Find both solutions of... | Mathematics 0580 | Paper 2 Mock 01 | Non-calculator (Extended)

Question 1 Report

Find both solutions of \(\dfrac{4}{x+1}+\dfrac{4}{x-1}=3\).

Answer

Answer Details

Clear both fractions by multiplying every term by the common denominator \((x+1)(x-1)\). This turns the equation into a quadratic.

Multiplying through gives \(4(x-1)+4(x+1)=3(x+1)(x-1)\) [M1].
The left side is \(4x-4+4x+4=8x\). The right side is a difference of two squares, \(3(x^{2}-1)=3x^{2}-3\), so \(8x=3x^{2}-3\) [M1].
Rearrange to the standard form: \(3x^{2}-8x-3=0\). Factorising needs two numbers multiplying to \(3\times(-3)=-9\) and adding to \(-8\), namely \(-9\) and \(+1\), which gives \((3x+1)(x-3)=0\) [M1] oe.
Setting each factor to zero gives \(x=3\) and \(x=-\dfrac{1}{3}\) [A1] oe.

Both values are valid, since neither makes a denominator zero; only \(x=1\) or \(x=-1\) would have to be rejected.

Check \(x=3\): \(\dfrac{4}{4}+\dfrac{4}{2}=1+2=3\), as required.

Find both solutions of \(\dfrac{4}{x+1}+\dfrac{4}{x-1}=3\).