Question

solve for x. \\(\frac{4}{x^2 + 2x - 3} = \frac{5}{x + 3} - \frac{1}{x - 1}\\)
0
1
3
5

Explanation:

Step1: Factor the denominator

Factor \(x^{2}+2x - 3\). We know that \(x^{2}+2x - 3=(x + 3)(x - 1)\) since \((x+3)(x - 1)=x^{2}-x+3x - 3=x^{2}+2x - 3\). So the equation becomes \(\frac{4}{(x + 3)(x - 1)}=\frac{5}{x + 3}-\frac{1}{x - 1}\).

Step2: Multiply through by the LCD

The least common denominator (LCD) of the fractions is \((x + 3)(x - 1)\). Multiply each term by \((x + 3)(x - 1)\) to eliminate the denominators:
\(4=5(x - 1)-1(x + 3)\)

Step3: Expand the right - hand side

Expand \(5(x - 1)-1(x + 3)\):
\(5(x - 1)=5x-5\) and \(-1(x + 3)=-x - 3\). So \(5(x - 1)-1(x + 3)=5x-5-x - 3\).
Combine like terms: \(5x-x-5 - 3=4x-8\).
Now our equation is \(4 = 4x-8\).

Step4: Solve for x

Add 8 to both sides of the equation: \(4 + 8=4x-8 + 8\), which gives \(12 = 4x\).
Divide both sides by 4: \(\frac{12}{4}=\frac{4x}{4}\), so \(x = 3\). But we need to check for extraneous solutions.

Step5: Check for extraneous solutions

We need to check if \(x = 3\) makes the original denominators zero.
For the denominator \(x + 3\), when \(x = 3\), \(x+3=6
eq0\).
For the denominator \(x - 1\), when \(x = 3\), \(x - 1=2
eq0\).
For the denominator \(x^{2}+2x - 3=(x + 3)(x - 1)\), when \(x = 3\), \((3 + 3)(3 - 1)=6\times2 = 12
eq0\). Wait, but let's re - check our calculation.
Wait, when we expanded \(5(x - 1)-1(x + 3)\):
\(5(x - 1)-1(x + 3)=5x-5-x - 3=4x-8\). Then the equation is \(4=4x - 8\).
Adding 8 to both sides: \(4 + 8=4x\), \(12 = 4x\), \(x = 3\). But let's substitute \(x = 3\) back into the original equation:
Left - hand side: \(\frac{4}{3^{2}+2\times3-3}=\frac{4}{9 + 6-3}=\frac{4}{12}=\frac{1}{3}\).
Right - hand side: \(\frac{5}{3 + 3}-\frac{1}{3 - 1}=\frac{5}{6}-\frac{1}{2}=\frac{5-3}{6}=\frac{2}{6}=\frac{1}{3}\). So \(x = 3\) is a valid solution. Wait, but let's check the other options.
Wait, maybe I made a mistake in the previous thought. Let's check \(x = 5\):
Left - hand side: \(\frac{4}{5^{2}+2\times5-3}=\frac{4}{25 + 10-3}=\frac{4}{32}=\frac{1}{8}\).
Right - hand side: \(\frac{5}{5 + 3}-\frac{1}{5 - 1}=\frac{5}{8}-\frac{1}{4}=\frac{5 - 2}{8}=\frac{3}{8}
eq\frac{1}{8}\).
Check \(x = 0\):
Left - hand side: \(\frac{4}{0^{2}+2\times0-3}=\frac{4}{-3}=-\frac{4}{3}\).
Right - hand side: \(\frac{5}{0 + 3}-\frac{1}{0 - 1}=\frac{5}{3}+1=\frac{5 + 3}{3}=\frac{8}{3}
eq-\frac{4}{3}\).
Check \(x = 1\):
Denominator \(x - 1=0\) and \(x^{2}+2x - 3=1 + 2-3 = 0\), so \(x = 1\) is extraneous.
Wait, but when we calculated \(x = 3\), it worked. But let's re - do the algebra:
Original equation: \(\frac{4}{(x + 3)(x - 1)}=\frac{5}{x + 3}-\frac{1}{x - 1}\)
Multiply both sides by \((x + 3)(x - 1)\):
\(4=5(x - 1)-(x + 3)\)
\(4=5x-5-x - 3\)
\(4=4x-8\)
\(4x=4 + 8=12\)
\(x = 3\). And when we substitute \(x = 3\) into the original equation, it holds.

Answer:

3 (corresponding to the option "3")