6.02c, 6.03a - Equations with algebraic fractions

6.02c, 6.03a - Equations with algebraic fractions

Multiplying away a denominator can make an equation much easier, but it can also conceal a forbidden value. Under 6.03a, you will solve equations containing algebraic fractions; under 6.02c, you will rearrange fractional formulae. You will keep the original restrictions and check the resulting answers.

Clearing denominators safely

If an equation turns into (x−2)(x−3)=0(x-2)(x-3)=0, must both values solve the original equation? Keep that question in mind while you learn to remove fraction bars safely.

Higher only. A fraction is defined only when its denominator is nonzero. Before multiplying, list the values that make any original denominator zero. On the remaining domain, multiplying by a common denominator is reversible because that multiplier is nonzero.

For 2/x−2/(x+1)=12/x-2/(x+1)=1, the restrictions are x≠0,−1x\ne0,-1. A common denominator is x(x+1)x(x+1). Multiply every term on both sides by it: the first numerator gains a factor x+1x+1, the second a factor xx, and the right side also changes.

Worked example 1: a difference of fractions

Question. Solve 2x−2x+1=1\frac2x-\frac2{x+1}=1.

Plan. Exclude zero denominators, clear them using x(x+1)x(x+1), and solve the quadratic.

x≠0,−1original restrictions2(x+1)−2x=x(x+1)multiply every term2=x2+xexpand and simplify(x+2)(x−1)=0rearrange and factorisex=−2orx=1solve both factors\begin{aligned} x&\ne0,-1 &&\text{original restrictions}\\ 2(x+1)-2x&=x(x+1) &&\text{multiply every term}\\ 2&=x^2+x &&\text{expand and simplify}\\ (x+2)(x-1)&=0 &&\text{rearrange and factorise}\\ x&=-2\quad\text{or}\quad x=1 &&\text{solve both factors} \end{aligned}

Answer. x=−2x=-2 or x=1x=1; both are allowed.

Check. At −2-2, the original left side is −1−(−2)=1-1-(-2)=1. At 11, it is 2−1=12-1=1. Neither original denominator is zero.

Change only subtraction to addition. The denominators and restrictions stay the same, but the resulting quadratic no longer has integer roots.

Worked example 2: a sum of fractions

Question. Solve 2x+2x+1=1\frac2x+\frac2{x+1}=1, giving exact answers.

Plan. Use the same common denominator, then use the quadratic formula.

x≠0,−1original restrictions2(x+1)+2x=x(x+1)clear denominatorsx2−3x−2=0expand and rearrangex=3±(−3)2−4(1)(−2)2quadratic formulax=3±172simplify\begin{aligned} x&\ne0,-1 &&\text{original restrictions}\\ 2(x+1)+2x&=x(x+1) &&\text{clear denominators}\\ x^2-3x-2&=0 &&\text{expand and rearrange}\\ x&=\frac{3\pm\sqrt{(-3)^2-4(1)(-2)}}2 &&\text{quadratic formula}\\ x&=\frac{3\pm\sqrt{17}}2 &&\text{simplify} \end{aligned}

Answer. x=(3+17)/2x=(3+\sqrt{17})/2 or x=(3−17)/2x=(3-\sqrt{17})/2. Neither is 00 or −1-1.

Check. For either root rr, r2=3r+2r^2=3r+2, so r(r+1)=4r+2r(r+1)=4r+2. The original left side is (4r+2)/(r(r+1))=1(4r+2)/(r(r+1))=1. Its denominator is nonzero: 00 and −1-1 do not satisfy r2−3r−2=0r^2-3r-2=0.

If a denominator is itself quadratic, factorise it first to find its zero values and a useful common denominator. A repeated factor needs only its highest required power: x(x+1)x(x+1) is already a common denominator of 1/[x(x+1)]1/[x(x+1)] and 2/(x+1)2/(x+1).

Watch out

Multiplying 2/x+2/(x+1)=12/x+2/(x+1)=1 by x(x+1)x(x+1) gives 2(x+1)+2x=x(x+1)2(x+1)+2x=x(x+1), not 2+2=12+2=1. Each term loses only its own denominator; the remaining factors still multiply it.

Keep exact roots and check the original exclusions.

Rearranging fractional formulae

A subject can appear in both the numerator and the denominator. Clear the denominator first, collect the terms containing the subject, then factor it out.

The original denominator restriction still applies. Any new division needs its own nonzero condition. If that condition fails, inspect the original formula instead of using a formula with a zero denominator.

Worked example 3: collecting a subject from two places

Question. Make aa the subject of p=(a+2)/(a−1)p=(a+2)/(a-1), stating the restrictions.

Plan. Exclude a=1a=1, multiply by a−1a-1, then collect and factor the aa-terms.

p(a−1)=a+2multiply, with a≠1pa−p=a+2expandpa−a=p+2collect subject termsa(p−1)=p+2factor out aa=p+2p−1divide if p≠1\begin{aligned} p(a-1)&=a+2 &&\text{multiply, with }a\ne1\\ pa-p&=a+2 &&\text{expand}\\ pa-a&=p+2 &&\text{collect subject terms}\\ a(p-1)&=p+2 &&\text{factor out }a\\ a&=\frac{p+2}{p-1} &&\text{divide if }p\ne1 \end{aligned}

Answer. a=(p+2)/(p−1)a=(p+2)/(p-1), for p≠1p\ne1; the original requires a≠1a\ne1. At p=1p=1, the original would imply a−1=a+2a-1=a+2, which is impossible.

Check. The result gives a−1=3/(p−1)a-1=3/(p-1), which is nonzero. Also a+2=3p/(p−1)a+2=3p/(p-1), so (a+2)/(a−1)=p(a+2)/(a-1)=p, as required.

Change only the sign of the constant in the numerator. The collected coefficient p−1p-1 stays the same.

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