(𝑥) ⋅ (3+ 2𝑥)
Expanding both sides, we get:
2𝑥^3 + 3𝑥^2 = (𝑥^2 - 𝑥)(3+ 2𝑥)
Next, distribute (𝑥^2 - 𝑥) to (3+ 2𝑥):
2𝑥^3 + 3𝑥^2 = 3𝑥^2 - 3𝑥 + 2𝑥^3 - 2𝑥^2
Combining like terms, we have:
0 = -5𝑥^2 - 3𝑥
Now, we set this equal to zero and factor out an 𝑥:
0 = 𝑥(-5𝑥 - 3)
Now, we have two possible solutions to consider:
1) 𝑥 = 0
2) -5𝑥 - 3 = 0
Solving -5𝑥 - 3 = 0:
-5𝑥 = 3
𝑥 = -3/5
Therefore, the possible solutions are 𝑥 = 0 and 𝑥 = -3/5. However, we need to check if these values satisfy the restrictions of the original equation.
The restrictions occur when the denominator becomes zero, so we set the two denominators equal to zero and solve for 𝑥:
For (𝑥−1)(𝑥) = 0:
𝑥−1 = 0 or 𝑥 = 0
𝑥 = 1
For 𝑥 = 0:
No restriction because 0 does not make any of the denominators zero.
For 𝑥 = -3/5:
No restriction here either.
Thus, the solution to the rational equation is 𝑥 = 0 and 𝑥 = -3/5, with the restrictions 𝑥 ≠ 1.